Secondary Mathematics 2 Table of Contents

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1 Secondar Mathematics Table of Contents Unit 1: Etending the Number Sstem Cluster 1: Etending Properties of Eponents (N.RN.1 and N.RN.)... 3 Cluster : Using Properties of Rational and Irrational Numbers (N.RN.3)... 9 Cluster : Performing Arithmetic Operations on Polnomials (A.APR.1)... 0 Unit : Quadratic Functions and Modeling Cluster 1 and : Interpreting and Analzing Functions (F.IF., F.IF.5, F.IF.7 and F.IF.9)... 5 Cluster 1 and 5: Quadratic Functions and Modeling (F.IF.6 and F.LE.3)... Factoring... 6 Cluster : Forms of Quadratic Functions (F.IF.8a, A.SSE.1a, and A.SSE.3a and b) Cluster 3 (Unit 6): Translating between descriptions and equations for a conic section (G.GPE.) Honors H.5.1 (G.GPE.3)... 7 Cluster 3: Building Functions that Model Relationships between Two Quantities (F.BF.1) Cluster : Transformations and Inverses (F.BF.3 and F.BF.) Unit 3: Epressions and Equations Cluster 1: Interpreting the Structure of Epressions (A.SSE.) Cluster 3 (Unit 1): Performing Arithmetic Operations with Comple Numbers (N.CN.1 and N.CN.) Honors H..1: (N.CN.3) Cluster and 5: Solving Equations in One Variable with Comple Solutions (A.REI. and N.CN.7) Honors (N.CN.8 and N.CN.9)... 1 Cluster 3: Writing and Solving Equations and Inequalities in One Variable (A.CED.1 and A.CED.) Honors H Cluster 3: Writing and Graphing Equations in Two Variables (A.CED.) Cluster 6: Solving Sstems of Equations (A.REI.7) Honors (A.REI.8 and A.REI.9) Cluster : Forms and Uses of Eponential Functions (F.IF.8b, A.SSE.1b, and A.SSE.3c)

2 Unit 1 Etending the Number Sstem

3 Unit 1 Cluster 1 (N.RN.1 & N.RN.): Etending Properties of Eponents Cluster 1: Etending properties of eponents Define rational eponents and etend the properties of integer eponents to rational eponents 1.1. Rewrite epressions, going between rational and radical form VOCABULARY If the eponent on a term is of the form m n, where n 0, then the number is said to have a rational eponent is an eample of a constant with a rational eponent. Properties of Eponents (All bases are non-zero) a b a b = + Properties of Rational Eponents (All bases are non-zero) p r p r q s q s = + Eamples = = a b = a b p q r s = p r q s = = 0 = 1 Students should have this propert memorized. a a a 0 = = a a However: = 1 a 0 Therefore: = 1 a = 1 a p q = 1 p q 5 = 1 5 n n n = p q p p q q = = m ( ) n = m n r p s p r q q s = = = = 3

4 m = m m m 1 m n = = n n m p q = p q p q m 1 m a n n b = = a a m b b n 3 = = = Practice Eercises A Simplif each epression using onl positive eponents / / ( ) 1/ / / /3 1/3 / / / /3 k 1. / 3 z 0 z 3 7 z 1/ Definition A radical can also be written as a term with a rational eponent. For eample, is an integer and n 0. In general, m n n m = where m and n are integers and 0 1 n n. n = where n m n = n m The denominator of the rational eponent becomes the inde of the radical. m n canalsobewrittenas n m

5 Rational Eponent Form a Radical Form a Practice Eercises B Rewrite each epression in radical form / 3. 5/9 a / k 6. 1/5 7. ( 6) 3 8. ( 8) ( ) 1/5 Practice Eercises C Rewrite each epression with rational eponents ( 10 ) r 7. 7 w k 9. 5 z Vocabular For an integer n greater than 1, if n a = k, then a is the nth root of k. A radical or the principal n th root of k: k, the radicand, is a real number. n, the inde, is a positive integer greater than one. 5

6 Properties of Radicals n n a = a n n n ab = a b n a = b n n a b Simplifing Radicals: Radicals that are simplified have: no fractions left under the radical. no perfect power factors in the radicand, k. no eponents in the radicand, k, greater than the inde, n. Vocabular A prime number is a whole number greater than 1 that is onl divisible b 1 and itself. In other words, a prime number has eactl two factors: 1 and itself. Eample: 5 = 1 5 prime 30 = 3 5 not prime Division Rules For a Few Prime Numbers A number is divisible b: If: Eample: The last digit is even 56 is (0,,, 6, 8) 55 is not (3+8+1=1 and 1 3=) Yes The sum of the digits is divisible b (3+8+3=1 and 1 3 = ) No 3 5 The last digit is 0 or 5 7 If ou double the last digit and subtract it from the rest of the number and the answer is: 0 Divisible b is 809 is not 67 (Double is, 67 = 63 and 63 7 = 9 ) Yes 905 (Double 5 is 10, = 80 and = 11 ) No 7 6

7 Simplifing Radicals Method 1: Find Perfect Squares Under the Radical 1. Rewrite the radicand as factors that are perfect squares and factors that are not perfect squares.. Rewrite the radical as two separate radicals. 3. Simplif the perfect square. Method : Use a Factor Tree 1. Work with onl the radicand.. Split the radicand into two factors. 3. Split those numbers into two factors until the number is a prime number.. Group the prime numbers into pairs. 5. List the number from each pair onl once outside of the radicand. 6. Leave an unpaired numbers inside the radical. Note: If ou have more than one pair, multipl the numbers outside of the radical as well as the numbers left inside. Method 3: Divide b Prime Numbers 1. Work with onl the radicand.. Using onl prime numbers, divide each radicand until the bottom number is a prime number. 3. Group the prime numbers into pairs.. List the number from each pair onl once outside of the radicand. 5. Leave an unpaired numbers inside the radical. Note: If ou have more than one pair, multipl the numbers outside of the radical as well as the numbers left inside. Eample: Eample: Eample:

8 Method : Use Eponent Rules Eample: Rewrite the eponent as a rational eponent.. Rewrite the radicand as factors that are perfect squares and factors that are not perfect squares. 3. Rewrite the perfect square factors with 1/ 75 ( 5 3) 1/ an eponent of. ( 5 3) 1/. Split up the factors, giving each the 1/ 1/ rational eponent. ( 5 ) ( 3) 5. Simplif. 5 ( 3) 1/ 6. Rewrite as a radical 5 3 Method with Variables: Eample: 1. Rewrite the eponent as a rational eponent.. Rewrite the radicand as two factors. One with the highest eponent that is divisible b the root and the other factor with an eponent of what is left over. 3. Split up the factors, giving each the 3 7 7/3 ( 6 ) 1/3 rational eponent. ( ) 1/3 6 1/3. Rewrite the eponents using eponent rules. 1/3 5. Simplif Rewrite as a radical 6/3 1/3 Practice D Simplif each radical epression p p u v m 3 n r m 8

9 Unit 1 Cluster (N.RN.3): Using Properties of Rational and Irrational numbers Cluster 1: Etending properties of eponents 1..1 Properties of rational and irrational numbers (i.e. sum of rational numbers is rational, sum of a rational and irrational number is irrational) Number Sstems Comple Numbers: all numbers of the form a + bi where a and b are real numbers i, i Real Numbers Rational Numbers consist of all numbers that can be written as the ratio of two integers 1, 0.15,, Integers are the whole numbers and their opposites (-3, -, -1, 0, 1,, 3, ) Natural Numbers (Counting Numbers) Whole Numbers include zero and the natural numbers Imaginar Numbers: are of the form of bi where i = 1-3i, i Irrational Numbers consist of all numbers that cannot be written as the ratio of two integers. π, 3, 5 9

10 Properties of Real Numbers Description Numbers Algebra Commutative Propert = a + b = b + a You can add or multipl real 7 numbers in an order without ( 11) = 11( 7) ab = ba changing the result. Associative Propert The sum or product of three or more real numbers is the same ( 5 + 3) + 7 = 5 + ( 3 + 7) ( ab) c = a ( bc) regardless of the wa the numbers are grouped. Distributive Propert When ou multipl a sum b a number, the result is the same 5( + 8) = 5 + 5( 8) a ( b + c) = ab + bc whether ou add and then multipl or whether ou multipl ( + 8) 5 = ( 5) + ( 8) ( b + c) a = ba + ca each term b the number and then add the products. Additive Identit Propert The sum of a number and 0, the additive identit, is the original number = 3 n + 0 = 0 + n = n Multiplicative Identit Propert The product of a number and 1, the multiplicative identit, is the original number. Additive Inverse Propert The sum of a number and its opposite, or additive inverse, is 0. Multiplicative Inverse Propert The product of a non-zero number and its reciprocal, or multiplicative inverse Closure Propert The sum or product of an two 1 = n 1 = 1 n = n ( 5) = 0 n + ( n) = = = 5 real numbers is a real number. 6 = 1 1 n = 1, n 0 n a + b R ab R 10

11 Wh can t I come in????? Sorr we are a CLOSED set! Closure When an operation is eecuted on the members of a set, the result is guaranteed to be in the set. Addition: If two integers are added together, Eample: + 5 = 3 the sum is an integer. Therefore, integers are closed under addition. Multiplication: If two integers are multiplied together, the product is an integer. Therefore, Eample: ( 6)( 7) = integers are closed under multiplication. Subtraction: If one integer is subtracted from another, the difference is an integer. Therefore, integers are closed under subtraction. Eample: ( 6) = Division: If one integer is divided b another 10 ( ) = 5 closed integer, the quotient ma or ma not be an integer. Therefore, integers are not closed 1 ( ) 10 = not closed under division. Eample: 5 You Decide 1. What number sstems are closed under addition? Justif our conclusions using the method of our choice.. What number sstems are closed under multiplication? Justif our conclusions using the method of our choice. 3. What number sstems are closed under subtraction? Justif our conclusions using the method of our choice.. What number sstems are closed under division? Justif our conclusions using the method of our choice. 11

12 Vocabular For an integer n greater than 1, if n a = k, then a is the nth root of k. A radical or the principal n th root of k: k, the radicand, is a real number. n, the inde, is a positive integer greater than one. Properties of Radicals n n a = a n n n ab = a b n a = b n n a b Simplifing Radicals: Radicals that are simplified have: no fractions left under the radical. no perfect power factors in the radicand, k. no eponents in the radicand, k, greater than the inde, n. Vocabular A prime number is a whole number greater than 1 that is onl divisible b 1 and itself. In other words, a prime number has eactl two factors: 1 and itself. Eample: = prime 30 = 3 5 not prime Division Rules For a Few Prime Numbers A number is divisible b: If: Eample: The last digit is even 56 is divisible b (0,,, 6, 8) 55 is not divisible b (3+8+1=1 and 1 3=) Yes The sum of the digits is divisible b = 383 (3+8+3=1 and 3 ) No 5 The last digit is 0 or is divisible b is not divisible b 5 If ou double the last digit 67 (Double is, and and subtract it from the rest 63 7 = 9 ) Yes 7 of the number and the answer is: 905 (Double 5 is 10, and 0 3 Divisible b = 11 7 ) No 1

13 Simplifing Radicals Method 1: Find Perfect Squares Under the Radical. Rewrite the radicand as factors that are perfect squares and factors that are not perfect squares. 5. Rewrite the radical as two separate radicals. 6. Simplif the perfect square. Method : Use a Factor Tree 7. Work with onl the radicand. 8. Split the radicand into two factors. 9. Split those numbers into two factors until the number is a prime number. 10. Group the prime numbers into pairs. 11. List the number from each pair onl once outside of the radicand. 1. Leave an unpaired numbers inside the radical. Note: If ou have more than one pair, multipl the numbers outside of the radical as well as the numbers left inside. Method 3: Divide b Prime Numbers 6. Work with onl the radicand. 7. Using onl prime numbers, divide each radicand until the bottom number is a prime number. 8. Group the prime numbers into pairs. 9. List the number from each pair onl once outside of the radicand. 10. Leave an unpaired numbers inside the radical. Note: If ou have more than one pair, multipl the numbers outside of the radical as well as the numbers left inside. Eample: Eample: Eample:

14 Method : Use Eponent Rules 7. Rewrite the eponent as a rational eponent. 8. Rewrite the radicand as factors that are perfect squares and factors that are not perfect squares. 9. Rewrite the perfect square factors with Eample: 75 1/ 75 ( 5 3) 1/ an eponent of. ( 5 3) 1/ 10. Split up the factors, giving each the 1/ 1/ rational eponent. ( 5 ) ( 3) 11. Simplif. 1/ Rewrite as a radical 5 3 Method with Variables: Eample: 7. Rewrite the eponent as a rational eponent. 8. Rewrite the radicand as two factors. One with the highest eponent that is divisible b the root and the other factor with an eponent of what is left over. 9. Split up the factors, giving each the 3 7 7/3 ( 6 ) 1/3 rational eponent. ( ) 1/3 6 1/3 10. Rewrite the eponents using eponent rules. 1/3 11. Simplif Rewrite as a radical 6/3 1/3 Adding and Subtracting Radicals To add or subtract radicals, simplif first if possible, and then add or subtract like radicals. Eample: 1. The both have the same term under the radical so the are like terms Add the coefficients of the radicals. ( + 5) Eample: 1. The both have the same term under the radical so the are like terms.. Subtract the coefficients of the radicals. ( 7) 3 ( 3) 3 1

15 Eample: 1. The are not like terms, but one of them can be simplified.. Rewrite the number under the radical Use the properties of radicals to write the factors as two radicals is a perfect square and the square root of it is Multipl the coefficients of the second radical. 6. Now the are like terms, add the coefficients (5 + 10) None of them are like terms. Simplif if ou can. Eample: Factor each number inside the radical Use the properties of radicals to simplif and 9 are perfect squares; their square roots are and Multipl the numbers outside of the radical Onl the terms with are like (10 9) + 3 terms. 7. Simplif The are not like terms, but the can be simplified.. Rewrite the epressions under the radical. 3. Use properties of radicals to rewrite the epressions.. The cube root of 8 is and the cube root of 7 is Multipl the coefficients of the last radical. 6. Add or subtract the coefficients of the like terms. 7. Simplif Eample:

16 1. Eample: The are not like terms, but one of them can be simplified. 3. Rewrite the epression under the radical Use properties of radicals to rewrite the epression and are perfect cubes. The cube 3 root of 8 is and is Multipl the coefficients of the first radical Now the are like terms, add the coefficients of each. ( ) Simplif Multipling Radicals Multipling radicals with the same inde 1. Multipl the coefficients and multipl the numbers under the radicand.. If possible, simplif. This is alread simplified. Eample: Eample: ( 3) ( 3 + )( 6 3) 1. Use the distributive method to multipl. 3 ( 6 ) + 3 ( 3) ( 3). Use properties of radicals to simplif Simplif an radicals Combined like terms if possible Eample: ( 5 1)( 5 ) 1. Use the distributive method to multipl ( ) + ( 1) 5 + ( 1)( ). Use properties of radicals to simplif Simplif and combine like terms. 5 + ( 1) 5 +. The square root of 5 is Combine like terms

17 Eample: ( 5 + 3)( 5 3) 1. Use the distributive method to multipl ( 3) + ( 3) 5 + 3( 3). Use properties of radicals to simplif Simplif and combine like terms. 5 + ( ) 9. The square root of is Simplif Multipling Radicals with Different Indices Note: In order to multipl radicals with different indices the radicands must be the same. Eample: Rewrite each radical using rational eponents Use properties of eponents to simplif. 3. Combine the fractions b finding a common denominator. 1. Rewrite each radical using rational eponents.. Use properties of eponents to simplif. 3. Combine the fractions b finding a common denominator. 1. Rewrite the inner radical using rational eponents.. Rewrite the outer radical using rational eponents Eample: Eample: Use properties of eponents to simplif.. Simplif b multipling fractions

18 Practice Eercises A Add or Subtract Practice Eercises B Multipl and simplif the result ( ) 5. ( 5) 6. 3 ( ) 7. ( 5 3)( 5 + 3) 8. ( 3 + )( 3 5) 9. ( 5 5 )( + 5 ) z 18

19 You Decide: 1. Add: +. Can ou write the result as the ratio of two numbers? (Use our graphing 5 calculator to change the sum from a decimal to a fraction b pushing the math button and select FRAC). Add: 1 +. Can ou write the result as the ratio of two numbers? 3 3. Add: Can ou write the result as the ratio of two numbers?. Add: Can ou write the result as the ratio of two numbers? 5. Add: + 3. Can ou write the result as the ratio of two numbers? 6. Add: π. Can ou write the result as the ratio of two numbers? 7. Write a rule based on our observations with adding rational and irrational numbers. 19

20 Unit 1 Cluster (A.APR.1): Polnomials Cluster : Perform arithmetic operations on polnomials 1..1 Polnomials are closed under addition, subtraction, and multiplication 1..1 Add, subtract, and multipl polnomials (NO DIVISION) VOCABULARY A term that does not have a variable is called a constant. For eample the number 5 is a constant because it does not have a variable attached and will alwas have the value of 5. A constant or a variable or a product of a constant and a variable is called a term. For eample 3 are all terms.,, or Terms with the same variable to the same power are like terms. and 7 are like terms. An epression formed b adding a finite number of unlike terms is called a polnomial. The variables 3 1 can onl be raised to positive integer eponents is a polnomial, while + 5 is not a polnomial. NOTE: There are no square roots of variables, no fractional powers, and no variables in the denominator of an fractions. A polnomial with onl one term is called a monomial( 6 ). A polnomial with two terms is called a binomial ( + 1 ). A polnomial with three terms is called a trinomial( 5 + 3). Polnomials are in standard (general) form when written with eponents in descending order and the 3 constant term last. For eample is in standard form. The eponent of a term gives ou the degree of the term. The term 3 has degree two. For a polnomial, the value of the largest eponent is the degree of the whole polnomial. The polnomial has degree. The number part of a term is called the coefficient when the term contains a variable and a number. 6 has a coefficient of 6 and has a coefficient of -1. The leading coefficient is the coefficient of the first term when the polnomial is written in standard 3 form. is the leading coefficient of Degree n General Polnomial: n f = a + a + + a + a + a n n 1 n Leading Coefficient a n Leading Term 0 Constant

21 CLASSIFICATIONS OF POLYNOMIALS Name Form Degree Eample Zero f ( ) = 0 None f ( ) = 0 Constant f = a, a 0 0 f ( ) = 5 Linear f = a + b 1 f = + 1 Quadratic Cubic = + + f a b c = f a b c d 1 7 f = f = 3 Practice Eercises A: Determine which of the following are polnomial functions. If the function is a polnomial, state the degree and leading coefficient. If it is not, eplain wh f = f = f = f ( ) = f = f = 5 Operations of Polnomials Addition/Subtraction: Combine like terms. Eample 1: Horizontal Method Vertical Method 3 3 ( 3 + 1) + ( ) 3 3 = ( + ) + ( 3 + ) + ( 5) + ( 1+ 3) = Eample : Horizontal Method ( + 3 ) ( ) 3 = = Vertical Method ( )

22 Multiplication: Multipl b a monomial Eample 3: Eample : ( + ) ( ) = = = Multiplication: Multipl two binomials ( 5 7)( + 9) Distributive (FOIL) Method ( 5 7)( + 9) = 5( + 9) 7( + 9) = * combinelike terms = Bo Method *combine terms on the diagonals of the unshaded boes(top right to lower left) Vertical Method = Multiplication: Multipl a binomial and a trinomial ( + 3)( 6 7 5) Distributive Method ( + 3)( 6 7 5) = ( 6 7 5) + 3( 6 7 5) 3 = ( ) + ( ) = * combine liketerms = Bo Method *combine terms on the diagonals of the unshaded boes(top right to lower left) Vertical Method =

23 Practice Eercises B: Perform the required operations. Write our answers in standard form and determine if the result is a polnomial. 1. ( 3 + 7) + ( ). ( 3 5) ( ) 3. ( 3 + 3) ( ). ( + 3) + ( ) 5. ( + 3) 6. ( + 3 ) 7. 3u ( u 1) 8. ( 3 )( 5) 9. ( + 7)( 3) 10. ( 3 5)( + ) 11. ( + 3)( + 1) 1. ( 3 )( 3 + ) 13. ( + 7) 1. ( 3 5) 15. ( 5 3 1) 16. ( 3 3)( 3 + 3) 17. ( + 3)( + ) 18. ( + 3 )( 3) 19. ( + 3)( + + 1) 0. ( 3 + 1) + ( + + 1) YOU DECIDE Are polnomials closed under addition, subtraction, multiplication? Justif our conclusion using the method of our choice. 3

24 Unit Quadratic Functions and Modeling

25 Unit Cluster 1 (F.1F., F.1F.5, F.1F.6) Unit Cluster (F.1F.7, F.1F.9) Interpret functions that arise in applications in terms of a contet Analzing functions using different representations Cluster 1:.1.1 Interpret ke features; intercepts, intervals where increasing and decreasing, intervals where positive and negative, relative maimums and minimums, smmetr, end behavior, domain and range, periodicit.1. Relate the domain of a function to its graph or a contet.1.3 Average rate of change over an interval: calculate, interpret, and estimate from a graph. Cluster :..1 Graph functions from equations b hand and with technolog showing ke features (square roots, cube roots, piecewise-defined functions including step and functions, and absolute value)...1 Graph linear and quadratic functions and show intercepts, maima, and minima..3 Compare properties (ke features) of functions each represented differentl (table, graph, equation or description) VOCABULARY The domain is the set of all first coordinates when given a table or a set of ordered pairs. It is the set of all -coordinates of the points on the graph and is the set of all numbers for which an equation is defined. The domain it is written from the least value to the greatest value. The range is the set of all second coordinates when given a table or a set of ordered pairs. It is the set of all -coordinates of the points on the graph. When modeling real world situations, the range is the set of all numbers that make sense in the problem. The range is written from the least value to the greatest value. Eample: Find the domain and range of f = + 3. Domain 1. Find an values for which the function is undefined. The square root function has real number solutions if the epression under the radicand is positive or zero. This means that + 0 therefore.. Write the domain in interval notation. The domain is [, ). 5

26 Range 1. Find all values for which the output eists. The square root function uses the principal square root which is a positive number or zero ( 0 ). However, the function has been shifted down three units so the range is also shifted down three units 3.. Write the range in interval notation. The range is [ 3, ) Eample: Find the domain and range of the function graphed to the right Domain 1. List all the -values of the function graphed. If ou were to flatten the function against the -ais ou would see something like this: The function is defined for all the -values along the -ais.. Write the domain in interval notation. The domain is (, ). Range 1. List all the -values of the function graphed. If ou were to flatten the function against the -ais ou would see something like this: The function is defined for all the -values greater than or equal to -.. Write the range in interval notation. The range is [, ). 6

27 Eample: The path of a ball thrown straight up can be modeled b the equation h t = 16t + 0t + where t is the time in seconds that the ball is in the air and h is the height of the ball. What is the real world domain and range for the situation? Domain 1. Find all the values that would make sense for the situation. The domain represents the amount of time that the ball is in the air. At t = 0 the ball is thrown and enters the air shortl afterwards so the domain must be greater than zero. The ball will hit the ground at 1.5 seconds. Once it is on the ground it is no longer in the air so the domain must be less than 1.5 seconds. The ball is in the air for 0 < t < 1.5 seconds.. Write the domain in interval notation. The domain is (0,1.5). Range 1. Find all the values that would make sense for the situation. The ball will not go lower than the ground so the height must be greater than zero. The ball will go no higher than its maimum height so the height must be less than or equal to 10.5 feet. The range will be 0 < h Write the range in interval notation. The range is (0,10.5]. Practice Eercises A: Find the domain and range. 1. f = f = Your cell phone plan charges a flat fee of $10 for up to1000 tets and $0.10 per tet over The parking lot for a movie theater in the cit has no charge for the first hour, but charges $1.50 for each additional hour or part of an hour with a maimum charge of $7.50 for the night. 7

28 VOCABULARY The -intercept is where a graph crosses or touches the -ais. It is the ordered pair ( a,0). Where a is a real number. The -intercept is where a graph crosses or touches the -ais. It is the ordered pair ( 0,b ). Where b is a real number. A relative maimum occurs when the -value is greater than all of the -values near it. A function ma have more than one relative maimum value. A relative minimum occurs when the -value is less than all of the -values near it. A function ma have more than one relative minimum value. Eample: =. Find the intercepts of the function f 1 -intercept 1. Substitute in for f(). = 1. Substitute 0 in for. 0 = 1 3. Solve for. 1 = 1 =. Write the intercept as an ordered pair. 1,0 -intercept 1. Substitute 0 in for. = (0) 1. Solve for. = 0 1 = 1 3. Write the intercept as an ordered pair. ( 0, 1) 8

29 Eample: f = 3 5. Find the intercepts of the function -intercept 1. Use our graphing calculator to graph the function.. Use the Calculate Menu ( nd, Trace, Zero) to find the -intercepts. (Zero is another name for the -intercept) -intercept The -intercepts are 1,0 3,0. and 1. To find the -intercept, replace each with 0. = Solve the equation for. = 0 0 = 3. Write the intercept as an ordered pair. ( 0, ) Eample: f = + +. Find the maimum of To find the maimum use our graphing calculator to graph the function. Then use the Calculate Menu ( nd, Trace, Maimum). Enter a number that is to the left of the maimum, for eample 0, then push enter. Then enter a number that is to the right of the maimum, for eample, then push enter. You can guess the value of the maimum or just push enter again and the maimum will be calculated. The maimum is (, 8) Eample: = + 3. Find the minimum of f ( ) To find the maimum use our graphing calculator to graph the function. Then use the Calculate Menu ( nd, Trace, Minimum). Enter a number that is to the left of the minimum, for eample -3, then push enter. Then enter a number that is to the right of the minimum, for eample -1, then push enter. You can guess the value of the minimum or just push enter again and the minimum will be calculated. The minimum is (-, -3)

30 Practice Eercises B Find the and -intercepts for each function = 10 f = +. 7 f = f = 3. f = f = Find the relative maimums or minimums of each function. 7. f ( ) = f ( ) = f = f ( ) = 5 f = f = VOCABULARY An interval is a set of numbers between two -values. An open interval is a set of numbers between two -values that doesn t include the two end values. Open intervals are written in the form ( 1, ) or 1 < <. A closed interval is a set of numbers between two -values that does include the two end values. Closed intervals are written in the form [, ]. 1 or 1 A function f is increasing when it is rising (or going up) from left to right and it is decreasing when it is falling (or going down) from left to right. A constant function is neither increasing nor decreasing; it has the same -value for its entire domain. A function is positive when f ( ) > 0 or the -coordinates are alwas positive. A function is negative when f ( ) < 0 or the -coordinates are alwas negative. Eample: Find the intervals where the function a. increasing b. decreasing c. constant d. positive e. negative f = + 3 is:

31 Increasing/Decreasing/Constant 1. Find the maimums or minimums. The minimum is (-1, -).. Determine if the function is rising, falling, or constant between the maimums and minimums. 3. Write the intervals where the function is increasing, decreasing, or constant using interval notation. To the left of the minimum the function is falling or decreasing. To the right of the minimum the function is rising or increasing. The function is increasing on the interval ( 1, ). The function is decreasing on the interval (, 1). The function is never constant. Positive/Negative 1. Find all the -intercepts of the function. The -intercepts are at (-3, 0) and (1, 0).. Determine if the function has positive or negative -values on the intervals between each -intercept b testing a point on the interval. 3. Write the intervals where the function is positive or negative using interval notation. < 3 3 < < 1 > 1 = f ( ) = 5 Positive = 0 f (0) = 3 Negative = f () = 5 Positive The function is positive on the intervals (, 3) and (1, ). The function is negative on the interval ( 3,1). Eample: Find the intervals where the function 3 a. increasing b. decreasing c. constant d. positive e. negative f = + 1 is: Increasing/Decreasing/Constant 1. Find the maimums or minimums. There are no maimums or minimums.. Determine if the function is rising, falling, or constant on its entire domain. 3. Write the intervals where the function is increasing, decreasing, or constant using interval notation. The function is rising from left to right so it is increasing on its entire domain. The function is increasing on the interval,. The function is never decreasing nor is it constant. 31

32 Positive/Negative 1. Find all the -intercepts of the function. The -intercept is (-1, 0).. Determine if the function has positive or negative -values on the intervals between each -intercept b testing a point on the interval. 3. Write the intervals where the function is positive or negative using interval notation. < 1 > 1 = = 0 f ( ) = 1 f (0) 0.6 Negative Positive The function is positive on the interval ( 1, ). The function is negative on the interval (, 1). Eample: Find the intervals where the function f a. increasing b. decreasing c. constant d. positive e. negative + + 3, 1 = is: 3, > Increasing/Decreasing/Constant 1. Find the maimums or minimums and an breaks in the domain.. Determine if the function is rising, falling, or constant between each maimum or minimum and each break in the graph. 3. Write the intervals where the function is increasing, decreasing, or constant using interval notation. There is a maimum at (-, 0) and a break in the domain at = -1. The function is rising (increasing) to the left of the maimum. It is falling (decreasing) to the right of the maimum. It is constant to the right of = -1. The function is increasing on the interval,. It is decreasing on the interval (, 1). It is constant on the interval ( 1, ). 3

33 Positive/Negative 1. Find all the -intercepts of the function and an places where there is a break in the domain.. Determine if the function has positive or negative -values on the intervals between each -intercept b testing a point on the interval. 3. Write the intervals where the function is positive or negative using interval notation. The -intercept is (-5, 0). There is a break in the domain at = -1. < 5 5 < < 1 > 1 = 6 f ( 6) = 1 Negative = 3 f ( 3) = Positive = 0 f (0) = 3 Positive The function is positive on the intervals ( 5, 1) and ( 1, ). The function is negative on the interval (, 5). Practice Eercises C Find the intervals where the function is: a. increasing b. decreasing c. constant d. positive e. negative 1 f = f = f = 3 f = f = + 1 f 6., < 0 = 1, > 0 33

34 VOCABULARY GRAPHICALLY ALGEBRAICALLY A function is smmetric with respect to the -ais if,, on the for ever point graph, the point (, ) is also on the graph. In other words, if ou substitute in for ever ou end up with the original function. When looking at the graph, ou could fold the graph along the -ais and both sides are the same f = + 5 f ( ) = + 5 f = f ( ) = + 5 A function is smmetric with respect to the origin, on if, for ever point the graph, the point (, ) is also on the graph. In other words, if ou substitute in for ever ou end up with the opposite of the original function. When looking at the graph, there is a mirror image in Quadrants 1 & 3 or Quadrants & f = 8 3 f ( ) = 8 f ( ) = f = An equation with no smmetr. If ou substitute in for ever ou end up with something that is neither the original function nor its opposite. When looking at the graph, ou could not fold the graph along the -ais and have both sides are the same or it doesn t reflect a mirror image. Eample: f = + f ( ) = ( ) + ( ) f = f f 3

35 Determine what kind of smmetr, if an, f = 1 has. Test for -ais Smmetr Replace with and see if the result is the same as the original equation. f = 1 f ( ) = 1 This is not the same as the original equation. Test for Origin Smmetr Replace with and see if the result is the opposite of the original equation. f = 1 f ( ) = 1 This is not the opposite of the original equation. The function f = 1 has no smmetr. Graph Eample: f = Determine what kind of smmetr, if an, Test for -ais Smmetr Replace with and see if the result is the same as the original equation. f = ( ) f ( ) = This is equal to the original equation. The function Test for Origin Smmetr Replace with and see if the result is the opposite of the original equation. f = ( ) f ( ) = This is not the opposite of the original equation. f = has -ais smmetr. Graph Eample: Determine what kind of smmetr, if an, the function graphed at the right has Pick a point (, ) Test for -ais Smmetr, on the graph and see if is also on the graph. The point (-, ) is on the graph but the point (, ) is not. The function does not have -ais smmetr. The function graphed has origin smmetr. Pick a point (, ) Test for origin smmetr, on the graph and see if is also on the graph. The point (-, ) is on the graph and the point (, -) is also on the graph. The function has origin smmetr. 35

36 VOCABULARY End behavior describes what is happening to the -values of a graph when goes to the far right. ( + ) or goes the far left End behavior is written in the following format: Right End Behavior: Left End Behavior: lim f = c lim f = c Eample: Find the end behavior of f 3 = +. As gets larger the function is getting more and more negative. Therefore, the right end behavior is lim f =. As gets smaller the function is getting more and more positive. Therefore the left end behavior is lim f = Eample: f = 3 1. Find the end behavior of As gets larger the function is getting more and more negative. Therefore, the right end behavior is lim f =. As gets smaller the function is getting more and more negative. Therefore the left end behavior is lim f = Eample: Find the end behavior of f = 1. As gets larger the function is getting more and more positive. Therefore, the right end behavior is lim f =. The domain is restricted to numbers greater than or equal to, therefore this graph has no left end behavior

37 Practice Eercises D Graph each function below and find the: a. Domain and Range b. Intercepts, if an c. Determine whether the function has an smmetr. d. List the intervals where the function is increasing, decreasing, or constant. e. List the intervals where the function is positive or negative. f. Find all the relative maimums and minimums. g. Find end behavior 1. f 5 =. f = 3 3. f = f = f = , 1 f = 1 +, 1< 1 3 VOCABULARY Periodicit refers to a function with a repeating pattern. The period of this function is 6 horizontal units. Meaning the pattern will repeat itself ever 6 horizontal units

38 You Decide Mr. Astro s phsics class created rockets for an end of the ear competition. There were three groups who constructed rockets. On launch da the following information was presented for review to determine a winner. Group A estimated that their rocket was easil modeled b the equation: = Group B presented the following graph of the height of their rocket, in feet, over time. 500 height (feet) time (seconds) Group C recorded their height in the table below. Time (seconds) Height (feet) Who should be the winner of the competition? Use mathematical reasons to support our conclusion. 38

39 Unit Cluster (F.IF.7b) Graphing Square Root, Cube Root, and Piecewise-Defined Functions, Including Step Functions and Absolute Value Functions Cluster : Analzing functions using different representations..1b Graph functions from equations b hand and with technolog showing ke features (square roots, cube roots, piecewise-defined functions including step functions, and absolute value) VOCABULARY There are several tpes of functions (linear, eponential, quadratic, absolute value, etc.). Each of these could be considered a famil with unique characteristics that are shared among the members. The parent function is the basic function that is used to create more complicated functions. Square Root Function Parent Function 1/ f = = Domain: [ 0, ) Range: [ 0, ) Intercepts: -intercept Ke Features 0,0, -intercept ( 0,0 ) Intervals of Increasing/Decreasing: increasing ( 0, ) Intervals where Positive/Negative: ( 0, ) Relative maimums/minimums: minimum at( 0,0 ) Smmetries: none End Behavior: right end behavior lim = ; left end behavior lim =

40 Cube Root Function Parent Function f = 3 = 1/ Ke Features Domain: (, ) Range: (, ) Intercepts: -intercept ( 0,0 ), -intercept ( 0,0 ) Intervals of Increasing/Decreasing: increasing (, ) Intervals where Positive/Negative: positive( 0, ), negative (,0) Relative maimums/minimums: none Smmetries: origin End Behavior: right end behavior lim behavior lim 3 = 3 = ; left end Absolute Value Function Parent Function f = Ke Features Domain: (, ) Range: [ 0, ) Intercepts: -intercept ( 0,0 ), -intercept ( 0,0 ) Intervals of Increasing/Decreasing: increasing ( 0, ), decreasing (,0) Intervals where Positive/Negative: positive,0 0, Relative maimums/minimums: minimum at ( 0,0 ) Smmetries: -ais smmetr End Behavior: right end behavior lim = ; left end behavior lim = Piecewise-Defined Functions A piecewise-defined function is a function that consists of pieces of two or more functions. For +, < eample f = 1, 0 is a piecewise-defined function. It has a piece of the + 5, > 0 0

41 function f = + but onl the piece where <. It also contains the function f ( ) = 1, but onl where 0 > 0.. Finall, it contains the function f 5 = + but onl where Piecewise-Defined Function +, < f = 1, 0 + 5, > Ke Features Domain: (, ) Range: (,5] Intercepts: -intercept (.5, 0 ), -intercept ( 0,1 ) Intervals of Increasing/Decreasing: increasing (, ), decreasing ( 5, ) Intervals where Positive/Negative: positive,0 and 0,.5, and.5, ( ), negative Relative maimums/minimums: none Smmetries: none lim + 5 = ; left End Behavior: right end behavior end behavior lim ( ) + = Step Functions are piecewise-defined functions made up of constant functions. It is called a step function because the graph resembles a staircase. Step Function f = int Ke Features Domain: (, ) Range: { is an integer} Intercepts: -intercept = [ 0,1 ) and = 0, -intercept ( 0,0 ) Intervals of Increasing/Decreasing: neither increasing nor decreasing Intervals where Positive/Negative: positive( 1, ), negative (,0) Relative maimums/minimums: none Smmetries: none End Behavior: right end behavior lim int behavior lim int = = ; left end 1

42 Unit Cluster 1(F.IF.6) and Cluster 5(F.LE.3) Quadratic Functions and Modeling Cluster 1: Interpret Functions that Arise in Applications in Terms of a Contet.1.3 Average rate of change over an interval: calculate, interpret, and estimate from a graph. Cluster 5: Constructing and comparing linear, quadratic, and eponential models; solve problems.5.1 Eponential functions will eventuall outgrow all other functions VOCABULARY The average rate of change of a function over an interval is the ratio of the difference (change) in over the difference (change) in. average rate of change = = 1 1 Eample: Find the average rate of change for f = on the interval [0, ]. First, find the value of the function at each end point of the interval. f 0 = (0) 3(0) + 1 f = () 3() + 1 f ( 0) = f ( 0) = 1 f = f = f = 3 5 ( 0,1) (,3) Net, find the slope between the two points (0, 1) and (, 3). 3 1 m = = = 1 0 The average rate of change of f = on the interval [0, ] is 1.

43 Eample: The per capita consumption of read-to-eat and read-to-cook breakfast cereal is shown below. Find the average rate of change from 199 to 1995 and interpret its meaning. Years since Cereal Consumption (pounds) The ear 199 is two ears since 1990 and 1995 is 5 ears since 1990, therefore the interval is [, 5]. Find the slope between the two points (, 16.6) and (5, 17.1) m = = = The average rate of change from 199 to 1995 is 0.16 pounds per ear. This means that each household increased their cereal consumption an average of 0.16 pounds each ear from 199 to Eample: Joe is visiting the Eiffel Tower in Paris. He accidentall drops his camera. The camera s height is graphed. Use the graph to estimate the average rate of change of the camera from to 7 seconds and interpret its meaning. At seconds the height of the camera is approimatel 650 feet. At 7 seconds the height of the camera is approimatel 100 feet. Find the slope between the points (, 650) and (7, 100). Height in feet m = = = Time in seconds The negative indicates that the camera is falling. The camera is picking up speed as it is falling. This means that for each second the camera is falling from to 7 seconds, it increases in speed an average of feet per second from to 7 seconds. 3

44 Practice Eercises A Find the average rate of change for each function on the specified interval. f = on [-1, 3] 1. f = + on [-, -] 3. f = on [-3, 0]. f = 6 on [-1, 0]. Find the average rate of change on the specified interval and interpret its meaning. 5. Man of the elderl are placed in nursing care facilities. The cost of these has risen significantl since Use the table below find the average rate of change from 000 to 010. Years since 1960 Nursing Care Cost (billions of $) The net sales of a compan are shown in the graph below. Estimate the average rate of change for 007 to The height of an object thrown straight up is shown in the table below. Find the average rate of change from 1 to seconds. Time (seconds) Height (feet) The graph below shows fuel consumption in billions of gallons for vans, pickups and SUVs. Estimate the average rate of change for 005 to 01. Net Sales in millions of $ Fuel Consumption Years since Years since 1980

45 Practice Eercises B Complete the tables. f = g = h( ) = Practice Eercises C Find the average rate of change for functions f, g,and h for the specified intervals. Determine which of the three functions is increasing the fastest. 1. [0, 1]. [3, 5] 3. [-, 5]. [0, 3] 5. [-, 0] 6. [0, 5] 5

46 Practice Eercises D 1. Graph the following functions on the same coordinate plane. a. 3 k 1 = b. p = 3 5 c. 0 = 3 7 r Find the average rate of change for functions k, p, and r for the specified intervals. Determine which of the three functions is increasing the fastest. a. [-, ] b. [3, 5] c. [0, 10] You Decide Use eercises C and D to help ou answer the following questions. 1. For each eercise, determine which function has the greatest average rate of change on the interval [ 0, )?. In general, what tpe of function will increase faster? Eplain our reasoning. 6

47 FACTORING (To be used before F.IF.8) VOCABULARY Factoring is the reverse of multiplication. It means to write an equivalent epression that is a product. Each of the items that are multiplied together in a product is a factor. An epression is said to be factored completel when all of the factors are prime polnomials, that is the cannot be factored an further. The greatest common factor is the largest epression that all the terms have in common. FACTOR OUT A COMMON TERM Eample: What is the largest factor that evenl divides,6, and 8? : 1 6 : : 1 The common numbers are 1and. Multipl them and the product is the greatest common factor ( + + ) 3 Divide each term b the greatest common factor. Rewrite with the common term on the outside of the parenthesis and the simplified terms inside the parenthesis. Eample: 3 8w 3w + 5w What is the largest factor that evenl divides 3 8 w,3 w, and 5w? 8 w : w w w w w : w w w w : w w 1 5 8w 3w 5w + w w w 3 ( ) w w w The common numbers are w and w. Multipl them and the product is the greatest common factor. Divide each term b the greatest common factor. Rewrite with the common term on the outside of the parenthesis and the simplified terms inside the parenthesis. 7

48 Eample: z z z z z z z 3 9 : : 1 3 z z z 15 z : z z z + z 3z 3z 3z ( + ) 3z 3z z 5 What is the largest factor that evenl divides 3 9 z,1 z, and 15z? The common numbers are 3 and z. Multipl them and the product is the greatest common factor. Divide each term b the greatest common factor. Rewrite with the common term on the outside of the parenthesis and the simplified terms inside the parenthesis. Practice Eercises A Factor out the greatest common factor FACTOR A TRINOMIAL WITH A LEADING COEFFICIENT OF 1 When factoring a trinomial of the form a + b + c,where a = 1, and b and c are integers, find factors of c that add to equal b. If p and q are the factors, the factored form looks like ( + p)( + q). Eample: Factors of 6 Sum (adds to be) ( )( 3) 8 Find factors of 6 that add to be 5. The factors are and This is the factored form.

49 Another wa to look at factoring is with an area model like the one pictured below The rectangular area represents the trinomial The width across the top is or + 3 and the length down the side is or +. To obtain the area of the rectangle, ou would multipl the length times the width or Notice that this result is the same as when we found factors of the constant term that added to the coefficient of the term. When factoring a trinomial of the form a b + c,where a = 1, and b and c are integers, find factors of c that add to equal b. If p and q are the factors, the factored form looks like ( p)( q). Eample: Factors of 6 Sum (adds to be) ( )( 3) Find factors of 6 that add to be -5. The factors are - and -3. This is the factored form. This eample can also be modeled with an area model as the picture below demonstrates The rectangular area represents the trinomial The width across the top is or 3 and the length down the side is 1 1 or. To obtain the area of the rectangle, ou would multipl the 3. length times the width or Notice that this result is the same as when we found factors of the constant term that added to the coefficient of the term. 9

50 When factoring a trinomial of the form a + b c,where a = 1, and b and c are integers, find factors of c that add to equal b. If p and q are the factors, the factored form looks like ( + p)( q). Eample: Factors of 6 Sum (adds to be) ( 1)( 6) Find factors of -6 that add to be 5. The factors are -1 and 6. + This is the factored form. This eample can also be modeled with an area model as the picture below demonstrates The rectangular area represents the trinomial The width across the top is or + 6 and the length down the side is 1. To obtain the area of Notice the rectangle, ou would multipl the length times the width or that this result is the same as when we found factors of the constant term that added to the coefficient of the term. When factoring a trinomial of the form a b c,where a = 1, and b and c are integers, find factors of c that add to equal b. If p and q are the factors, the factored form looks like ( + p)( q). 50

51 Eample: 5 6 Factors of 6 Sum (adds to be) ( 1)( 6) Find factors of -6 that add to be -5. The factors are 1 and This is the factored form. This eample can also be modeled with an area model as the picture below demonstrates The rectangular area represents the trinomial 5 6. The width across the top is or 6 and the length down the side is + 1. To obtain the area of Notice the rectangle, ou would multipl the length times the width or that this result is the same as when we found factors of the constant term that added to the coefficient of the term. Practice Eercises B Factor each epression

52 VOCABULARY A perfect square is a number that can be epressed as the product of two equal integers. For eample: 100 is a perfect square because = 100 and is a perfect square because =. FACTOR USING THE DIFFERENCE OF TWO SQUARES When something is in the form a b, where a and b are perfect square epressions, the a b a + b. factored form looks like Eample: 9 = and 9 are both perfect squares and ou are finding the difference between them, so ou can use the difference of two squares to factor. 9 = 7 7 Therefore: a = and b = This is the factored form. Eample: = = 6 6 ( 5 6)( 5 6) 5 and 36 are both perfect squares and ou are finding the difference between them, so ou can use the difference of two squares to factor. Therefore: a = 5 and b = 6 + This is the factored form. 5

53 Practice Eercises C Factor each epression FACTOR BY GROUPING When factoring a trinomial of the form a + b + c,where a, b, and c are integers, ou will need to use the technique of factoring b grouping. Eample: 6 15 (6)(15) = 90 Factors of Multipl the leading coefficient and the constant. Choose the combination that will either give the sum or difference needed to result in the coefficient of the term. In this case the difference should be -1, so 9 and -10 will give ou the desired result. Or in other words, when ou combine 9 and 10 ou will end up with ( ) + ( ) ( + ) ( + ) ( + 3) ( + 3)( 3 5) Rewrite the equation using the combination in place of the middle term. Group the first two terms and the last two terms together in order to factor. Factor the greatest common factor out of each group. Write down what is in the parenthesis (the should be identical). This is one of the factors. Add the left-overs to obtain the second factor. 53

54 Eample: (1)(10) = 10 Factors of ( 1 8 ) + ( ) ( ) + ( ) ( 3 ) ( 3 )( + 5) Multipl the leading coefficient and the constant. Choose the combination that will either give the sum or difference needed to result in the coefficient of the term. In this case the difference should be 7, so -8 and 15 will give ou the desired result. Or in other words, when ou combine 8 and 15 ou will end up with 7. Rewrite the equation using the combination in place of the middle term. Group the first two terms and the last two terms together in order to factor. Factor the greatest common factor out of each group. Write down what is in the parenthesis (the should be identical). This is one of the factors. Add the left-overs to obtain the second factor. Eample: 5 ()(5) = 100 Factors of Multipl the leading coefficient and the constant. Choose the combination that will either give the sum or difference needed to result in the coefficient of the term. In this case the difference should be 0, so -10 and 10 will give ou the desired result. Or in other words, when ou combine 10 and 10 ou will end up with 0. 5

55 ( 10 ) + ( 10 5 ) ( ) + ( ) ( 5) ( 5)( + 5) Rewrite the equation using the combination in place of the middle term. Group the first two terms and the last two terms together in order to factor. Factor the greatest common factor out of each group. Write down what is in the parenthesis (the should be identical). This is one of the factors. Add the left-overs to obtain the second factor. Practice Eercises D Factor the epression FACTORING GUIDELINES #1: Alwas look for a greatest common factor. Then factor it out if there is one. #: Count the number of terms. If there are two terms, determine if ou can use the difference of two squares. If ou can, factor. If not, proceed to #3. #3: If there are three terms, check the leading coefficient. If it is 1, then find factors of the constant term that add to the coefficient of the -term. If not, proceed to #. #: If the leading coefficient is not 1, factor b grouping. Mied practices E Factor the epression

56 Unit Cluster (F.IF.8), Unit 3 Cluster 1 (A.SSE.1a) and Unit 3 Cluster (A.SSE.3a,b) Forms of Quadratic Functions Cluster : Analzing functions using different representations.. Writing functions in different but equivalent forms (quadratics: standard, verte, factored) using the processes of factoring or completing the square to reveal and eplain different properties of functions. Interpret these in terms of a contet. Cluster 1: Interpret the structure of epressions 3.1.1a Interpret parts of an epression, such as terms, factors, and coefficients Cluster : Writing epressions in equivalent forms and solving 3..1 Choose an appropriate from of an equation to solve problems (factor to find zeros, complete the square to find maimums and minimums VOCABULARY Forms of Quadratic Functions Standard Form: f = a + b + c, where a 0. Eample: f = Verte Form: f = a( h) + k, where a 0. Eample: f = ( + 3) + 5 Factored Form: f = a( p)( q ), where a 0. Eample: f = ( )( + 7) A zero of a function is a value of the input that makes the output f ( ) equal zero. The zeros of a function are also known as roots, -intercepts, and solutions of a b c + + = 0. The Zero Product Propert states that if the product of two quantities equals zero, at least one of the quantities equals zero. If ab = 0 then a = 0 or b = 0. Finding Zeros (Intercepts) of a Quadratic Function When a function is in factored form, the Zero Product Propert can be used to find the zeros of the function. If f = a ( p) then a( p) = 0 can be used to find the zeros of If 0 = a( p) then either a = 0 or ( p) = 0. Therefore, either = 0 or = p. f. 56

57 Eample: Find the zeros of f = ( + 7) f = ( + 7) ( + 7) = 0 Substitute zero in for f(). = 0 or + 7 = 0 Use the zero product propert to set each factor equal to zero. = 0 or = 7 Solve each equation. The zeros are ( 0,0 ) and ( 7,0) Write them as ordered pairs. If f = ( p)( q) then ( p)( q) 0 f ( ). = can be used to find the zeros of If ( p)( q) = 0 then either ( p) = 0 or ( q) 0 Therefore, either = p or = q. Eample: Find the zeros of f = ( 5)( + 9) f = ( 5)( + 9) ( 5)( 9) 0 =. + = Substitute zero in for f(). 5 = 0 or + 9 = 0 = 5 or = 9 The zeros are ( 5,0 ) and ( 9,0) Use the zero product propert to set each factor equal to zero. Solve each equation. Write them as ordered pairs. NOTE: If a quadratic function is given in standard form, factor first then appl the Zero Product Propert. Eample: Find the zeros of f = = 0 ( )( ) 8 3 = 0 8 = 0 or 3 = 0 f = 11 + Substitute zero in for f(). Factor the trinomial. (See factoring lesson in Unit for etra help.) Use the zero product propert to set each factor equal to zero. = 8 or = 3 Solve each equation. The zeros are ( 8,0 ) and ( 3,0 ) Write them as ordered pairs. 57

58 f = 15 Eample: Find the zeros of f = = Substitute zero in for f(). ( )( ) = 0 Factor the trinomial. 5 = 0 or + 3 = 0 = 5 = 3 5 or 3 = = 5 The zeros are,0 and 3,0 Use the zero product propert to set each factor equal to zero. Solve each equation. Write them as ordered pairs. Practice Eercises A Find the zeros of each function. f = f = ( 1)( 3) f = f = f = ( 6) f = f = f = f = ( + 13)( ) f = 6. f = 1 9. f = COMPLETING THE SQUARE To complete the square of b ± b, add. In other words, divide the coefficient b two and square the result. + b + b + b + b b + + b ( + 3)( + 3) ( + 3)

59 An area model can be used to represent the process of completing the square for the epression The goal is to arrange the pieces into a square. The pieces are divided evenl between the + 3 long. two sides so that each side is However, there is a large piece of the square that is missing. In order to complete the square ou need to add 9 ones pieces. To complete the square of b + b, add. In other words, divide the coefficient b two and square the result. To complete the square of a + b, factor out the leading coefficient, a, giving b ou a +. Now add a b, which is the square a of the coefficient of divided b two. + b + b + b + b b + + a a b + + b + b + + a b b a + + a a b b a + + a a b a + a ( + ( 1) ) 3( 1)( 1) ( 1)

60 Practice Eercises B For each epression complete the square Finding Maimum/Minimum (the verte) Points of a Quadratic Function VOCABULARY Remember when a quadratic function is in verte form = ( ) + h, k is the verte of the parabola. The value of a determines whether the parabola opens up or down. f a h k the point The verte of a parabola that opens up, when a > 0, is the minimum point of a quadratic function. The verte of a parabola that opens up, when a < 0, is the maimum point of a quadratic function. Eample: Find the verte of f ( ) or minimum point. = + 3, then determine whether it is a maimum ( ) f = + 3 ( ) f = + 3 Rewrite the equation so it is in the general verte form f = a( h) + k. Verte: (-, -3) h = and k = 3 The verte is a minimum. The leading coefficient is 1, which makes a > 0 60

61 Eample: Find the verte of f ( ) or minimum point. ( ) f = ( ) f = = 5 8 +, then determine whether it is a maimum Verte: (8, ) h = 8 and k = The verte is a maimum. This equation is alread in the general verte form f = a( h) + k. The leading coefficient is -5, which makes a < 0. Practice Eercises C Find the verte and determine whether it is a maimum or minimum point. 1. f ( ) = 5 3. f = ( + 6). f ( ) = f ( ) = f = f ( ) = 7 1 NOTE: If a quadratic function is given in standard form, complete the square to rewrite the equation in verte form. Eample: Find the verte of f = , then determine whether it is a maimum or minimum point. f = f = f = ( + + ( 6) ) ( 6) f = Collect variable terms together inside parenthesis with constant term outside the parenthesis. b Complete the square b adding inside the parenthesis. Now subtract b outside the parenthesis to maintain equalit. In other words ou are reall adding zero to the equation. 61

62 f = Simplif ( ) f = Factor and combine like terms. Verte: (-6, -9) h = 6 and k = 9 The verte is a minimum. The leading coefficient is 1, which makes a > 0 Eample: Find the verte of f = , then determine whether it is a maimum or minimum point. f = f = f = f = ( + ( 3) ) 3 ( 3) f = f = ( ) Collect variable terms together inside parenthesis with constant term outside the parenthesis. Factor out the leading coefficient. In this case 3. b Complete the square b adding inside the parenthesis. Notice that everthing in the parenthesis is multiplied b 3 so we need to subtract b 3 outside the parenthesis to maintain equalit. In other words ou are reall adding zero to the equation. Simplif f = Factor and combine like terms. Verte: (-3, -9) h = 3 and k = 9 The verte is a minimum. The leading coefficient is 3, which makes a > 0 6

63 Eample: Find the verte of f = 8 + 3, then determine whether it is a maimum or minimum point. f = f = f = f = ( ) ( + ) 3 f = f = Simplif ( ) Collect variable terms together inside parenthesis with constant term outside the parenthesis. Factor out the leading coefficient. In this case -. b Complete the square b adding inside the parenthesis. Notice that everthing in the parenthesis is multiplied b - so we need to subtract b outside the parenthesis to maintain equalit. In other words ou are reall adding zero to the equation. f = Factor and combine like terms. Verte: (-1, 7) h = 1 and k = 7 The verte is a maimum. The leading coefficient is -, which makes a < 0 Practice Eercises D Find the verte of each equation b completing the square. Determine if the verte is a maimum or minimum. 1. f = f = f = f = + 1. f = f = 9 63

64 The ais of smmetr is the vertical line that divides a parabola in half. The zeros will alwas be the same distance from the ais of smmetr. The verte alwas lies on the ais of smmetr. When completing the square we end up with Eample: Ais of smmetr b f = a + + k a b f = a + k a f = a h + k k f = h = = = ( 3) 6 = f () = 3() + 1() 1 = 11 b Notice the -coordinate of the verte is. a The -coordinate can be found b evaluating b the function at. a The point (, 11) is the verte. Since 3 < 0, (, 11) is the maimum point of the function. Therefore, another method for finding the verte (h, k) from a standard form equation is to use h = b a and b k = f a. Practice Eercises E Identif the verte of each function. Then tell if it is a maimum or minimum point. 1. f = f = f = f =

65 YOU DECIDE A model rocket is launched from ground level. The function h( t) = 16t + 160t models the height h (measured in feet) of the rocket after time t (measured in seconds). Find the zeros and the verte of the function. Eplain what each means in contet of the problem. Practice Eercises F Solve 1. The height h(t), in feet, of a weeping willow firework displa, t seconds after having been launched from an 80-ft high rooftop, is given b h( t) = 16t + 6t When will it reach its maimum height? What is its maimum height?. The value of some stock can be represented b V = , where is the number of months after Januar 01. What is the lowest value V() will reach, and when did that occur? 3. Suppose that a flare is launched upward with an initial velocit of 80 ft/sec from a height of h t = 16t + 80t +. How long ft. Its height in feet, h(t), after t seconds is given b will it take the flare to reach the ground?. A compan s profit can be modeled b the equation p = where is the number of units sold. Find the maimum profit of the compan. 5. The Rainbow Bridge Arch at Lake Powell is the world s highest natural arch. The height of an object that has been dropped from the top of the arch can be modeled b the equation h( t) = 16t + 56, where t is the time in seconds and h is the height in feet. How long does it take for the object to reach the ground? 6. The amount spent b U.S. companies for online advertising can be approimated b 1 a( t) = t t + 8, where a(t) is in billions of dollars and t is the number of ears after 010. In what ear after 010 did U.S. companies spend the least amount of mone? 65

66 Unit 6 Cluster 3 (G.GPE.): Parabolas as Conics Cluster 3: Translating between descriptions and equations for a conic section 6.3. Find the equation of a parabola given the focus and directri parallel to a coordinate ais. VOCABULARY A parabola is the set of all points,, P, in a plane that are an equal distance from both a fied point, the focus, and a fied line, the directri. 66

67 Equation Direction Standard Form for the Equation of a Parabola Verte at (0, 0) Verte at (h, k) 1 = k = 1 ( h) p p Opens upward if p > 0 Opens upward if p > 0 Opens downward if p < 0 Opens downward if p < 0 Focus (0, p) (h, k + p) Directri = p = k p Graph Eample 1: Use the Distance Formula to find the equation of a parabola with focus ( 0,3 ) and directri = 3. PF = PD ( ) + ( ) = ( ) + ( ) ( 3) ( ) ( 3) A point P(, ) on the graph of a parabola is the same distance from the focus F ( 0,3) and a point on the directri D (, 3). + = + + Substitute in known values. + ( 3) = + 3 Simplif. ( + ( 3) ) = ( + 3) + ( 3) = ( + 3) Square both sides of the equation and use the properties of eponents to simplif. 67

68 = ( + 3) ( 3) ( 6 9) ( 6 9) = = 1 1 = 1 Solve for. Eample: Use the Distance Formula to find the equation of a parabola with focus (-5, 3) and directri = 9. PF = PD ( ) + ( ) = ( ) + ( ) 1 1 ( ) 5 ( 3) ( ) ( 9) A point P(, ) on the graph of a parabola is the same distance from the focus F ( 5,3) and a point on the directri D(,9). + = + Substitute in known values. ( 5 ) ( 3) ( 9) + + = Simplif. ( 5 ) ( 3) ( = ) ( 3) = ( 9) ( + 5) = ( ) ( 6 + 9) ( ) + 5 = ( 9) ( 3) + 5 = ( + 5) = 1( 6) 1 ( + 5) = 6 1 Square both sides of the equation and use the properties of eponents to simplif. Combine the terms on one side of the equation and the terms on the other side of the equation. Practice Eercises A Use the distance formula to find the equation of parabola with the given information. 1. focus ( 0, 5) directri = 5. focus (,6 ) directri = 8. focus ( 0,7 ) directri = 7 5. focus ( 3, ) directri = 1 3. focus ( 0, 3) directri = 6 6. focus ( 3,3 ) directri = 7 68

69 Equation Direction Standard Form for the Equation of a Parabola Verte at (0, 0) Verte at (h, k) 1 h = 1 ( k ) p p Opens to the right if p > 0 Opens to the right if p > 0 Opens to the left if p < 0 Opens to the left if p < 0 Focus ( p,0) ( h + p, k ) Directri = p = h p D(-p, ) P(, ) D(h - p, ) P(, ) Graph (h, k) F(p, 0) (h, k) F(h + p, k) = - p = h - p Eample: Use the Distance Formula to find the equation of a parabola with focus (,0 ) and directri =. PF = PD ( ) + ( ) = ( ) + ( ) 1 1 ( 0) ( ) ( ) A point (, ) on the graph of a parabola is the same distance from the focus (,0 ) and a point on the directri (, ). + = + + Substitute in known values. ( ) ( ) + = + Simplif. ( ) ( + = + ) + = ( + ) Square both sides of the equation and use the properties of eponents to simplif. 69

70 1 8 = + ( ) ( ) = = 8 = Solve for. Eample: Use the Distance Formula to find the equation of a parabola with focus (,3 ) and directri = 6. PF = PD ( ) + ( ) = ( ) + ( ) 1 1 ( ) ( 3) ( 6) ( ) A point (, ) on the graph of a parabola is the same distance from the focus (,3 ) and a point on the directri ( 6, ). + = + Substitute in known values. ( ) ( 3) ( 6) + = Simplif. ( ) ( 3 ) ( 6 + = ) ( ) + ( 3) = ( 6) ( 3) = ( 6) ( ) ( 3) = ( ) ( ) 3 = + 0 ( 3) = ( 5) 1 ( 3) = 5 Square both sides of the equation and use the properties of eponents to simplif. Combine the terms on one side of the equation and the terms on the other side of the equation. Practice Eercises B Use the distance formula to find the equation of parabola with the given information. 1. focus (,0) directri =. focus (, 3) directri = 5. focus ( 5,0 ) directri = 5 5. focus (, ) directri = 6 3. focus ( 3,0 ) directri = 3 6. focus ( 1,1) directri = 5 70

71 Practice Eercises C Determine the verte, focus, directri and the direction for each of the following parabolas. 1. 1( + 1) = ( 3). 6( 3) = ( + 1). ( + ) = 6( 1) 5. ( + 3) = 1( ) 3. ( 1) = ( + 5) 6. ( 6) = 16( ) You Decide A parabola has focus (-,1) and directri = -3. Determine whether or not the point (,1) is part of the parabola. Justif our response. 71

72 Unit 6 Cluster 3 Honors (G.GPE.3) Deriving Equations of Ellipses and Hperbolas Cluster 3: Translate between the geometric description and the equation for a conic section H.5.1 Derive the equations of ellipses and hperbolas given the foci, using the fact that the sum or difference of distances from the foci is constant. VOCABULARY An ellipse is the set of all points in a plane the sum of whose distances from two fied points (called foci), F 1 and F, is constant. The midpoint of the segments connecting the foci is the center of the ellipse. P(, ) d1 F1(-c, 0) d F(c, 0) An ellipse can be elongated horizontall or verticall. The line through the foci intersects the ellipse at its vertices. The segment whose endpoints are the vertices is called the major ais. The minor ais is a segment that is perpendicular to the major ais and its endpoints intersect the ellipse. 7

73 Deriving the Standard Equation of an Ellipse PF1 + PF = a + c c + 0 = a ( + c) + ( 0) = a ( c) + ( 0) ( ( + c) + ( 0) ) = ( a ( c) + ( 0) ) ( + c) + ( 0) = a a ( c) + ( 0) + ( c) + ( 0) + c + c + = a a c c + c + c = a a c + 0 c a = a c + 0 c a ( 0) = a c + ( 0) c a a c + = + ( 0) a c = a c + ( a c) = a ( c) + ( 0) a a c + c = a c + 0 a a c + c = a c + c + a a c + c = a a c + a c + a a + c = a + a c + a a a c = a c + a a a c = a c + a a b = b + a Let The sum of the distance from a point P, on the ellipse to each foci, c,0 and c,0, is equal to a. Use the distance formula and substitute in known values. Isolate one of the radicals. Square each side then simplif. Isolate the radical again. Square each side then simplif. Combine all the terms containing and on one side. b = a c. 73

74 a b b + a = a b a b 1 = + a b Divide b a b. Equation Major Ais Minor Ais Standard Form for the Equation of an Ellipse Centered at (0, 0) Horizontal Ellipse + = 1, a > b a b Along the -ais Length: a Along the -ais Length: b Vertical Ellipse + = 1, a > b b a Along the -ais Length: a Along the -ais Length: b Foci ( c,0 ) and ( c,0) ( 0, c) and ( 0, c) Vertices ( a,0 ) and ( a,0) ( 0, a) and ( 0, a) Pthagorean Relation a = b + c a = b + c Basic Graph Eample: Locate the vertices and foci for the ellipse 5 + = 100. Graph the ellipse. 5 + = = = 1 5 The standard equation of an ellipse is equal to 1. Divide each side of the equation b 100 and simplif. 7

75 + = 1 5 a = 5 = 5 b = = 5 = + c 1 = c 1 = c Vertices: ( 0, 5 ) and ( 0,5) Foci: ( 0, 1 ) and ( 0, 1) Identif a and b. Remember a > b. Note: a and b are lengths therefore the positive square root will ALWAYS be used. Use a and b to find c. Remember a = b + c. The vertices are ( 0, a) and ( 0, a) foci are ( 0, c) and ( 0, c) ellipse is vertical. and the because the Begin graphing the ellipse b plotting the center which is at (0, 0). Then plot the 0, 5 0,5. vertices which are at ( ) and Use the length of b to plot the endpoints of the minor ais. b = so the endpoints are units to the left and right of the,0 and center (0, 0). The are at (,0 ) Connect our points with a curve. 75

76 Eample: Write an equation in standard form for an ellipse with foci located at (,0 ) and (,0) and vertices located at ( 6,0 ) and ( 6,0). a b + = 1 6 = b + The ellipse is horizontal because the foci and vertices are along the -ais. Use the standard equation for a horizontal ellipse. a = 6 36 = b + c = 3 = b Find b using a = b + c. + = 1 Substitute in known values Practice Eercises A Locate the vertices and foci of the ellipse, then graph = = = = = 6. = 1 Write an equation in standard form for the ellipse that satisfies the given conditions. 7. Foci: ( 5,0 ) and ( 5,0) Vertices: ( 8,0 ) and ( 8,0) 8. Foci: ( 0, ) and ( 0, ) Vertices: ( 0, 7 ) and ( 0, 7) 9. Foci: ( 0, 3 ) and ( 0,3) Vertices: ( 0, ) and ( 0, ) 11. Major ais endpoints: ( 0, ± 6) Minor ais length Foci: ( 6,0 ) and ( 6,0) Vertices: ( 10,0 ) and ( 10,0) 1. Endpoints of aes are ( ± 5,0) and ( 0, ± ) 76

77 Ellipses Centered at (h, k) Standard Form for the Equation of an Ellipse Centered at (h, k) Equation ( h) ( k ) ( h) ( k ) a + = a > b b 1, b + = a > b a 1, Center ( h, k ) ( h, k ) Major Ais Minor Ais Parallel to the -ais Length: a Parallel to the -ais Length: b Parallel to the -ais Length: a Parallel to the -ais Length: b Foci ( h c, k ) and ( h + c, k ) ( h, k c) and ( h, k + c) Vertices ( h a, k ) and ( h + a, k ) ( h, k a) and ( h, k + a) Pthagorean Relation Eample: a = b + c a = b + c Locate the center, the vertices and the foci of the ellipse ( ) ( ) the ellipse. ( ) ( ) ( + ) ( ) = = ( + 3) ( ) + = 1 16 ( + 3) ( ) + = 1 16 a = 16 = b = = 16 = + c 1 = c 1 = c 3 = c Center: ( 3,) ( 3, ) and ( 3 +, ) Vertices: ( 7, ) and ( 1, ) Foci: ( 3 3, ) and ( 3 + 3, ) = 16. Graph The standard equation of an ellipse is equal to 1. Divide each side of the equation b 16 and simplif. Identif a and b. Remember Use a and b to find c. Remember a = b + c. a > b. h = 3 and k = The ellipse is horizontal, therefore the h a, k and h + a, k and vertices are the foci are ( h c, k ) and ( h c, k ) +. 77

78 Begin graphing the ellipse b plotting the 3,. Then plot the center of the ellipse vertices ( 7, ) and ( 1, ) Use the length of b to plot the endpoints of the minor ais. b = so the endpoints are units above and below the center 3,. 3,0 3,. ( ) The are at ( ) and Connect our points with a curve. Eample: Write an equation in standard form for an ellipse with foci at (,1 ) and (,5) vertices at (, 1 ) and (, 7). ( h) ( k ) b + = 1 = b + b 16 = + 1 = b a = Center:, (,3) ( + ) ( 3) and The ellipse is vertical because the foci and vertices are parallel to the -ais. Use the standard equation for a horizontal ellipse. a = 7 ( 1) c = 5 1 a = 8 a = Find b using c = c = a = b + c. The center is the midpoint of the vertices. + = 1 Substitute in known values

79 Practice Eercises B Locate the center, vertices and foci of the ellipse, then graph. 1. ( ) ( 1) + = 1. 9 ( ) ( + ) + = ( 3) ( + 1) + = ( 3) + 9( + ) = ( 1) + ( + 3) = ( ) ( ) = Write an equation in standard form for the ellipse that satisfies the given conditions. 7. Foci: ( 1, ) and ( 5, ) Vertices: ( 0, ) and ( 6, ) 9. Foci: (, ) and ( 6, ) Vertices: (, ) and ( 8, ) 11. Vertices: ( 5, ) and ( 3, ) Minor ais length is Foci: ( 3, 6 ) and ( 3, ) Vertices: ( 3, 7 ) and ( 3,3) 10. Foci: ( 1, 0 ) and ( 1, ) Vertices: ( 1,1 ) and ( 1, 5) 1. Vertices: 0, and 6, Minor ais length is. 79

80 VOCABULARY A hperbola is the set of all points in a plane whose distances from two fied points in the plane have a constant difference. The fied points are the foci of the hperbola. The line through the foci intersects the hperbola at its vertices. The segment connecting the vertices is called the transverse ais. The center of the hperbola is the midpoint of the transverse ais. Hperbolas have two oblique asmptotes that intersect at the center. Deriving the Standard Equation of a Hperbola PF1 PF = ± a The difference of the distance from a point P, on the hperbola to each foci, c,0 and c,0, is equal to ± a. 80

81 c c + 0 = ± a c + + c + = ± a c + = ± a + + c + ( ) c + = ± a + + c + c + = a ± a + c c + c + + c = a ± a + c + c + c + c + c = a ± a + c + c a = ± a + c + c a c = a ± a + c + = ± a + c + c a = ± a + c + c a = ± a + c + c + a c + a = a + c + c + a c + a = a + c + c + c + a c + a = a + a c + a c + a c + a = a + a c + a c a a = a c a ( ) = ( ) c a a a c a b a = a b Let b a a b = a b a b = 1 a b Use the distance formula and substitute in known values. Isolate one of the radicals. Square each side then simplif. Isolate the radical again. Square each side then simplif. Combine all the terms containing and on one side. b = c a. Divide b a b. 81

82 Equation Transverse Ais Conjugate Ais Standard Form for the Equation of a Hperbola Centered at (0, 0) Opens Left and Right = 1 a b -ais Length: a -ais Length:b Opens Up and Down = 1 a b -ais Length: a -ais Length:b Foci ( c,0 ) and ( c,0) ( 0, c) and ( 0, c) Vertices ( a,0 ) and ( a,0) ( 0, a) and ( 0, a) Pthagorean Relation Asmptotes c = a + b b = ± a c = a + b a = ± b Basic Graph Eample: Find the vertices, foci and asmptotes of the hperbola hperbola =. Then graph the 9 = = = 1 9 The standard equation of an ellipse is equal to 1. Divide each side of the equation b 36 and simplif. 8

83 = 1 9 a = 9 = 3 b = = c c c = = 9 + = Vertices: ( 3,0 ) and ( 3,0) Foci: ( 13, 0 ) and ( 13, 0) Asmptotes: = and 3 = 3 Identif a and b. Use a and b to find c. Remember c = a + b. This hperbola opens left and right so the a,0 a,0 and the foci vertices are ( ) and are ( c,0 ) and ( c,0). b The asmptotes are = ±. a Begin graphing the hperbola b plotting the center at (0, 0). Then plot the vertices 3,0 and 3,0. at Use the length of b to plot the endpoints of the conjugate ais. b = so the endpoints are units above and below the center 0, 0,. ( 0,0 ). The are at ( ) and Construct a rectangle using the points. 83

84 Draw the asmptotes b drawing a line that connects the diagonal corners of the rectangle and the center Use the asmptotes to help ou draw the hperbola. The hperbola will open left and right and pass through each verte. Eample: Write an equation in standard form for the hperbola with foci ( 0, 3) and conjugate ais has length. 0,3 whose a b b = = 1 b = 3 = a + = a = a The foci are along the -ais so the hperbola s branches open up and down. The conjugate ais is length. Use it to solve for b. Use b = and c = 3 to solve for Remember c = a + b. a. = 1 Substitute in known values. 5 8

85 Practice Eercises C Locate the center, vertices, foci and asmptotes of the hperbola, then graph. 1. = = = = = 6 6. = 16 Write an equation in standard form for the hperbola that satisfies the given conditions. 7. Foci: ( 0, ) and ( 0, ) Vertices: ( 0, 1 ) and ( 0,1) 9. Foci: ( 0, 7 ) and ( 0, 7) Vertices: ( 0, 5 ) and ( 0,5) 11. Vertices: (,0 ) and (,0) Conjugate ais length is Foci: ( 5,0 ) and ( 5,0) Vertices: ( 3,0 ) and ( 3,0) 10. Foci: ( 10,0 ) and ( 10,0) Vertices: ( 6,0 ) and ( 6,0) 1. Vertices: ( 0, 3 ) and ( 0,3) Conjugate ais length is 6. Equation Transverse Ais Conjugate Ais Standard Form for the Equation of a Hperbola Centered at (h, k) Opens Left and Right ( h) ( k ) = 1 a b Parallel to -ais Length: a -ais Length:b Opens Up and Down ( k ) ( h) = 1 a b Parallel to -ais Length: a -ais Length:b Foci ( h c, k ) and ( h + c, k ) ( h, k c) and ( h, k + c) Vertices ( h a, k ) and ( h + a, k ) ( h, k a) and ( h, k + a) Pthagorean Relation c = a + b b a c = a + b Asmptotes k = ± ( h) k = ± ( h) a b 85

86 Eample: Find the center, vertices, foci and asmptotes of the hperbola ( + ) ( 5 ) Then graph the hperbola. = ( + ) ( 5) = ( + ) ( 5) = a = 9 = 3 b = 9 = 7 c c c = = = Center: (,5) ( 3, 5 ) and ( + 3, 5 ) Vertices: ( 5,5 ) and ( 1,5 ) Foci: ( 58,5 ) and ( + 58, 5) 7 = 3 + and 7 5 = ( + ) 3 Asmptotes: 5 ( ) Identif a and b. Use a and b to find c = a + b. c. Remember that The hperbola s branches open left and h a, k and right so the vertices are ( h + a, k ). The foci are ( h c, k ) ( h + c, k ). and Begin graphing the hperbola b plotting the center at (-, 5). Then plot the 5,5 and 1,5. vertices at 86

87 Use the length of b to plot the endpoints of the conjugate ais. b = 7 so the endpoints are 7 units above and below the,5., and center ( ) The are at (,1). 1 8 Construct a rectangle using the points Draw the asmptotes b drawing a line that connects the diagonal corners of the rectangle and the center Use the asmptotes to help ou draw the hperbola. The hperbola will open left and right and pass through each verte. 87

88 Eample: Write an equation in standard form for the hperbola whose vertices are (, 1) ( 8, 1) and whose conjugate ais has length 8. and ( k ) ( h) = 1 a b , = 3, 1 a = 8 ( ) Center: a = 10 a = 5 b = 8 b = ( 3) ( + 1) = The foci are parallel to the -ais so the hperbola s branches open left and right. The midpoint of the vertices is the center of the hperbola. The vertices are at (, 1) and ( 8, 1). Use the distance between them to find a. The conjugate ais is length 8. Use it to solve for b. Substitute in known values. Practice Eercises D Locate the center, vertices, foci and asmptotes of the hperbola, then graph. 1. ( 5) ( 6) = = ( + 1) ( 3) = ( + ) ( + 6) = ( 6) 5( ) = ( ) ( ) = 8 Write an equation in standard form for the hperbola that satisfies the given conditions. 7. Foci: ( 1,9 ) and ( 1,1 ) Vertices: ( 1,7 ) and ( 1,3 ) 9. Foci: ( 8, ) and (, ) Vertices: ( 7, ) and ( 3, ) 11. Vertices: ( 3,6 ) and ( 3, ) Minor ais length is Foci: (, 5 ) and ( 8, 5) Vertices: ( 0, 5 ) and ( 6, 5) 10. Foci: ( 3,5 ) and ( 3, 11) Vertices: ( 3,1 ) and ( 3, 7) 1. Vertices: 7, and 3, Minor ais length is 6. 88

89 Unit Cluster 3 (F.BF.1) Building Functions That Model Relationships Between Two Quantities Cluster 3: Building functions that model relationships between two quantities.3.1 Focus on quadratics and eponentials to write a function that describes a relationship between quantities ( nd difference for quadratics).3.1 Determine an eplicit epression or steps for calculation from contet..3.1 Combine functions using arithmetic operations. Vocabular A function is a relation for which each input has eactl one output. In an ordered pair the first number is considered the input and the second number is considered the output. If an input has more than one output, then the relation is not a function. For eample the set of ordered pairs {(1,), (3,5),(8,11)} is a function because each input value has an output value. The set {(1, ) (1, 3), (6, 7)} does not represent a function because the input 1 has two different outputs and 3. Linear Function- a function that can be written in the form = m + b,where m and b are constants. The graph of a linear function is a line. A linear function can be epressed in two different was: Linear notation: = m + b Function notation: f = m + b f = + 1 Linear functions can model arithmetic sequences, where the domain is the set of positive integers, because there is a common difference between each successive term. The common difference can also be called the first difference. Linear functions can model an pattern where the first difference is the same number. 1, 3, 5, 7, st difference 89

90 Eponential Function- a function of the form f = ab where a and b are constants and a 0, b > 0, and b 1. Eponential functions are most easil recognized b the variable in the eponent. The values of f() are either increasing (eponential growth) if a > 0 and b > 1 or decreasing (eponential deca) if a > 0 and 0 < b < Eponential functions can model geometric sequences, where the domain is the set of positive integers, because each successive term is multiplied b the same number called the common ratio. Eponential functions can model an pattern where the net term is obtained b multipling each successive term b the same number. f ( ) = 1, 3, 9, 7, common ratio Quadratic Function- a function that can be written in the form f = a + b + c where a 0. Quadratic functions are most easil recognized b the term. The graph is a parabola. A quadratic function can be formed b multipling two linear functions. The quadratic function to the right can also be written as ( 3)( 1) f = To determine if a pattern or a sequence can be modeled b a quadratic function, ou have to look at the first and second difference. The second difference is the difference between the numbers in the first difference. If the first difference is not the same number but the second difference is, then the pattern or sequence can be modeled b a quadratic function. f = 5 3 1,, 9, 16, st difference nd difference 90

91 Eample: Determine if the pattern 1, 3, 9, 19, would be modeled b a linear function, an eponential function, or a quadratic function. Answer: Check the first difference to see if it is the same number each time. For this pattern, it is not the same, so it will not be modeled b a linear function. Check to see if each term is being multiplied b the same factor. For this pattern, it is not the same, so it will not be modeled b an eponential function. Check the second difference to see if it is the same number each time. For this pattern, it is the same, so the pattern can be modeled b a quadratic function. 1, 3, 9, 19, , 3, 9, 19, , 3, 9, 19, Conclusion: The pattern can be modeled b a quadratic function. + + Practice Eercises A Determine if the pattern would be modeled b a linear function, an eponential function, or a quadratic function , 18, 8, , 7, 9, 6. 8, 16,, 91

92 Eample: Using a graphing calculator determine the quadratic function modeled b the given data f() Input the data into a TI-83 or TI-8 calculator list Enter the information into our lists b pushing STAT followed b Edit. If ou have values in our lists alread, ou can clear the information b highlighting the name of the list then pushing CLEAR and ENTER. Do not push DEL or it will delete the entire list. Enter the values into L1 and the f() values into L. Push nd MODE to get back to the home screen. Make a scatter plot Push nd Y= to bring up the STAT PLOT menu. Select Plot1 b pushing ENTER or 1. Turn Plot1 on b pushing ENTER when ON is highlighted. Make sure that the scatter plot option is highlighted. If it isn t, select it b pushing ENTER when the scatter plot graphic is highlighted. The Xlist should sa L1 and the Ylist should sa L. If it doesn t, L1 can be entered b pushing nd 1 and L b nd. To view the graph ou can push GRAPH. If ou want a nice viewing window, first push ZOOM arrow down to option 9 ZOOMSTAT and either push ENTER or push 9. Creating a quadratic regression equation You do not have to graph a function to create a regression, but it is recommended that ou compare our regression to the data points to determine visuall if it is a good model or not. From the home screen push STAT, arrow right to CALC and either push 5 for QuadReg or arrow down to 5 and push ENTER. (To do an eponential regression, push 0 for EpReg or arrow down to 0 and push ENTER.) Tpe nd 1, (the comma is located above 7) nd, VARS arrow right to Y-VARS select FUNCTION and Y1 then push ENTER. f = 3 1. It has been The quadratic regression is pasted into Y1 so that ou can push GRAPH again and compare our regression to the data. 9

93 Practice Eercises B Find the regression equation. Round to three decimals when necessar. 1. Given the table of values use a graphing calculator to find the quadratic function f() Use a graphing calculator to find a quadratic model for the data f() From 197 to 1998 the U.S. Fish and Wildlife Service has kept a list of endangered species in the United States. The table below shows the number of endangered species. Find an appropriate eponential equation to model the data. Year Number of species. The cell phone subscribers of the small town of Herriman are shown below. Find an eponential equation to model the data. Year Subscribers 85 80,59 6,360 17,90 Eample: When doctors prescribe medicine, the must consider how much the effectiveness of the drug will decrease as time passes. The table below provides information on how much of the drug remains in a person s sstem after t hours. Find a model for the data. t (hours) Amount (mg) Answer: Sometimes it is helpful to look at the graph of the points. For this particular eample, it is difficult to determine if this should be modeled b an eponential or a quadratic function 93

94 from the graph. Therefore, consider the contet of the eample. The amount of the drug will continue to decrease 00 unless more is given to the patient. If the patient does not 100 receive more medication, at some point there will onl be trace amounts of the drug left in the patient s sstem. This would suggest a function that continues to decrease until it reaches a leveling off point. An eponential model would be better suited for this situation Use the regression capabilities of our graphing calculator to find an eponential model for the data. Follow the instructions for the previous eample but make sure that ou select option 0: EpReg. The function that models the data is: f ( ) = 9.977( 0.950) Practice Eercises C Determine if the data is best modeled b an eponential or quadratic function. Then find the appropriate regression equation. 1. The pesticide DDT was widel used in the United States until its ban in 197. DDT is toic to a wide range of animals and aquatic life, and is suspected to cause cancer in humans. The half-life of DDT can be 15 or more ears. Half-life is the amount of time it takes for half of the amount of a substance to deca. Scientists and environmentalists worr about such substances because these hazardous materials continue to be dangerous for man ears after their disposal. Write an equation to model the data below. Year Amount of DDT (in grams) Use a graphing calculator to find a model for the data f()

95 3. The table shows the average movie ticket price in dollars for various ears from 1983 to 003. Find the model for the data. Years since 1983, t Movie ticket price, m The table below shows the value of car each ear after it was purchased. Find a model for the data. Years after purchase Value of car,000 0,160 16,93 1,5 11,99 10,037 Combining functions using arithmetic operations Let f and g be an two functions. A new function h can be created b performing an of the four basic operations on f and g. Operation Definition Eample: Addition h = f + g Subtraction h = f g Multiplication h = f g Division f h g f = 5 +, g = 3 = ( 3 ) = + h = 5 + ( 3 ) = 8 + h h = (5 + ) ( 3 ) = = ( 5 + ) h = = = The domain of h consists of the -values that are in the domains of both f and g. Additionall, the domain of a quotient does not include -values for which g() = 0. Adding and Subtracting Functions Eample: Let f = + 1 and the domain of the new function. g = + 3. Perform the indicated operation and state a. h = f + g b. h = g f c. h = f g 95

96 Answer: a. h = f + g ( 1) ( 3 ) h = Replace f with 1 + and g ( ) with + 3. h = Remove the parentheses because we can add in an order. h = Combine the like terms. ( + 3), (1 + ) g The domain is (, ) b. h = g f. The domain for both f ( ) and is (, ). just like g ( ) so it has the same ( 3 ) ( 1) h domain. = + + Replace h a quadratic function g with f ( ) with and h = Distribute the negative throughout the second term and remove the parentheses. h = + 5 Combine the like terms. The domain is (, ). The domain for both f ( ) and g ( ) is (, ). just like g ( ), so it has the same c. h = f g ( 1) ( 3 ) h domain. = + + Replace h a quadratic function, f with 1 + and g ( ) with + 3. h = Distribute the two through the first term and distribute the negative through the second term. h = Combine like terms. The domain is (, ). The domain for both f ( ) and g ( ) is (, ). just like g ( ) so it has the same domain. h a quadratic function 96

97 Multipling Functions Eample: Answer: Let f = + 1 and the domain of the new function. g = + 3. Perform the indicated operation and state a. h = f g b. h = g g C. h = f f a. h = f g ( 1) ( 3 ) h = + + Replace f with 1 + and g ( ) with h = Multipl using our method of choice. (See Unit 1 Cluster lesson) g The domain is (, ). The domain for both f ( ) and is (, ). h a polnomial function just like g ( ) so it has the same domain. b. h = g g ( 3 ) ( 3 ) h = + + Replace 3 g with + 3 h = Multipl using our method of choice. (See Unit 1 Cluster lesson),. The domain is (, ). The domain for g ( ) is h a polnomial function just like g ( ) so it has the same domain. c. h = f f h = ( + 1) ( + 1) Replace h 1 f with + 1. = + + Multipl using our method of choice. (See Unit 1 Cluster lesson),. The domain is (, ). The domain for f ( ) is h a quadratic function so it has the same domain. 97

98 Dividing Functions VOCABULARY p A rational function is a function of the form r = where p ( ) and q q are polnomials and q 0. The domain of a rational function includes all real numbers ecept for those that would make q ( ) = 0. A rational epression is in simplified form if the numerator and the denominator have no common factors other than 1 or -1. Eample: Answer: Let f = + 1 and the domain of the new function. f a. h h g a. h g = + 3. Perform the indicated operation and state = b. = g f C. h ( ) = f g = h h = ( 1)( + ) The domain is,,1 1,. and f = f Replace f ( ) with + 1 and g ( ) with + 3 Factor the numerator and the denominator to see if the function can be simplified. (See Unit Cluster (F.IF.8) for help with factoring) h is a rational The new function function. The domain cannot include an numbers for which the denominator is zero. The denominator is zero when = and = 1. b. h h h = g f + 3 = + 1 ( 1)( + ) = + 1 Replace f ( ) with + 1 and g ( ) with + 3 Factor the numerator and the denominator to see if the function can be simplified. (See Unit Cluster 98

99 The domain is 1 1,,. (F.IF.8) for help with factoring) h is a rational The new function function. The domain cannot include an numbers for which the denominator is zero. The denominator is zero when 1 =. c. h h h f ( + ) f = 1 = + 1 ( + 1) = + 1 Replace f ( ) with + 1. Factor the numerator and the denominator to see if the function can be simplified. (See Unit Cluster (F.IF.8) for help with factoring) h = Divide out the factors and simplif the epression. The domain is Although the simplified form of h( ) 1 1,,. is not a rational function, it started out as a rational function and the same restrictions appl on the simplified form. The denominator is zero when 1 =. Practice Eercises E If f = + 3, g = 3 and domain of the new function. h = 1 + 6, find the following. State the 1. f + h. g f g h. f + g 5. g h 8. f h 3. h 3g 6. h g + 9. h h 10. f h 11. g h 1. f g 99

100 Evaluating Combined Functions Eample: Let f = + 5 and g = Evaluate each epression. a. f + g ( 1) b. f ( 3) g ( 0) c. g f ( 1) ( 1) Answer: a. f f f g + g ( 1) = + 5 = 7 1 = (1) + 8(1) + 5 g 1 = g 1 = g 1 = = f + g 1 = This epression tells ou to find the value of f at = and the value of g at = 1 and add the results. Find the value of f at =. Find the value of g at = 1. Add the results. b. f f f g g g g ( 3) g ( 0) ( 3) = 3+ 5 ( 3) = ( 0) = ( 0) = ( 0) = 5 0 = = 10 f 3 g 0 = 10 This epression tells ou to find the value of f at = 3 and the value of g at = 0 and multipl the results. Find the value of f at = 3. Find the value of g at = 0. Multipl the results. 100

101 c. g f g g g g f f ( 1) ( 1) ( 1) = ( 1) = ( 1) = 1 ( 1) = 1+ 5 ( 1) = 1 = g ( 1) 1 = or -0.5 f 1 This epression tells ou to find the value of g at = 1 and the value of f at = 1 and divide the results. Find the value of g at = 1. Find the value of f at = 1. Divide the results. Practice Eercises F If f = and g = 3 5, find the value of each epression f. f ( 1) g( 5) 7. f ( ) g. g ( 1) + f ( 3) 5. f g 8. f ( 1) 3g ( 1) 3. ( 3) ( 3) f f 6. g f ( 1) 9. g ( 0) f 0 101

102 Practice Eercises G 1. A compan estimates that its cost and revenue can be modeled b the functions C = , 000 and R = where is the number of units produced. The compan s profit, P, is modeled b R C and determine the profit when 1,000,000 units are produced.. Find the profit equation 1 p = + 30; 0 50 where p represents the price 15 and the number of units sold. Write an equation for the revenue, R, if the revenue is the price times the number of units sold. What price should the compan charge to have maimum revenue?. Consider the demand equation 3. The average Cost C of manufacturing computers per da is obtained b dividing the cost function b the number of computers produced that da,. If the cost function is C = , find an equation for the average cost of manufacturing. What 3 is the average cost of producing 100 computers per da?. The service committee wants to organize a fund-raising dinner. The cost of renting a facilit is $300 plus $5 per chair or C = , where represents the number of people attending the fund-raiser. The committee wants to charge attendees $30 each or R 30 How man people need to attend the fund-raiser for the event to raise $1,000? =. 10

103 Unit Cluster (F.BF.3 and F.BF.): Transformations and Inverses Cluster : Building New Functions from Eisting Functions..1 Transformations, odd and even graphicall and algebraicall.. Find inverse functions (simple) focus on linear and basic restrictions for quadratics, introduce one-to-one and horizontal line test VOCABULARY There are several tpes of functions (linear, eponential, quadratic, absolute value, etc.). Each of these could be considered a famil. Each famil has their own unique characteristics that are shared among the members. The parent function is the basic function that is used to create more complicated functions. The graph of a quadratic function is in the shape of a parabola. This is generall described as being u shaped. The maimum or minimum point of a quadratic function is the verte. When a quadratic function is written in verte form, f = a ( h) + k, then the verte, ( h, k ), is highlighted. The ais of smmetr is the vertical line that divides the graph in half, with each half being a reflection of the other. The equation for the ais of smmetr is = h. Quadratic parent function f = f = ( 3) = 9 ( ) = ( 1) = 1 (0) = 0 (1) = 1 () = (3) = The ais of smmetr is the line = 0. The verte is the point (0, 0). The domain is the set of, 0,. all real numbers ( ). The range is the set of positive real numbers including zero [ ) 103

104 f = Vertical Shift: f = ( 3) = 7 ( ) = ( 1) = 1 (0) = (1) = 1 () = 3 (3) = 7 Effect on the graph: The parabola has been shifted down units. f = + 1 Vertical Shift: Ais of smmetr: = 0 Verte: (0, -) Domain: (, ) Range: [, ) f = + 1 ( 3) + 1 = 10 ( ) + 1 = 5 ( 1) + 1 = (0) + 1 = 1 (1) + 1 = () + 1 = 5 3 (3) + 1 = 10 Effect on the graph: The parabola has been shifted up 1 unit Ais of smmetr: = 0 Verte: (0, 1) Domain: (, ) Range: [ 1, ) Horizontal Shift: f = ( ) f = ( 1 ) = 9 (0 ) = (1 ) = 1 ( ) = 0 (3 ) = 1 ( ) = 5 (5 ) = 9 Effect on the graph: the parabola has been shifted units to the right Ais of smmetr: = Verte: (, 0) Domain: (, ) Range: [ 0, ) 10

105 Horizontal Shift: f = ( + 3) f = + 3 ( 6 + 3) = 9 ( 5 + 3) = ( + 3) = 1 ( 3 + 3) = 0 ( + 3) = 1 ( 1+ 3) = 0 (0 + 3) = 9 Effect on the graph: the parabola has been shifted three units to the left. Reflection: f = f = ( 3) = 9 = ( 1) = 1 (0) = 0 (1) = 1 () = 3 (3) = 9 Effect on the graph: the parabola has been reflected over the -ais. f = Vertical Stretch: f = ( 3) = 18 ( ) = 8 ( 1) = (0) = 0 (1) = () = Ais of smmetr: = 3 Verte: (-3, 0) Domain: (, ) Range: [ 0, ) Ais of smmetr: = 0 Verte: (0, 0) Domain: (, ) Range: (,0] Ais of smmetr: = 0 Verte: (0, 0) Domain: (, ) Range: [ 0, ) 3 (3) = 18 Effect on the graph: the -coordinates of the parabola have been multiplied b

106 f = 1 Vertical Shrink: f = ( 3) = ( 1) = (1) = ( 3) = Effect on the graph: the -coordinates of the parabola have been divided Ais of smmetr: = 0 Verte: (0, 0) Domain: (, ) Range: [ 0, ) Horizontal Shift f = a( h) + k Vertical Stretch or Reflection Vertical Shift Eample: Describe the transformations performed on f = to make it f = ( 1) + 5. Then graph the function and identif the ais of smmetr, the verte, the domain and the range. Transformations: 6 Ais of smmetr: = 1 reflected over the -ais Verte: (1, 5) shifted one unit to the right Domain: (, ) shifted up three units - -6 Range: (,5] 106

107 Eample: Describe the transformations performed on f = to make it f = 3 ( + ). Then graph the function and identif the ais of smmetr, the verte, the domain and the range. Transformations: 6 Ais of smmetr: = -coordinates multiplied b 3/ Verte: (-, -) shifted two units to the left shifted down four units Domain: (, ) Range: [, ) Practice Eercises A Describe the transformations performed on f = to make it the following: 1. f = + 6. = ( + 5) 7 3. = 3( ) f f Graph each function and identif the ais of smmetr, the verte, the domain and the range. 1 f = 6 f = f = f ( ) = f = f = 3 ( ) 3 VOCABULARY The absolute value function is actuall a piecewise-defined function consisting of two linear equations., if 0 f = =, if < 0 Absolute value is often defined as the distance from zero. Therefore, the output is positive. The point where the two linear equations meet is called the verte. It is also the minimum or h, k, can easil be identified when the absolute value maimum of the function. The verte, function is represented in the form f = a h + k. 107

108 Absolute Value parent function f = f = -3 3 = 3 - = -1 1 = = = 1 = 3 3 = The verte is the point (0, 0). The domain is the set of all real numbers (, ). The range is the set of positive real numbers including zero [ 0, ). Vertical Shift: f = f = -3 3 = 1 - = -1 1 = = 1 1 = 3 = 3 3 = Verte: (0, -) Domain: (, ) Range: [, ) Effect on the graph: The function has been shifted down units. 108

109 Horizontal Shift: f = 1 f = 1-1 = = = = 0 1 = = 1 = Verte: (1, 0) Domain: (, ) Range: [ 0, ) Effect on the graph: The function has been shifted right 1 unit. Reflection: f = f = -3 3 = 3 - = -1 1 = = = 1 = 3 3 = Verte: (0, 0) Domain: (, ) Range: (,0] Effect on the graph: The function has been reflected over the -ais. Vertical Stretch: f = 3 f = = 9-3 = = = = 3 3 = = Verte: (0, 0) Domain: (, ) Range: [ 0, ) Effect on the graph: The -coordinates of the function have been multiplied b

110 Horizontal Shift f = a h + k Vertical Stretch or Reflection Vertical Shift Eample: Describe the transformations performed on f = to make it f 3 5 = +. Then graph the function and identif the ais of smmetr, the verte, the domain and the range. Transformations: reflected over the -ais -coordinates multiplied b shifted three units to the right shifted up five units Verte: (3, 5) Domain: (, ) Range: (,5] -6 Practice Eercises B Describe the transformations performed on f = to make it the following: 1. f = + 5. f = f = Graph each function and identif the verte, the domain and the range. f = + f = f = f = 5 8. f = f =

111 Vocabular The words even and odd describe the smmetr that eists for the graph of a function. A function is considered to be even if, for ever number in its domain, the number is also in the f = f. Even functions have -ais smmetr. domain and A function is considered to be odd if, for ever number in its domain, the number is also in the f = f. Odd functions have origin smmetr. domain and Even Function The function graphed at the left is even because (, 1) is a point on the graph and (-, 1) is also a point on the graph. Notice that - is the opposite of, but both inputs give the f = f, i.e. opposite inputs same output. Therefore, generate the same output Odd Function The function graphed at the left is odd because (, ) is a point on the graph and (-, -) is also a point on the graph. Notice that the input - is the opposite of, and gives the opposite f = f, i.e. opposite output from. Therefore, inputs generate outputs that are opposites of each other Neither Even nor Odd The function graphed at the left is neither even nor odd. It is not even because the point (, ) is on the graph, but (-, ) is not. Similarl, it is not odd because the point (-, -) is not a point on the graph. 111

112 Practice Eercises C Determine if the following graphs represent functions that are even, odd or neither Determining Even and Odd Algebraicall It is possible to graph functions and visuall determine whether the function is even or odd, but there is also an algebraic test that can be applied. It was previousl stated that if a function is f = f. even, then evaluating the function at and should produce the same output or If a function is odd, then evaluating the function at and should produce outputs that are f = f. opposite or Eample: Is the function 3 f = an even function, an odd function, or neither? Original function. f = 3 Substitute in for each in the function. f ( ) = 3( ) Simplif. f ( ) = 3 Compare the output to the original function. 3 3 If the are the same, then the function is even. If the are opposite, then the function is odd. If the are anthing else, then the are neither. f ( ) f f ( ) = f Conclusion: f = 3 is an odd function. 11

113 Eample: Is the function g = + an even function, an odd function or neither? Original function. g = + Substitute in for each in the function. g Simplif. = + g = + + Compare the output to the original function If the are the same, then the function is even. If the are opposite, then the function is odd. If the are anthing else, then the are neither. g ( ) g g ( ) g Eample: Conclusion: g = + is neither an even nor an odd function. Is the function h = + an even function, an odd function, or neither? Original function. h = + Substitute in for each in the function. h = + Simplif. h = + Compare the output to the original function. + = + If the are the same, then the function is even. If the are opposite, then the function is odd. If the are anthing else, then the are neither. h( ) = h h( ) h Conclusion: h = + is an even function. Practice Eercises D Determine algebraicall if the function is even, odd, or neither f = f 3 = f = 3 f = 5 1. f = 5 5. f = + 6.

114 VOCABULARY The graph of an inverse relation is the reflection of the graph of the original relation. The line of reflection is =. The original relation is the set of ordered pairs: {(-, 1), (-1, ), (0, 0), (1, -), (, -)}. The inverse relation is the set of ordered pairs: {(1, -), (, -1), (0, 0), (-, 1), (-, )}. Notice that for the inverse relation the domain () and the range () reverse positions. Original Relation Domain: {-, -1, 0, 1, } Range: {-, -, 0, 1, } The points are reflected over the line =. Notice that each point is the same distance awa from the line, but on the opposite side of the line. Inverse Relation Domain: {-, -, 0, 1, } Range: {-, -1, 0, 1, } Practice Eercises E Find the inverse relation. 1. {(1, -1), (, -), (3, -3), (, -), (5, -5)}. {(-, ), (-, 1), (0, 0), (, 1), (, )} 3. {(-10, 6), (3, -9), (-1, ), (-7, 1), (6, 8)}. {(7, 6), (, 9), (-3, -), (-7, 1), (8, 10)} VOCABULARY If no vertical line intersects the graph of a function f more than once, then f is a function. This is called the vertical line test. If no horizontal line intersects the graph of a function f more than once, then the inverse of f is itself a function. This is called the horizontal line test. The inverse of a function is formed when the independent variable is echanged with the dependent variable in a given relation. (Switch the and with each other.) A function takes a starting value, performs some operation on this value, and creates an output answer. The inverse of a function takes the output answer, performs some operation on it, and arrives back at the 1 original function's starting value. Inverses are indicated b the notation f. 11

115 A function is a one-to-one function if and onl if each second element corresponds to one and onl one first element. In order for the inverse of a function to be a function, the original function must be a one-to-one function and meets the criteria for the vertical and horizontal line tests. Not all functions meet the criteria to have an inverse which is also a function. However, if the domain is restricted, or in other words onl part of the domain is used, then the inverse will be a function. This eample is not one-to-one. It is a function because the vertical line intersects the graph onl once. However, the horizontal line intersects the graph twice. There is an inverse to this eample, but the inverse will not be a function. This eample is one-to-one. It is a function because the vertical and horizontal lines intersect the graph onl once. The inverse will be a function. Eample: Find the inverse of f = 3 1. Original function f = 3 1 Replace f ( ) with. = 3 1 Replace with and with. = 3 1 Isolate. + 1 = 3 The inverse of f = 3 1 is f = = =

116 Eample: 1 Find the inverse of f = +. Original function 1 f = + Replace f ( ) with. 1 = + Replace with and with. 1 = + Isolate. 1 = = ( ) 8 = 1 1 The inverse of f = + is f = 8. Eample: 1 Original function 1 f =, 0 Replace f ( ) with. 1 = Find the inverse of the function f = ; Domain (,0] and Range [ 0, ) Replace with and with. 1 = Isolate. Simplif the radical (Unit 1 Cluster1:N.RN.) = ± = ± = Determine whether to use the positive or negative answer b referring back to the restricted domain. The domain of the original function is restricted to the negative real numbers including zero, therefore, the range of the inverse function must also be the same. This leads us to choose the negative square root. = 1 f Domain [ 0, ) and Range (,0] Notice the domain and range have switched from the original function s domain and range. 116

117 Eample: Find the inverse of the function f = 3( + 1) 5 ; Domain [ 0, ) and Range [ 5, ) Original function Replace f with. f = , 0 = Replace with and with. = Isolate. Simplif the radical (Unit 1 Cluster1:N.RN.) ( ) + 5 = = + 3 ( 1) + 5 ± = ± 1 = 3 Determine whether to use the positive or negative answer b referring back to the restricted domain. The domain of the original function is restricted to the positive real numbers including zero, therefore, the range of the inverse function must also be the same. This leads us to choose the positive square root. + = f ( ) Domain [ 5, ) and Range [ 0, ) Notice the domain and range have switched from the original function s domain and range. Practice Eercises F Find the inverse of the following. State the domain and range of the inverse. For problems 7 9 restrict the domain before finding the inverse. 1. f = 3 +. f = f = f = f = f = 3 7. f 3 5 = 8. f = ( + ) 9. f ( ) =

118 Unit 3 Epressions and Equations 118

119 Unit 3 Cluster 1 (A.SSE.): Interpret the Structure of Epressions Cluster 1: Interpret the structure of epressions 3.1. Recognize functions that are quadratic in nature such as 6 = 3 + VOCABULARY A quadratic pattern can be found in other tpes of epressions and equations. If this is the case, we sa these epressions, equations, or functions are quadratic in nature. Recall the standard form of a quadratic epression is a + b + c, where a, b, and c are real numbers and a 0. The following are eamples of epressions that are quadratic in nature: Epression: Notice: Rewritten: + ( 3 ) = 6 3( 3 ) + 5( 3 ) = and ( ) = 3 3 ( ) 5( )( ) 1( ) 1/ 1/ + + ( 1/ ) = 1/ 9( 1/ ) + 1( 1/ ) ( ) = ( ) ( ) ( + 1) ( + 1) 15 ( + 1) = ( + 1 ) ( ) ( ) = 16 and ( 5) 5 = 5 ( ) ( 5) 6 = and ( 3 ) = 6 3 ( ) ( ) Practice eercise A Determine if the epression is quadratic in nature /3 1/ ( ) ( ) / 1/ 6. ( + ) 1 119

120 FACTORING REVIEW 1. COMMON TERM a) What number will go into all of the numbers evenl b) Common variable use the common variable with the lowest power EXAMPLE ( + ) 3. DIFFERENCE OF TWO SQUARES Looks like: a b = a + b a b a) Terms must be perfect squares b) Must be subtraction c) Powers must be even 3. PERFECT SQUARE TRINOMIALS Looks like: a + ab + b = a + b or a + b a + b or a ab + b = a b or a b a b a) First and last term must be perfect squares b) Middle term is equal to ab c) Sign in the parenthesis is the same as the first sign. GROUPING a) Group terms that have something in common b) Factor out common term in each parenthesis c) Write down what is in the parenthesis, the should be identical d) Then add the left-overs 5. FACTOR TRINOMIALS BY GROUPING a + b + c a) Multipl a and c b) Find all the factors of the answer EXAMPLE 5 EXAMPLE ( + 5)( 5) Does the middle term equal ab? a = 3 and b = 5 so 3 5 = 30 Therefore Yes it does! ( 3 5) + + factors to: + or ( 3 + 5)( 3 + 5) EXAMPLE ( 15 1) + ( 10 1) 3( 5 7) + ( 5 7) ( 5 7) ( 5 7)( 3 + ) EXAMPLE 6 15 (6)(15) = 90 1 and 90 and 5 3 and 30 10

121 5 and 18 6 and 15 9 and 10 c) Choose the combination that will either give the sum or difference needed to result in b. In this case the difference d) Rewrite the equation using the combination in place of the middle term e) Now group in order to factor f) Factor out the common term in each parenthesis 9 and ( ) ( ) Notice that when factoring out a -1 it changes the sign on c ( + ) ( + ) ( + 3) ( + 3)( 3 5) g) Write down what is in the parenthesis, the should be identical h) Add the left-overs to complete the answer The same strategies used to factor quadratic epressions can be used to factor anthing that is quadratic in nature. (For more information on factoring, see the factoring lesson in Unit.) Epression: Rewritten: Factor: + ( 3 ) ( ) 1 ( 3 )( + 3) 3 3 ( ) 5( )( ) 1( ) ( )( 3 ) ( 3 1/ + )( 3 1/ + ) + + 9( 1/ ) + 1( 1/ ) + or ( 3 1/ + ) / 1/ ( ) ( ) ( 3)( 3) or ( 3) ( ) ( 5) ( + 5)( 5) 6 3 ( ) ( ) ( + 3 )( 3 ) 11

122 Practice set B Factor each quadratic in nature epression /5 1/ /3 1/ Sometimes rewriting an epression makes it easier to recognize the quadratic pattern. Eample: ( + 1) ( + 1) 15 u = ( + 1) You can use a new variable to replace ( 1) Now replace ever ( + 1) in the epression u u 15 ( u 5 )( u + 3 ) ( ) ( ) with u. Notice how this is quadratic in nature. We can use quadratic factoring techniques to factor this epression You must remember to replace the u with + 1. Eample: 1/3 1/ /6 1/6 u = You can use a new variable to replace. 1/6 3u 8u + Now replace ever in the epression with u. Notice how this is quadratic in nature. ( 3u )( u ) We can use quadratic factoring techniques to factor this epression ( 3 1/6 )( 1/6 ) You must remember to replace the u with 1/6. Practice Eercises C Identif the u in each epression, then factor using u substitution. Write the factored form in terms of ( 3) + 15( 3) ( ) ( ) ( ) ( ) ( + ) + 11( + ) 1 1/ 1/ ( + ) 1 9 1

123 Unit 1 Cluster 3 (N.CN.1 and N.CN.) Performing Arithmetic Operations With Comple Numbers Cluster 3: Performing arithmetic operations with comple numbers i = 1, comple number form a + bi 1.3. Add, subtract, and multipl with comple numbers VOCABULARY The imaginar unit, i, is defined to be i = 1. Using this definition, it would follow that i = 1 because i = i i = 1 1 = 1. The number sstem can be etended to include the set of comple numbers. A comple number written in standard form is a number a + bi, where a and b are real numbers. If a = 0, then the number is called imaginar. If b = 0 then the number is called real. Simplifing Radicals with i Etending the number sstem to include the set of comple numbers allows us to take the square root of negative numbers. Eample: Simplif i 3 3i Rewrite the epression using the properties of radicals. Remember that i = 1. Eample: Simplif i 6 i 6 Rewrite the epression using the properties of radicals. Remember that i = 1. 13

124 Practice Eercises A Simplif each radical Performing Arithmetic Operations with Comple Numbers You can add, subtract, and multipl comple numbers. Similar to the set of real numbers, addition and multiplication of comple numbers is associative and commutative. Adding and Subtracting Comple Numbers Eample: Add ( 3+ i) + ( 5 i) ( 3+ i) + ( 5 i) 3 + i + 5 i Remove the parentheses i i Group like terms together. 8 i Combine like terms. Eample: Subtract ( 7 5i) ( + 6i) ( 7 5i) ( + 6i) 7 5i + 6i 7 + 5i 6i 9 11i Distribute the negative and remove the parentheses. Group like terms together. Combine like terms. 1

125 Eample: Simplif 8 ( 5i ) + ( + 5i) ( i) ( i) i + + 5i i + 5i Distribute the negative and remove the parentheses. Group like terms together i Combine like terms. Multipling Comple Numbers Eample: Multipl 3( 7 + 6i) 3( 7 + 6i) 1 18i Distribute the negative three to each term in the parentheses. Eample: Multipl i ( + 9i) ( + i) i 9 8i + 36i 8i i i Distribute the i to each term in the parentheses. B definition i = 1 so substitute -1 in for i. Write the comple number in standard form. 15

126 Eample: Multipl ( + 9i)( 3 10i) Distributive (FOIL) Method Bo Method Vertical Method ( + 9i)( 3 10i) = ( 3 10i) + 9i ( 3 10i) = 6 + 0i 7i 90i *combine like terms = 6 7i 90i *remember that i = 1 = 6 7i 90 1 = 6 7i + 90 = 96 7i 9i 3 6 7i 10i 0i 90i *combine terms on the diagonals of the unshaded boes(top right to lower left) = 6 7i 90i *remember that i = 1 = 6 7i 90 1 = 6 7i + 90 = 96 7i + 9i 3 10i 6 7i 0i 90i 6 7i 90i *remember that 1 = 6 7i 90 1 = 6 7i + 90 = 96 7i i = Eample: Multipl ( 5 + i)( 5 i) Distributive (FOIL) Method ( 5 + i)( 5 i) = 5( 5 i) + i ( 5 i) = 5 10i + 10i i *combine like terms = 5 i *remember that 1 = 5 1 = 5 + = 9 i = Bo Method 5 i i i 10i i *combine terms on the diagonals of the unshaded boes(top right to lower left) = 5 i *remember that i = 1 = 5 1 = 5 + = 9 Vertical Method 5 + i 5 i 10i i i 5 + 0i i *remember that 1 = 5 1 = 5 + = 9 i = 16

127 Practice Eercises B Simplif each epression. 1. ( 10 i) + ( 3 6i). ( 9 + 6i) ( i) 3. i ( 7 5i ) + ( 10 7i). 7 + ( + 8i) ( 10 + i) 5. 5( 8 i) 6. 11i ( 9i) 7. i( i ) 8. ( 6 7i)( 1i) 9. ( + i)( 7 5i) 10 ( 1+ i)( i ) 11. ( 10 i)( 7 + 5i ) 1. ( 10 + i)( 10 i) 17

128 Unit 1 Cluster 3 Honors (N.CN.3) H.1 Find the conjugate of a comple number; use conjugates to find quotients of comple numbers. VOCABULARY The conjugate of a comple number is a number in the standard comple form a + bi, where the imaginar part bi has the opposite sign of the original, for eample a bi has the opposite sign of a + bi. Conjugate pairs are an pair of comple numbers that are conjugates of each other such as 3 + i and 3 i. The product of conjugate pairs is a positive real number. ( a + bi) ( a bi) a abi + abi b i a a b + b ( 1) ( 3+ i) ( 3 i) 9 1i + 1i 16i This propert will be used to divide comple numbers. Eample: Find the conjugate of the following comple numbers. a. i b. + 5i c. 3 i d. 7 i a. The opposite of i is i. The conjugate of i is i. b. The opposite of 5i is 5i. The conjugate of + 5i is 5i. c. The opposite of i is i. The conjugate of 3 i is 3 + i. d. The opposite of i is i. The conjugate of 7 i is 7 + i. Practice Eercises A Find the conjugate of the following comple numbers i. 8 9i i. 1+ 7i 18

129 Divide b an imaginar number bi If there is an imaginar number in the denominator of a fraction, then the comple number is not in standard comple form. In order to write it in standard comple form, ou must multipl the numerator and the denominator b the conjugate of the denominator. This process removes the imaginar unit from the denominator and replaces it with a real number (the product of conjugate pairs is a positive real number) without changing the value of the comple number. Once this is done, ou can write the number in standard comple form b simplifing the fraction. Eample: Write in standard comple form: 8i 8i 8i 8i 16i 6i 16i i 6 16 i i 1 = = i 16 The conjugate of 8i is 8i. Multipl the numerator and the denominator b the conjugate. Remember i = 1. Simplif. Eample: Write in standard comple form: i 9i 6 + 8i 9i 9i 9i 5i 7i 81i 5i 7 1 ( ) i i i 9 3 The conjugate of 9i is 9i. Multipl the numerator and the denominator b the conjugate. Remember i = 1. Rewrite numerator in standard comple form a + bi. Rewrite whole solution in comple form a + bi, reducing as needed. 19

130 Practice Eercises B Write in standard comple form i. 6 i i i 6i i 5i 6. 3i i Dividing Comple Numbers in Standard Form a + bi To divide comple numbers, find the comple conjugate of the denominator, multipl the numerator and denominator b that conjugate, and simplif. Eample: Divide 10 + i 10 + i 10 i + i i 10( i) ( + i)( i) Multipl the numerator and denominator b the conjugate of +i, which is i 0 10i Distribute i + i i 0 10i Simplif i 0 10i ( 1) Note that i = i i = i Simplif and write in standard comple form. 130

131 Eample: Divide 7 i 5i 7i 5i 7i + 5i 5i + 5i ( 7i)( + 5i) ( 5i )( + 5i) i 8i 35i i 0i 5i i 35i 16 5i ( ) i i i i i Multipl b the conjugate of the denominator. Distribute Combine like terms. Remember that i = 1. Combine like terms again. Simplif and write in standard comple form. Eample: Divide 6 + i 1 i 6 + i 1 i 6 + i 1+ i 1 i 1+ i Multipl b the conjugate of the denominator. 131

132 ( 6 + i)( 1+ i) ( 1 i)( 1+ i) 6 + 1i + i + i 1+ i i i 6 + 1i + i 1 i 6 + 1i + 1 ( ) Distribute. Combine like terms. 1 1 Remember i = i 1+ Combine like terms again. + 1i i Put in standard comple form a + bi. Practice Eercises C Divide each comple rational epression and write in standard comple form. 1. 5i 6i. 8i 1+ 3i i i 3+ i i 6 + 6i i i 13

133 Unit 3 Cluster (A.REI.) and Unit 3 Cluster 5 (N.CN.7): Solve Equations and Inequalities in One Variable Cluster : Solving equations in one variable 3..1a Derive the quadratic formula b completing the square. 3..1b Solve equations b taking the square root, completing the square, using the quadratic formula and b factoring (recognize when the quadratic formula gives comple solutions and write them as a ± bi ) Solve with comple numbers VOCABULARY The square root of a number is a value that, when multiplied b itself, gives the number. For eample if r = a, then r is the square root of a. There are two possible values for r; one positive and one negative. For instance, the square root of 9 could be 3 because 3 = 9 but it could also be 3 because 3 = 9. A perfect square is a number that can be epressed as the product of two equal integers. For eample: 100 is a perfect square because = 100 and is a perfect square because =. Solving Equations b Taking the Square Root When solving a quadratic equation b taking the square root, ou want to isolate the squared term so that ou can take the square root of both sides of the equation. Eample: Solve the quadratic equation =. = In this eample the squared term is alread isolated. and it is = Take the square root of each side of the equation. ( ) 1/ / = ± = ± = ± = or = Using the properties of rational eponents ou can simplif the left side of the equation to. The number 16 is a perfect square because =. 133

134 Eample: Solve the quadratic equation + = 16 3 = 1 3 = =. In this eample the squared term is and it needs to be isolated. Use a reverse order of operations to isolate. = 18 Take the square root of each side of the equation. ( ) 1/ / = ± 9 = ± 9 = ± 3 = 3 or = 3 Using the properties of rational eponents ou can simplif the left side of the equation to. Using the properties of radical epressions ou can simplif the right side of the equation. (See Unit 1 Cluster for help with simplifing) Eample: Solve the quadratic equation 3( ) + = 5. ( ) ( ) ( ) 3 + = 5 3 = 8 = 16 In this eample the squared term is ( ) and it needs to be isolated. Use a reverse order of operations to isolate ( ). ( ) = 16 Take the square root of each side of the equation. ( ) ( ) 1/ = ± 16 / = ± = ± = or = = or = = 6 or = Using the properties of rational eponents ou can simplif the left side of the equation to. The number 16 is a perfect square because = 16. You still need to solve each equation for. 13

135 Practice Eercises A Solve each quadratic equation. 1. = 5. = 8 3. = = = = 9 7. ( 3) + = ( 3) = = 5 Solving Quadratic Equations b Completing the Square Sometimes ou have to rewrite a quadratic equation, using the method of completing the square, so that it can be solved b taking the square root. Eample: Solve 10 = = = = = 5 5 = ( ) 5 = Complete the square on the left side of the equation. Factor the epression on the left side. ( 5) = Take the square root of each side. 5 = ± Simplif. 5 = or 5 = = 5 + or = 5 Solve for. 135

136 Eample: Solve + = 6. + = = = = = 1 ( ) + = 1 Complete the square on the left side of the equation. Factor the epression on the left side. ( + ) = 1 Take the square root of each side. + = ± i Simplif. Remember that the 1 = i + = i or + = i = + i or = i Solve for. Eample: Solve =. 3 5 = = = = = = = + 3 Collect the terms with variables on one side of the equation and the constant term on the other side. Complete the square on the left side of the equation. 136

137 = = = 6 36 Factor the epression on the left side of the equation. Isolate the squared term. 5 9 = Take the square root of each side = ± Simplif = or = = + or = Solve for = or = 3 Practice Eercises B Solve the quadratic equations b completing the square. 1. = = = = = = 0 VOCABULARY b ± b ac The quadratic formula, =, can be used to find the solutions of the quadratic a equation a + b + c = 0, when a 0. The portion of the quadratic equation that is under the radical, b ac, is called the discriminant. It can be used to determine the number and tpe of solutions to the quadratic equation a + b + c =

138 Using the Discriminant to Determine Number and Tpe of Solutions If b ac > 0, then there are two real solutions to the quadratic equation. Eample: Determine the number and tpe of solutions for the equation + 6 = = 0 a = 1, b = 1, and c = 6 b ac 1 (1)( 6) 1 ( 6) 1 ( ) > 0 Identif a, b, and c. The quadratic equation that the function crosses the -ais twice. Substitute the values of a, b, and c into the discriminant formula. Simplif using order of operations. Determine if the result is greater than zero, equal to zero, or less than zero. + 6 = 0 has two real solutions. You can see from the graph If b ac = 0, then there is one real solution to the quadratic equation. Eample: Determine the number and tpe of solutions for the equation + + = = 0 a = 1, b =, and c = b ac (1)() 16 () = 0 Identif a, b, and c. Substitute the values of a, b, and c into the discriminant formula. Simplif using order of operations. The quadratic equation that the function touches the -ais onl once. Determine if the result is greater than zero, equal to zero, or less than zero. + + = 0 has one real solution. You can see from the graph

139 If b ac < 0, then there no real, but two imaginar solutions to the quadratic equation. Eample: Determine the number and tpe of solutions for the equation + 1= = 0 a = 1, b = 0, and c = 1 b ac 0 (1)(1) 0 (1) 0 < 0 Identif a, b, and c. Substitute the values of a, b, and c into the discriminant formula. Simplif using order of operations. Determine if the result is greater than zero, equal to zero, or less than zero. + 1= 0 has no real solutions, but it has two imaginar The quadratic equation solutions. You can see from the graph that the function never crosses the -ais Practice Eercises C Determine the number and tpe of solutions that each quadratic equation has = = 6. + = 139

140 Solving Quadratic Equations b Using the Quadratic Formula Eample: Solve = 0 using the quadratic formula = 0 a = 3, b = 5, and c = ± = (3) 5 5 (3) 5 ± 5 1 = 6 5 ± 5 8 = 6 = = Eample: Solve 5 ± ± = and 6 3 = 0 Make sure all the terms are on the same side and that the equation equals 0. Identif a, b, and c. Substitute the values for a, b, and c into the quadratic formula. Use order of operations to simplif = These are actuall two different solutions. 6 using the quadratic formula. 3 = 0 a =, b = 1, and c = 3 = 1 ± ( 1) ()( 3) () Make sure all the terms are on the same side and that the equation equals 0. Identif a, b, and c. Substitute the values for a, b, and c into the quadratic formula. 10

141 1± 1 8( 3) = 1± 1 ( ) = 1± 1+ = 1± 5 = 1± 5 = = = = and 1 5 = = = 1 Use order of operations to simplif. Simplif each answer. Eample: Solve = using the quadratic formula = = 0 a = 5, b = 10, and c = (5)(1) ± = (5) 10 ± (1) = 50 = 10 ± ± 0 = ± 0 = = 50 1 = 5 Make sure all the terms are on the same side and that the equation equals 0. Identif a, b, and c. Substitute the values for a, b, and c into the quadratic formula. Use order of operations to simplif. Simplif the answer. Notice that we onl got one answer this time because the discriminant was 0. 11

142 Eample: Solve = using the quadratic formula. + 1 = = 0 a =, b = 1, and c = 0 ± = () 1 1 ()(0) 1 ± 1 8(0) = = 1 ± ± 16 = 1 ± i = = 3 ± i = 3 + i or = 3 i Make sure all the terms are on the same side and that the equation equals 0. Identif a, b, and c. Substitute the values for a, b, and c into the quadratic formula. Use order of operations to simplif. Simplif the answer. Notice that we got two imaginar answers this time because the discriminant was less than 0. Practice Eercises D Solve the quadratic equation using the quadratic formula = = = = = 6 6. = = = 9. = 3 1

143 Practice Eercises E Solve each of the following equations using the method of our choice = = = 0. ( 3) = 9 5. ( 1) = = = 8. ( + 1) = = = = = 3 13

144 Unit 3 Cluster 5 (N.CN.8, N.CN.9-Honors): Use comple numbers in polnomial identities and equations 3.5. Etend polnomial identities to the comple numbers Know the Fundamental Theorem of algebra; show that it is true for quadratic polnomials. The Fundamental Theorem of Algebra states that ever polnomial of degree n with comple coefficients has eactl n roots in the comple numbers. Note: Remember that ever root can be written as a comple number in the form of a + bi. For instance = 3 can be written as = 3 + 0i. In addition, all comple numbers come in conjugate pairs, a + bi and a bi. Eample: f = f 3 = Degree: Comple Roots: Degree: 3 Comple Roots: 3 Polnomial Identities 1. a + b = a + ab + b. a + b c + d = ac + ad + bc + bd 3. a b = ( a + b)( a b). + ( a + b) + AB = ( + a)( + b) 5. If a b c + + = 0 then ± = b b ac a 1

145 Eample: Find the comple roots of f 6 = + and write in factored form. = = 6 = ± 8i = ( )( + ) = 8i 8 i = f 8i 8 i f 1. Set equal to zero to find the roots of the function. Solve.. Recall factored form is p q = f. Substitute the zeros in for p and q. Eample: Find the comple roots of f = and write in factored form. ( ) ( 1) 16 ± = = 16 ± ± = 16 ± i = = 8 ± i ( ) ( + )( + + ) = 8 + i 8 i = f 8 i 8 i f 1. Use the quadratic formula to find the roots of the function.. Recall factored form is p q = f. Substitute the zeros in for p and q. 15

146 Eample: Find the comple roots of f = and write in factored form. f = Factor the quadratic in nature function. + = = 0 = = 6 = ± i = ± i 6 f = ( i)( ( i)) i 6 i 6 f = ( i)( + i) i 6 + i 6. Set each factor equal to zero to find the roots. 3. Recall factored form is p q = f. Substitute the zeros in for p and q. Practice Eercises A Find the comple roots. Write in factored form

147 Unit 3 Cluster 3 (A.CED.1, A.CED.) Writing and Solving Equations and Inequalities Cluster 3: Creating equations that describe numbers or relationships Write and solve equations and inequalities in one variable (including linear, simple eponential, and quadratic functions) Solve formulas for a variable including those involving squared variables Writing and Solving Quadratic Equations in One Variable When solving contetual tpe problems it is important to: Identif what ou know. Determine what ou are tring to find. Draw a picture to help ou visualize the situation when possible. Remember to label all parts of our drawing. Use familiar formulas to help ou write equations. Check our answer for reasonableness and accurac. Make sure ou answered the entire question. Use appropriate units. Eample: Find three consecutive integers such that the product of the first two plus the square of the third is equal to 137. First term: Second term: + 1 Third term: + ( + 1) + ( + ) = 137 ( + 1) + ( + )( + ) = = = The first number is. Since the are consecutive numbers, the second term is one more than the first or + 1. The third term is one more than the second term or = +. Multipl the first two together and add the result to the third term squared. This is equal to 137. Multipl and combine like terms. + = Make sure the equation is equal to = 0 ( ) ( ) = 0 ( ) ( ) ( 7)( + 19) = = 0 Factor. 17

148 7 = 0 = 7 First term: 7 Second term: 8 Third term: = 0 = 19 = 19 Use the Zero Product Propert to solve for. The numbers are integers so has to be 7. Eample: A photo is 6 in longer than it is wide. Find the length and width if the area is 187 in. width = length = + 6 = = ( + 6) = ( + ) A wl The width is the basic unit, so let it equal. The length is 6 inches longer than width or + 6. A photo is rectangular so the area is equal to the width times the length. The area is 187 square inches = + Multipl the right side = + Make sure the equation equals 0. ( )( ) 0 = = 0 = = 0 = 17 The width is 11 inches and the length is 17 inches. Factor the epression on the right side of the equation. Use the Zero Product Propert to solve for. The length of a photo cannot be negative. Therefore, must be 11. The length is + 6 = = 17. Note: Often problems will require information from more than one equation to solve. For eample, ou might need the perimeter equation to help ou write the area equation or vice versa. The primar equation is the equation ou solve to find the answer ou are looking for. The secondar equation is the equation ou use to help set up our primar equation. Eample: Find two numbers that add to 150 and have a maimum product. What is the maimum product? Secondar equation: + = 150 One number is. The other number is. The sum is 150. Write an equation for this. 18

149 = 150 Primar Equation: P = 150 P = 150 P = ( 150 ) P = P = + + P ( ) 565 Use the sum equation to solve for. Write an equation for the product in terms of. Simplif the right side of the equation. = ( 75) Find the maimum. Remember 150 = ( 1) = 75 = 75 P = 150 = 75 The two numbers are 75 and = 565 The maimum product is 565. P P P ( 75) = 150( 75) ( 75) ( 75) = ( 75) = 565 the maimum is the verte. Using the method of our choice from Unit Lesson F.IF.8. The second number is 150 or 75. Find the maimum product. Eample: Jason wants to fence in a rectangular garden in his backard. If one side of the garden is against the house and Jason has 8 feet of fencing, what dimensions will maimize the garden area while utilizing all of the fencing? Garden First draw a picture of the house and garden. Label the sides of our garden. The amount of fence used is the distance around the garden ecluding the side net to the house. This is the same as the perimeter. 19

150 + = 8 = 8 = lw = ( 8 ) A A P = 8 A = + 8 A = ( ) A = + + A ( 1) 88 ( 1) 88 The length of the garden is and the width is. The perimeter is 8. Write an equation for this. Use the sum equation to solve for. Write an equation for the product in terms of. Simplif the right side of the equation. = + Find the maimum area. 8 = ( ) = 1 = 1 A = 8 P P P ( 1) = 8( 1) ( 1) ( 1) = ( 1) = 88 = The length is 1 feet and the width is feet. 1 = 88 The maimum area is 88 ft. Remember the maimum is the verte. Using the method of our choice from Unit Lesson F.IF.8. The second number is 8 ( 1) or. Find the maimum area. Practice Problems A Solve 1. The maimum size envelope that can be mailed with a large envelope rate is 3 inches longer than it is wide. The area is180 in. Find the length and the width. 3. The base of a triangular tabletop is 0 inches longer than the height. The area is 750 in. Find the height and the base.. A rectangular garden is 30 ft. b 0 ft. Part of the garden is removed in order to install a walkwa of uniform width around it. The area of the new garden is one-half the area of the old garden. How wide is the walkwa?. Find two numbers that differ b 8 and have a minimum product. 150

151 5. Alec has written an award winning short stor. His mother wants to frame it with a uniform border. She wants the finished product to have an area of 315 in. The writing portion occupies an area that is 11 inches wide and 17 inches long. How wide is the border? 6. Britton wants to build a pen for his teacup pig. He has 36 feet of fencing and he wants to use all of it. What should the dimensions of the pen be to maimize the area for his pig? What is the maimum area? 7. The product of numbers is 76. One number is 6 more than twice the first number. Find the two numbers. 8. Find three consecutive integers such that the square of the second number plus the product of the first and third numbers is a minimum. VOCABULARY Objects that are shot, thrown, or dropped into the air are called projectiles. Their height, measured from the ground, can be modeled b a projectile motion equation. The object is alwas affected b gravit. The gravitational constant is different depending on the units of measurement. For eample, the gravitational constant in feet is 3 ft/sec and in meters it is 9.8 m/sec. Similarl, the projectile motion equation for an object shot or thrown straight up or down is different depending on the units of measurement. Feet: h( t) = 16t + v t + h 0 0 Meters: h( t) =.9t + v0t + h0 h( t ) represents the height at an time t. The time is measured in seconds. The initial velocit, v 0, is the speed at which the object is thrown or shot. It is measured in ft/sec or m/sec. The initial height, h 0, is the height that the object is shot or thrown from. It is measured in feet or meters. Eample: The Willis Tower (formerl Sears Tower) in Chicago, Illinois is the tallest building in the United States. It is 108 stories or about 1,51 feet high. (Assume that each floor is 13 feet high.) a. A window washer is 8 floors from the top and he drops a piece of equipment, how long will it take for the equipment to reach the ground? b. How far from the ground is the piece of equipment after 5 seconds? c. When does the equipment pass the 16 th floor? 151

152 a. b. h( t) = 16t + v0t + h0 h t t t = 16 + (0) h t = 16t h t = 16t = 16t = 16t t = 100 t = 65 t = ± 65 t = 65 sec. or 8.06 sec. h t 16t 100 = + ( 5) = 60 h 5 = h h = The building is measured in feet so use the projectile motion equation for feet. The equipment was dropped making the initial velocit 0 ft./sec. The building has 108 floors, but he stopped 8 short of the top floor. Each floor is 13 feet high; multipl the number of floors b the height of each floor to get the initial height. We want to know when the equipment hits the ground making the final height zero. Solve for t. Negative time means ou are going back in time. Therefore, time is positive. We want to know when the height of the equipment at 5 seconds. The equipment is 60 feet from the ground after 5 seconds. c. h t = + 16t = 16t = 16t = 16t = 16t 5 = t 5 = t ± 13 = t t = 13 sec. or 7.11 sec. We want to know when the equipment passes the 16 th floor. The equation is the same equation written in part a. The 16 th floor is 08 feet above the ground. Solve for t. Negative time means ou are going back in time. Therefore, time is positive. 15

153 Eample: The Salt Lake Bees are planning to have a fireworks displa after their game with the Tacoma Rainiers. Their launch platform is 5 feet off the ground and the fireworks will be launched with an initial of 3 feet per second. How long will it take each firework to reach their maimum height? a. h( t) = 16t + v0t + h0 h t t t = 16 + (3) + 5 h t t t = h t = t + t = 16( ) + 5 h t t t h t = 16 t t ( t ) h t = b t = a 3 t = = 1 16 ( ) The height of the fireworks is measured in feet so use the projectile motion equation for feet. The fireworks were launched with an initial velocit 3 ft./sec. The launch platform is 5 feet off the ground. Using the method of our choice from Unit Lesson F.IF.8. Find the amount of time it will take to reach the maimum height. The t coordinate of the verte indicates WHEN the firework will reach its maimum height. The firework will reach its maimum height after 1 second. Practice Eercises B Solve. 1. A bolt falls off an airplane at an altitude of 500 m. How long will it take the bolt to reach the ground? 3. How far will an object fall in 5 seconds if it is thrown downward at an initial velocit of 30 m/sec from a height of 00 m? 5. A coin is tossed upward with an initial velocit of 30 ft/sec from an altitude of 8 feet. What is the maimum height of the coin? 7. A water balloon is dropped from a height of 6 feet. How long before it lands on someone who is 6 feet tall?. A ball is thrown upward at a speed of 30 m/sec from an altitude of 0 m. What is the maimum height of the ball?. A ring is dropped from a helicopter at an altitude of 6 feet. How long does it take the ring to reach the ground? 6. What is the height of an object after two seconds, if thrown downward at an initial velocit of 0 ft/sec from a height of 175 feet? 8. A potato is launched from the ground with an initial velocit of 15 m/sec. What is its maimum height? 153

154 Solving Quadratic Inequalities in One Variable Eample: Solve 3 > 0. Find where the epression on the left side of the inequalit equals zero. 3 = = 0 3 = 0 or + 1 = 0 = 3 or = 1 We are asked to find where the epression is greater than zero, in other words, where the epression is positive. Determine if the epression is positive or negative around each zero. Select a value in the interval and evaluate the epression at that value, then decide if the result is positive or negative. < 1 1 < < 3 > 3 = = 0 = ( ) Positive Negative Positive There are two intervals where the epression is positive: when < 1 and when > 3., 1 3,. The answer could be Therefore, the answer to the inequalit is represented on a number line as follows: Look at the graph of the function function is positive. f = 3. Determine the intervals where the 15

155 The function is positive (, 1) ( 3, ). Notice that this interval is the same interval we obtained when we tested values around the zeros of the epression. Eample: Solve Find where the epression on the left side of the inequalit equals zero = 0 ( ) 3 5 = 0 ( )( ) = = 0 or = 0 3 = 1 or = 1 = or = 3 We are asked to find where the epression is less than or equal to zero, in other words, where the epression is negative or zero. Determine if the epression is positive or negative around each zero. Select a value in the interval and evaluate the epression at that value, then decide if the result is positive or negative. 1 < < < > 1 3 = 1 + ( ) Negative 3 = Positive = Negative 155

156 1 There are two intervals where the epression is negative: when < 3 and when >. 1 It equals zero at = 3 and =. Therefore, the answer to the inequalit is (, 1 3 [, ). The answer could be represented on a number line as follows: Eample: Solve < 0. + Find where the denominator is equal to zero. + = 0 = We are asked to find where the epression is less than zero, in other words, where the epression is negative. Determine if the epression is positive or negative around where the denominator is equal to zero. Select a value in the interval and evaluate the epression at that value, then decide if the result is positive or negative. < > = 5 = Negative Positive The function is negative when <. Therefore, the answer to the inequalit is,. The answer could be represented on a number line as follows: 156

157 Eample: Solve Find where both the numerator and the denominator are equal to zero. 5 = = 0 5 = 3 = We are asked to find where the epression is greater than or equal to zero, in other words, where the epression is positive. Determine if the epression is positive or negative around where the numerator and denominator are equal to zero. Select a value in the interval and evaluate the epression at that value, then decide if the result is positive or negative. < 3 = ( ) Positive 5 3 < < = Negative 5 > = Positive 5 There are two intervals where the epression is positive: when < 3 and when >. The function is onl equal to zero when = 3 because the denominator cannot equal 5 zero. Therefore, the answer to the inequalit is (, 3 ],. The answer could be represented on a number line as follows: 157

158 Eample: A rocket is launched with an initial velocit of 160 ft/sec from a foot high platform. How long is the rocket at least 60 feet? Write an inequalit to represent the situation. The initial velocit, v 0, is 160 ft/sec and the initial height, h 0, is feet. The height will be at least (greater than or equal to) 60 ft. 16t + v t + h h t 160t You need to determine the real world domain for this situation. Time is the independent variable. It starts at 0 seconds and ends when the rocket hits the ground at 10 seconds. Move all the terms to one side of the inequalit so the epression is compared to zero. 16t + 160t 56 0 Find where the epression is equal to zero. 16t + 160t 56 = 0 ( t t ) = 0 ( t )( t ) 16 8 = 0 t = 0 or t 8 = 0 t = or t = 8 Determine if the epression is positive or negative around each zero. Select a value in the interval and evaluate the epression at that value, then decide if the result is positive or negative. 0 < t < < t < 8 8 < t < 10 t = 1 t = 3 t = 9 16t + 160t 56 16t + 160t 56 16t + 160t negative Positive negative The rocket is at or above 60 feet from t = seconds to t = 8 seconds. The difference is 6, so the rocket is at least 60 feet for 6 seconds. 158

159 Practice Eercises C Solve > < < < > A bottle of water is thrown upward with an initial velocit of 3 ft/sec from a cliff that is 190 feet high. For what time does the height eceed 190 feet? 11. A compan determines that its total profit function can be modeled b P = ,000. Find all values of for which it makes a profit. 1. A rocket is launched with an initial velocit of m/sec from a platform that is 3 meters high. The rocket will burst into flames unless it stas below 5 meters. Find the interval of time before the rocket bursts into flames. 159

160 Solving for a Specified Variable Sometimes it is necessar to use algebraic rules to manipulate formulas in order to work with a variable imbedded within the formula. Given the area of a circle, solve for the radius. A r = π Given: A = πr Solve for r: A ± = r π Surface area of a right clindrical solid with You ma have to use the quadratic formula. radius r and height h Solve for r: A = π r + π rh A = π r + π rh 0 = π r + π rh A a = π b = π h c = A ( ) ( π ) π h ± π h π A r = π h ± π h + 8π A r = π π h ± π h + π A r = π π ± π + π r = π h h A Practice Eercises D Solve for the indicated variable a + b = c ; solve for b. S = hl + hw + lw;solve for h A 6 s ; solve for s =. A = A 1 r ; solve for r 0 5. N k 3 k Gm1m = ; solve for k 6. F = ; solve for r r 1 N n n ; solve for n = = r ; solve for 160

161 Unit 3 Cluster 3 Honors: Polnomial and Rational Inequalities Cluster 3: Creating equations that describe numbers or relationships H.1. Solve polnomial and rational inequalities in one variable. Eample: Solve < < 0 Find where = = 0 ( )( ) 10 3 = 0 10 = 0 = 10 or 3 = 0 = 3 This epression is quadratic in nature. Factor the epression using that technique and use the zero product propert to solve for each factor. = ± 10 = ± 3 Test around each zero to determine if the epression is positive or negative on the interval. Test Epression evaluated Interval Positive/Negative Point at point < 10 = 10 < < 3 = 3 < < 3 = 0 3 < < 10 = > 10 = Positive Negative Positive Negative Positive The epression is less than zero when it is negative. The epression is negative on the intervals 10 < < 3 and 3 < < 10. The answer can also be written as ( 10, 3) ( 10, 3). 161

162 The answer could also be represented on a number line. Looking at the graph, the intervals that satisf this inequalit are the parts of the function below the -ais. Notice the intervals are the same Eample: Solve Find where = 0 ( ) + 6 = = 0 Factor the epression and use the zero product = = 0 propert to solve for each factor. or = 0 = 6 Test around each zero to determine if the epression is positive or negative on the interval. Test Epression evaluated Interval Positive/Negative Point at point < 6 = 10 6 < < 0 = 1 > 0 = 5 ( 10) + 6( 10) Negative Positive Positive 16

163 The epression is greater than zero when it is positive. The epression is positive on the intervals 6 < < 0 and < 0. Keep in mind it is also equal to zero so the endpoints are also included. The intervals that satisf the inequalit are 6 0 and 0. The interval would be written 6. The answer can also be written as[ 6, ). The answer could also be represented on a number line. Looking at the graph, the intervals that satisf this inequalit are the parts of the function above the -ais, including the values on the -ais. Notice the intervals are the same Eample: ( 1)( + )( ) 0 Find where ( )( )( ) ( 1)( + )( ) = 0 1 = 0 = 1 or + = 0 = or = 0 = 1 + = 0 Use the zero product propert to solve for each factor. 163

164 Test around each zero to determine if the epression is positive or negative on the interval. Interval Test Point < = 5 < < 1 = 0 1 < < = > = 6 Epression evaluated at point ( ) ( ) Positive/Negative Negative Positive Negative Positive The epression is less than zero when it is negative. The epression is negative on the intervals < and1 < <. Keep in mind it is also equal to zero so the endpoints are also included. The intervals that satisf the inequalit are and 1., 1,. The answer can also be written ( ] [ ] The answer could also be represented on a number line. Looking at the graph, the intervals that satisf this inequalit are the parts of the function below the -ais, including the values on the -ais. Notice the intervals are the same

165 Eample: Compare the inequalit to zero ( 1 Make sure the denominators (the bottoms) are ) 0 the same Combine the numerators (the tops) = = 0 or Set all of the factors in the numerator and = 6 = 1 denominator equal to zero, and then solve. Test around each zero to determine if the epression is positive or negative on the interval. Epression Interval Test Point Positive/Negative evaluated at point = < 6 = 10 Positive = < < 1 = 3 Negative = > 1 = Positive 6 The epression is greater than zero when it is positive. The epression is positive on the intervals < 6 and > 1. Keep in mind it is also equal to zero so the endpoints are also included ecept for the denominator which cannot be equal to zero. The intervals that satisf the inequalit are 6 and > 1., 6 1,. The answer can also be written as ( ] The answer could also be represented on a number line. 165

166 Looking at the graph, the intervals that satisf this inequalit are the parts of the function above the -ais, including the values on the -ais. Notice the intervals are the same Eample: + 3 > > > > > > 0 3 = 0 3 = 0 or = 0 = 3 Compare the inequalit to zero. Make sure the denominators (the bottoms) are the same. Combine the numerators (the tops). Set all of the factors in the numerator and denominator equal to zero, and then solve. 166

167 Test around each zero to determine if the epression is positive or negative on the interval. Interval Test Point < 0 = 5 0 < < 3 = 1 > 3 = 5 Epression Positive/Negative evaluated at point ( 5) 10 = ( 5) 3 8 Positive 5 ( 1) = 1 3 Negative 1 ( 5 ) = Positive 5 The epression is greater than zero when it is positive. The epression is positive on the intervals < 0 and > 3. Keep in mind the denominator cannot be equal to zero.,0 3,. The answer can also be written as The answer could also be represented on a number line. Looking at the graph, the intervals that satisf this inequalit are the parts of the function above the -ais, including the values on the -ais. Notice the intervals are the same

168 Eample: 3 < < < < ( ) ( ) < < 0 ( )( + 1) + 10 ( )( + 1) < 0 Compare the inequalit to zero. Make sure the denominators (the bottoms) are the same. Combine the numerators (the tops) = 0 = = 0 Set all of the factors in the numerator and or or = 10 = = 1 denominator equal to zero, and then solve. Test around each zero to determine if the epression is positive or negative on the interval. Interval Test Point < 10 = < < 1 = 5 1 < < 0 > = 5 Epression evaluated at point = = = Positive/Negative Negative Positive Negative Positive 168

169 The epression is less than zero when it is negative. The epression is negative on the intervals < 10 and 1 < <. Keep in mind the denominator cannot be equal to zero., 10 1,. The answer can also be written as The answer could also be represented on a number line. Looking at the graph, the intervals that satisf this inequalit are the parts of the function below the -ais. Notice the intervals are the same Practice Eercises A Solve. 1. ( 3)( + )( + 1) > 0. ( )( )( ) > 6. 3 < ( + 3 )( ) ( 1) > <

170 You Decide Carter s spaceship is trapped in a gravitational field of a newl discovered Class M planet. Carter will be in danger if his spaceship s acceleration eceeds 500 m/h/h. If his acceleration can be 500 modeled b the equation A( t) = m/h/h, for what range of time is Carter s spaceship 5 t below the danger zone? 170

171 Unit 3 Cluster 3 (A.CED.) Writing and Graphing Equations in Two Variables Cluster 3: Creating equations that describe numbers or relationships 3.3. Write and graph equations in or more variables with labels and scales Writing Quadratic Functions Given Ke Features A quadratic function can be epressed in several was to highlight ke features. Verte form: f = a( h) + k highlights the verte ( h, k ). Factored from: f = a ( p)( q) highlights the -intercepts ( p, 0) and ( q ) It is possible to write a quadratic function when given ke features such as the verte or the -intercepts and another point on the graph of the parabola. Eample: Write a quadratic equation for a parabola that has its verte at the point ( 1,6 ).,0., and passes through Answer: f = a( h) + k f = a + 6 = a = a = a = a 1+ 6 = a + = a f = + You are given the verte which is a ke feature that is highlighted b the verte form of a quadratic function. Use this form to help ou write the equation for the parabola graphed.,. h = and k =. Substitute The verte is these values into the equation and simplif if necessar. 1,6 to help ou find the value of Use the point a. The value of the function is 6 when = 1 so substitute 1 in for and 6 in for f(). Use order of operations to simplif the epression on the right side of the equation then solve for a. Rewrite the epression substituting in the value for a. 171

172 Eample: Write an equation for the parabola graphed below. verte ( 3, 1) ( 5, 5) - -6 Answer: f = a( h) + k = ( + 3) 1 f = a ( 3) + ( 1) f a 5 = a = a 1 5 = a 1 5 = a 1 = a 1 = a f = You are given the verte which is a ke feature that is highlighted b the verte form of a quadratic function. Use this form to help ou write the equation for the parabola graphed. 3, 1. h = 3 and k = 1. The verte is Substitute these values into the equation and simplif if necessar. 5, 5 to help ou find the value Use the point of a. The value of the function is -5 when = -5 so substitute -5 in for and -5 in for f(). Use order of operations to simplif the epression on the right side of the equation then solve for a. Rewrite the epression substituting in the value for a. 17

173 Eample: Write an equation for a parabola with -intercepts ( 3,0) and the point ( 3, 3). 5,0 and passes through Answer: f = a ( p)( q) = ( ( 3) )( 5) = ( + 3)( 5) f a f a 3 = a ( ) 3 = a = a 6 3 = 1a 3 = a 1 1 = a 1 f = You are given the -intercepts which are a ke feature that is highlighted b the factored form of a quadratic function. Use this form to help ou write the equation for the parabola graphed. 3,0 so p = 3. The other One -intercept is -intercept is ( 5,0) q = 5. Substitute these values into the equation and simplif if necessar. 3, 3 to help ou find the value Use the point of a. The value of the function is -3 when = 3 so substitute 3 in for each and -3 in for f(). Use order of operations to simplif the epression on the right side of the equation then solve for a. Rewrite the epression substituting in the value for a. 173

174 Eample: Write an equation for the parabola graphed below Answer: f = a ( p)( q) = ( ( ) )( ) = ( + )( ) f a f a 6 = a = a = a 6 = a 6 = a 3 = a 3 f = ( + )( ) You are given the -intercepts which are a ke feature that is highlighted b the factored form of a quadratic function. Use this form to help ou write the equation for the parabola graphed.,0 so p =. The other One -intercept is -intercept is (,0) q =. Substitute these values into the equation and simplif if necessar. 0,6 to help ou find the value of Use the point a. The value of the function is 6 when = 0 so substitute 0 in for each and 6 in for f(). Use order of operations to simplif the epression on the right side of the equation then solve for a. Rewrite the epression substituting in the value for a. 17

175 Practice Eercises A Write a quadratic equation for the parabola described. 1. Verte: (,3 ) Point: ( 0,7 ). Intercepts:,0 (,0 ) Point: ( 1,3 ). Verte: ( 1, ) Point: ( 1,8 ) 5. Intercepts: ( 1,0 ) ( 7,0 ) Point: ( 5, 1) 3. Verte: ( 3, 1) Point: (,0) 6. Intercepts: ( 5,0) (,0 ) Point: ( 3,8) Write a quadratic equation for the parabola graphed

176 Graphing Quadratic Equations Graphing from Standard Form f = a + b + c Eample: f = b = a ( ) ( 1) = = 1 = f f f = 3 1 = 1 1 Find the verte. The verte is (1, -3) Plot the verte f 0 = 0 0 f (0) = Find the -intercept. The -intercept is (0, -) Plot the -intercept Use the ais of smmetr to find another point that is the reflection of the -intercept Connect the points, drawing a smooth curve. Remember quadratic functions are U shaped. 176

177 Graphing from Verte Form Eample: f = 1 ( + 5) 5 f = a h + k f = 1 ( + 5) 5 The verte is (h, k). f = a( h) + k Find the verte. The verte is (-5, -) Plot the verte f ( 0) = 1 ( 0 + 5) 5 f (0) = 3 Find the -intercept. The -intercept is (0, 3) Plot the -intercept Use the ais of smmetr to find another point that is the reflection of the -intercept Connect the points, drawing a smooth curve. Remember quadratic functions are U shaped. 177

178 Graphing from Factored Form f = a( p) ( q) Eample: f 1 = ( + )( 3) 1 f = ( + )( 3) + = 0 3 = 0 = = 3 Find the -intercepts. The -intercepts are (-, 0) and (3, 0) Plot the -intercepts Find the -coordinate between the two = intercepts Use the -coordinate f = + 3 to find the - 6 coordinate f = The verte is , - f = = Plot the verte Connect the points, drawing a smooth curve. Remember quadratic functions are U shaped. 178

179 Practice Eercises B Graph the following equations. 1. f = f = f 3 = f = ( 1) 3 5. f = ( + ) f ( ) = f = ( + )( 5) 8. f = ( + 1)( + 5) 9. f = ( 1)( 3)

180 Unit 3 Cluster 6 (A.REI.7): Solve Sstems of Equations Solve simple sstems containing linear and quadratic functions algebraicall and graphicall. Recall solving sstems of equations in Secondar 1. We are looking for the intersection of the two lines. There were three methods used. Below are eamples of each method. = Solve: 3 = 1 Graphing Graph the two equations and find the intersection. The intersection is (-10, -7). Substitution 1. Solve for in the first equation. = + ( + ) 3 = 1. Substitute the solution for in the second equation = = 1 3. Solve for = 7 = + = 7 + = 1 + = 10 The solution is (-10, -7). Substitute back into the first equation to solve for. 5. The solution is the intersection. 180

181 Elimination = 3 = 1 ( ) = 3 = 1 + = 8 3 = 1 = 7 = 7 = + 1 = = 10 The intersection is (-10, -7) 1. In order to eliminate the s, multipl the top equation b -.. Combine the two equations. 3. Substitute = 7 into either original equation in order to solve for. The solution is the intersection. We will use these methods to help solve sstems involving quadratic equations. Eample: Find the intersection of the following two equations: = = Graphing 6 Graph the two equations and find the intersection(s) The intersections are (-, 0) and (3, 5). Substitution ( ) = 1. The first equation is alread solved for ; substitute for in the second equation. 181

182 + + = + + = + = = 0 6 = 0 ( )( ) 3 + = 0 = 3 or = = = = 3 = = 9 = = 5 = 0 The solutions are (3, 5) and (-, 0) Elimination + 0 = + = + 0 = 1( + = ). Simplif and write in standard polnomial form. 3. Solve for using the method of our choice. Substitute the values back into the first equation to solve for. 5. The solutions are the intersections. 1. Line up like variables. + 0 = =. Multipl the second equation b -1 then combine the two equations. 6 = 0 6 = 0 ( )( ) 3 + = 0 = 3 or = = = = 3 = = 9 = = 5 = 0 The solutions are (3, 5) and (-, 0) 3. Solve using the method of our choice.. Substitute the values back into the first equation to solve for. 5. The solutions are the intersections. 18

183 Eample: Using the method of our choice, find the intersection between the following equations: + = + = 1 Substitution = 1 1. Solve for in the second equation. ( ) + 1 = = = 0 ± = 5 = = = = 5 3 ± ± ± ± = or = or 1.7 = 1 = = 1 = = = Substitute 1 for in the first equation. 3. Simplif and solve for.. Substitute the values back into the first equation to solve for. 183

184 The solutions are, , 5 5 and 5. The solutions are the intersections. Or approimatel (0.7, -1.9) or (-1.7, 1.5) Practice Eercises A Solve each of the sstems of equations. + = 6 1. = = 1 = = =. = 3 = + 5. = + 3 = = = 6 7. = = + = 1 8. = 9. + = 89 = = 5 = = 1 = 1. ( ) ( ) = = 18

185 Unit 3 Cluster 6 Honors (A.REI.8 and A.REI.9) Solving Sstems of Equations with Vectors and Matrices H.3.1 Represent a sstem of linear equations as a single matri equation in a vector variable. H.3. Find the inverse of a matri if it eists and use it to solve sstems of linear equations (using technolog for matrices of dimension 3 3 or greater). In Secondar Mathematics 1 Honors ou learned to solve a sstem of two equations b writing the corresponding augmented matri and using row operations to simplif the matri so that it had ones down the diagonal from upper left to lower right, and zeros above and below the ones. 1 0 g 0 1 h When a matri is in this form it is said to be in reduced row-echelon form. The process for simplifing a matri to reduced row-echelon form is called Gauss-Jordan elimination after the two mathematicians, Carl Friedrich Gauss and Wilhelm Jordan. This process, using row operations, can be used for sstems of two or more variables. Row operations are listed below using the original matri Note: R indicates row and the number following indicates which row. For instance, R1 indicated row one. Interchange Rows We want to have ones along the diagonal. We can switch rows so that the one ends up in the correct position. Multipl a row b a scalar We want positive ones down the diagonal. Multipling each element b a scalar k, ( k ) 0, allows us to change values in one row while preserving the overall equalit. Smbol: R1 R All of the elements of Row 1 will switch positions with all the elements of Row. Smbol: k R R1 = The matri is now:

186 Combine Two Rows You can add or subtract two rows to replace a single row. This enables ou to get ones along the diagonal and zeros elsewhere. R : R1 + R Multipl b a scalar, then combine rows Sometimes it is necessar to multipl a row b a scalar before combining with another row to get a one or a zero where needed. R3 : R + R3 [ ] + [ ] [ ] R: R: The matri is now: [ ] + [ ] [ ] + [ ] [ ] R3: R3: R3: The matri is now: Practice Eercises A Perform the row operations on the given matri. 1. R R R R: 1 R3+ R Using the matri below, write appropriate row operation(s) to get the desired results. (Recall that a indicates the element of the matri is located in row m and column n.) mn a 11 = 1 5. a 1 = 0 6. a 3 = 0 186

187 Eample: Solve the following sstems of equations using the Gauss-Jordan elimination method. + z = z = 1 z = The solution is (, 1,3) Rewrite the equations in matri form. R : 3R1 + R [ ] + [ ] [ ] R : R : R3: R1 + R3 [ ] + [ ] [ ] R3: R3: R1: R + R1 [ ] + [ ] [ ] R1: R1: R3: R + R [ ] + [ ] [ ] R3: R3: R1:3R3 + R1 [ ] + [ ] [ ] R1: R1: R : R3+ R [ ] + [ ] [ ] R : R : Rewriting the answer in equation form ou end up with: = = 1 z = 3 187

188 Practice Eercises B Using Gauss-Jordan Elimination solve the following sstems of equations = 19 + = z = z = z = Technolog can be used to help solve sstems of equations. The following sstem of equations can be solved using our graphing calculator. z = 11 + z = z = Write the augmented matri. Enter the matri into our calculator. Note: [A] is the default matri Push nd 1 Arrow over to choose EDIT Push ENTER Tpe the dimensions. After each number push ENTER. For this eample: 3 ENTER ENTER Tpe each element in the first row. Push ENTER after each number. Continue until ever row has been entered. Push nd MODE to return to the home screen Push nd 1 Then ENTER ENTER This will give ou a chance to check our matri for accurac. Push nd 1 arrow over to MATH. Either arrow down to option B or push ALPHA APPS to get to rref( This is row reduced echelon form. 188

189 Push nd 1 ENTER ) Or select the matri in which our equations are stored. Push ENTER to obtain the answer. The answer is (-1,, 7) Rewriting the answer in equation form ou end up with: = 1 = z = 7 Practice Eercises C Solve each of the following sstems using technolog z = z = z = z = 0 + = 5 3z = z = 3 5 z = 3z = 6 Not ever sstem has a single point as the solution. The following situations ma also occur. Consistent and Independent + z = z = z = 13 A single point solution (, 6, 5) 189

190 A line solution defined b one variable Consistent and Dependent 6 z = z = z = Rewriting the answer in equation form ou end up with: 7 z = z = = 0 this is alwas true Note: z is an independent variable and can take on an real value The solution is written as: ( 7 + z, + 1 z, z) A plane solution defined b two variables Consistent and Dependent + 3z = 1 z w = + z w = 3 Rewriting the answer in equation form ou end up with: z w = z + w = 1 0 = 0 this is alwas true Note: z and w are independent variables and can take on an real value The solution is written as: ( + w + z, 1 w + z, z, w) All three planes coincide. 190

191 Inconsistent + + z = 6 z = z = 0 No solution There are no intersections common to all three planes or the three planes are parallel The equations would be: = 0 + z = 0 0 = 1 this is not true or No Solution Practice Eercises D Solve the following sstems. Indicate if the sstem is consistent or inconsistent z = 0 + 3z = 3 z = z = z = z = 8 + z = z = z = 1 + 3z = 5. z = 3 + 6z = z = w = 3 3 w = 3 + z + 3w = 1 z + w = z + w = z = w = 3 191

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