COUNCIL ROCK HIGH SCHOOL MATHEMATICS A Note Guideline of Algebraic Concepts Designed to assist students in A Summer Review of Algebra [A teacher prepared compilation of the 7 Algebraic concepts deemed most important for mastery by the student entering Accelerated/Honors Analysis revised 04]
Solving Equations in One Variable A linear equation in one variable has a single unknown quantity called a variable represented by a letter. Eg:, where is always to the power of. This means there is no ² or ³ in the equation. To solve linear equations, you add, subtract, multiply and divide both sides of the equation by numbers and variables, so that you end up with a single variable on one side and a single number on the other side. EX. Solve for. 7 8 Subtract Divide by 8 8 48 4 Subtract Multiply by -8 Solving Inequalities in One Variable 8 4 4 4 8 4 4 8 4 4 Multiply by Distribute - Combine like terms Subtract 8 Divide by 4 A. Inequality Symbols > means "is greater than" < means "is less than" means "is greater than or equal to" means "is less than or equal to" B. Steps for Solving Inequalities with One Variable Perform the distributive property on each side. Combine like terms on each side. Add or subtract to get the variable terms on the same side. Add or subtract to move the number terms to the other side. Multiply or divide to move the coefficient. Graph the solution Remember: If you multiply or divide both sides of an inequality by a negative number, the inequality symbol changes direction.
C. Steps for Graphing the Solutions to Inequalities with One Variable ** Make sure the variable is on the LEFT in all solutions. A solution with >: Graph an open circle on the number. (The number is not part of the solution.) A dark bar with an arrow goes to the right of the circle. A solution with <: Graph an open circle on the number. (The number is not part of the solution.) A dark bar with an arrow goes to the left of the circle. A solution with : Graph a solid circle on the number. (The number is part of the solution.) A dark bar with an arrow goes to the right of the circle. A solution with : Graph a solid circle on the number. (The number is part of the solution.) A dark bar with an arrow goes to the left of the circle. EX. Solve and Graph - - 7n 8 add -7n divide by -7 7n 7 7 n - change inequality sign - -4 - - - - + 4 > ( - 8) distribute - + 4 > subtract - + 4 > - subtract 4 - > -0 n 0 divide by - 0 n < change inequality sign 4 7 8 ( - ) < + 8 distribute 4 < + 8 substract 4 < 8 variables drop out True statement All Real Numbers ( ) + 7 distribute + 7 subtract - 7 variables drop out False statement No Solution - - 0 - - 0
Solving Combined Inequalities Solving combined inequalities is just like solving normal inequalities, but with two separate inequalities combined into one. The main concept to know is the difference between the words "AND" and "OR," which control or limit the answers. "AND" means the solution will be an intersection where both inequalities hold true. "OR" means the solution will be a union where one OR the other inequality holds true. A. Combined Inequalities with AND EX. Solve and Graph - < + < 8 - < + AND + < 8 Separate the combined inequality into two inequalities -8 < AND < Solve each inequality for the variable. -4 < AND < -4 < < Write the answer in combined inequality form. Graph the answers on a number line -4 0 Use open dots at -4 and and shade between them since all solutions lie BETWEEN -4 and B. Combined Inequalities with "OR" EX. Solve and Graph - > + or + > 8 - > + OR + > 8 Combined inequality with OR are already separated. -8 > OR > Solve each inequality for the variable. -4 > OR > < -4 OR > Write the answer in combined inequality form. Graph the answers on a number line -4 0 Use an open dot at -4 and a closed dot at and shade beyond them with arrow pointing outward to infinity. There are no solutions between -4 and.
Solving Absolute Value Equations Absolute Value The absolute value of a number, written, is the number s distance from zero on a number line. The absolute value equation a + b = c, where c is positive, is equivalent to the compound statement a + b = c or a + b = c. E: Solve and Graph = 9. Step : Set up equations using OR. - = 9 or = 9 Step : Solve each equation. = 4 = -4 = 7 or = - Step : Graph the solutions. - 0 7 Solving Absolute Value Inequalities The inequality a + b < c, where c is positive, means that a + b is between c and c. This is a compound and inequality. It is equivalent to c < a + b < c. E: Solve and Graph + 7 <. Step : Set up a compound inequality. -< + 7 < Step : Solve. -8 < < 4-9 < < Step : Graph the solutions. -9 0 The inequality a + b > c, where c is positive, means that a + b is beyond c and c. This is an or compound inequality. It is equivalent to a + b < -c or a + b > c. Alert: Make sure to isolate the absolute value on one side of the equation or inequality before setting up the equations/inequalities. EX: Solve and Graph 4 + 0 Step. Isolate the absolute value. 4 + 0 by subtracting 4, then dividing by. Step : Set up the compound inequality: or Step : Solve 8 or 8/ or / Step 4: Graph the solutions. 0 4
The Slope of a Line Slope of a line, m = rise run y y y Horizontal lines have 0 slope. Vertical Lines have no slope. y = = - Graphs of Linear Equations in Two Variables Slope Intercept Form: y = m + b (where m represents the slope and b represents the y intercept) To graph a line:.) Write the equation in slope intercept form.) Plot the y-intercept.) Use the slope to plot at least more points 4.) Draw a line through the points EX: Graph y = Solution.) Solve for y. y = Subtract from both sides -y = - + Divide by each term by - y.) Plot (0, -).) Plot at least more points using the slope (Up, Right )
Finding the Equation of a Line Point-Slope Form: y y= m( ) To write an equation of a line:.) Substitute the point in for and y and the slope in for m..) Solve for y. EX: Write the equation of the line that passes through (, -4) and has a slope of..) y y m ) ( Substitute (, -4) in for (, y) and in for m.) y ( 4) ( ) Distribute.) y 4 Subtract 4 from both sides 4.) y Standard Form: A + By = C A, B and C must be integers (no fractions) A must be positive EX: Rewrite the answer in the previous eample in standard form..) y Multiple each term by.) y = Subtract from both sides.) - + y = - Divide each term by - 4.) y = Parallel lines have the same slope. E: y & y 4are parallel lines Perpendicular lines have slopes that are opposite reciprocals. E: y & y are perpendicular lines EX: Write the equation in standard form that is parallel to y and passes through (-, ).) y y m ) Substitute (-, ) in for (, y) and - in for m (parallel lines ( have the same slope).) y ( ( )) Simplify.) y ( ) Distribute 4.) y Add.) y Add.) y
Algebraic Solving of Systems of Linear Equations in Two Variables To solve a system of equations algebraically, use substitution or elimination methods. EX: Solve the system using elimination. 8 y 4 y 8 Multiply first equation by -4 and second equation by y 0 y 4 Add to eliminate 7 4 Solve for Substitute into original equation to find y 8( ) y Solve for y y y y 7 Solution is an ordered pair Answer (-, 7) EX: Solve the system using substitution. + y = 7 = y Solve the bottom equation for : = -y + 7 Substitute epression into top equation for : (-y + 7) + y = Solve for y. -9y + + y = -7y = -9 Sometimes the values aren t pretty y = 9/7 Substitute the value for y into = -y + 7 to find. = - (9/7) + 7 Get common denominators = -7/7 + 49/7 Solution is an ordered pair = /7 Answer (/7, 9/7) Relations and Functions Relation: Set of ordered pairs Domain (D) of the function is the set of values. Range (R) of the function is the set of y values. Function: A relation where there is correspondence between the two sets, D and R, that assigns to each member of D eactly one element of R. EX: Given f ( ) and D =, 0, find the range. Is the relation a function? f ( ) 4 f (0) 0 0 f () 4 Range: R =, Relation is,, 0,, This is a function since each value of is paired with eactly one value of y. 7
Graphs of Linear Inequalities in Two Variables Linear Inequalities may have infinite solutions. To show these infinite solutions, give a graph as your solution. To graph a linear inequality: Replace the inequality sign with = and solve for y The graph of the EQUALITY will serve as your boundary line for your inequality. Decide if this boundary should be DASHED or SOLID: > or < => DASHED or => SOLID Choose a TEST POINT (0,0) is a great choice unless of course it s on the boundary line you may never choose a test point on the boundary line. SUBSTITUTE your test point into the original INEQUALITY. Simplify both sides and evaluate the statement: TRUE = SHADE the portion of the graph that DOES include the test point FALSE = SHADE the portion of the graph that does NOT include the test point EX. Graph the solutions of + y + 4 > 0 Replace the inequality sign with = and solve for y + y + 4 = 0 y = - 4 y = -/ y Graph the boundary line using dashed line (>) Choose a test point not on the boundary line. Let s use (0, 0). Substitute into the original inequality. + y + 4 > 0 (0) + (0) + 4 > 0 4 > 0 True, so shade to include (0, 0). y 8
Graphs of Systems of Linear Inequalities in Two Variables To graph the solution to a system of linear inequalities, graph each inequality on the same coordinate plane, and shade the area where the two inequality graphs overlap. EX. Graph the solutions of the system: + y + 4 > 0 To graph + y + 4 > 0, remove the inequality sign and solve for y to get the dashed boundary line. Choose a test point and shade. (see last note section for complete work) To graph, 4, remove the inequality sign and graph the solid vertical line =. Shade to the left for all of the value of less than. y y Darken the area of overlap for the solution to the system of inequalities. y 9
Addition and Scalar Multiplication of Matrices A matri is a rectangular array of numbers. The dimensions of the matri are given by the number of rows (horizontal) and the number of columns (vertical). EX: is a matri ( rows by columns). To add matrices, the matrices must have the same dimensions. Then, add the corresponding elements. EX: + = To subtract matrices, change to addition of the opposite of each element in the second matri. Then, add the corresponding elements. 0 7 0 7 add the opposite 7 4 Remember, the dimensions of the matrices must match in order to add or subtract. EX: + = Not Possible because the dimensions are not the same. Scalar Multiplication is when a real number is outside a matri. To find the scalar product, multiply each element in the matri by the scalar. EX: = Sometimes more than one operation is performed. Follow order of operations. Scalar multiplication first, then add/subtraction second. EX: = + = Two matrices are equal if and only if their dimensions are the same and each corresponding element is the same. EX: 4 9 y 4 can solve and get = and y = - means that =9 and = -y, so we 0
Laws of Eponents When multiplying terms with like bases, add the eponents. a m a n a mn E. = + + = 9 E. = + = = When dividing terms with like bases, subtract the eponents. 7 E. 4 = 4 = 7 4 E. 4 a a m n a mn m mn When raising a term to a power, multiply the eponents. E. ( ) = E. ( ) = = 4 n a a When raising a term composed of different bases to a power, m m m multiply the eponents for each term. ab a b E. ( y ) = y 0 E. ( ) = = 4 9 = 7 When raising a fraction to a power, multiply the eponents for each term in the m a a numerator and the denominator., b 0 m b b 4 E. y y 4 m E. 4 Any term to zero power equals. a o, a 0 E. (y ) 0 7 0 = E. n Any term to a negative power means reciprocal. a, a 0 n a z E. y z E. y
Factoring A. GCF: Factor out what is common to all terms. Write as a product of the GCF and a polynomial. EX. m 9 n p m 7 n p + 4m p m p (m 4 n + m n ) B. Difference of two Squares: a b = (a + b) (a b) EX. 4y ( + y ) ( y ) C. Sum of two Cubes: a + b = (a + b) (a ab + b ) Difference of two Cubes: a b = (a b) (a + ab + b ) EX. 4 8 + 7 ( 4)( + 4 + ) ( + )(4 + 9) D. Perfect Square Trinomials: a + ab + b = (a + b) a ab + b = (a b) EX. p 0p + t + t + 49 (p ) (4t + 7) E. General Trinomials: Guess and check by FOIL-ing STEP # Try pairs that multiply to be the first term a : a a There s only one pair since is prime. (a ) (a ) e. EX. a 4ab 4b (a ) (a ) (a + b) (a b) +ab ab STEP # Try all combinations checking inner and outer terms of your factors to see if they add to be the middle term of the trinomial: -4ab STEP # Try pairs that multiply to be the last term 4b : -4b b 4b -b -b b
F. Factor by Grouping: Group terms into two groups. The groups must be added together. You may re-arrange the erms To factor out the common quantities, the quantities must be EX. 4a 4b + by 0ay group (4a 4b) + (by 0ay) factor out GCF for each group (a 7b) + y (7b a) make y and (7b a) opposites (a 7b) y (a 7b) factor out GCF quantity (a 7b) ( y) EX. a ab + a b a + a ab b rearrange(optional) (a + a) + ( ab b) group a(a + ) b (a + ) factor out GCF for each group (a + ) (a b) factor out GCF quantity The quantities are not the same but, they are opposites. So, change y to -y and (7b a) to (a 7b) so the quantity factors match G. Factor Completely: Follow the flow chart to factor completely. Factor out the GCF Two terms in the quantity GCF(binomial) Three terms in the quantity GCF(trinomial) Four terms in the quantity GCF(polynomial) Difference of two squares Perfect Square Trinomial Grouping Sum or Difference of two cubes General Trinomial If there was no GCF, and it s none of the above, it s prime If there was no GCF, and it s none of the above, it s prime If there was no GCF, and it s none of the above, it s prime
Solve/Find the Roots by Factoring. The directions will be solve or find the roots. One method is factoring. When an equation has a variable with an eponent higher than one, you must get all terms on one side of the equation so that it is equal to zero. Then, you can factor. Set each individual factor equal to zero and solve. The answers are called solutions or roots. The number of solutions/roots you get is equal to the degree (highest variable eponent) of the equation (e. Leading term is two solutions eist, three solutions eist, etc) Since - = 0 is false, it does not give you a solution EX. -a 70a = 4 -a 70a 4 = 0 Set equation equal to zero - (a + 4 a + 49) = 0 Factor out GCF - (a + 7) = 0 Factor perfect square trinomial - = 0 or a+7 = 0 Set each factor equal to zero a = -7 Final solution: a = -7 (dbl rt) Notice that a is to the second power. Therefore, there are solutions. Since a+7 was squared, you are really getting the solution a = 7 twice No need to do the solving twice, but it counts as two solutions we call it a double root. EX. + = 0 + = 0 Already set equal to zero ( + ) = 0 Factor out GCF ( )( + ) = 0 Factor the general trinomial = 0 or = 0 or + = 0 Set each factor equal to zero = ½ a = - Final solution: = -, 0, ½ Notice that is to the third power. Therefore, there are solutions. 4
Find the Zeros by Factoring Sometimes the equation has two variables, and y. The directions will be find the zeros. Zeros are the values of when y = 0. So, simply substitute zero in for y and solve for as above. EX. y = ( + ) ( ) 40 0 = ( + ) ( ) 40 Substitute zero for y and solve for 0 = 9 40 Foil and combine like terms 49 = 0 ( + 7) ( 7) = 0 Factor difference of two squares ( + 7) = 0 or ( 7) = 0 Set each factor equal to zero = 7 = 7 Solve The roots are: = ±7 Sometimes the equation in two variables is written in function form: f() instead of y. Since f() is y, simply substitute zero in for f() and solve for as above. EX. f() = ( 4) 0 = ( 4) Substitute zero for f() and solve for This is already factored 4 = 0 Since the three factors are the same, set the factor equal to zero = 4 Solve for the triple root The roots are: = 4 (triple root)
Simplifying Rational Algebraic Epressions/Determining Domain Domain: the set of values for which a function is defined. EX. Simplify and state its domain. To simplify, factor the numerator and dominator completely. Then cancel out identical factors. ( )( ) To determine the domain, remember the original denominator 0. 0 So, This means the domain is all real numbers ecept - or D:,. 9 EX. Simplify f and state its domain. To simplify, factor the numerator and dominator completely. Then cancel out identical factors. 9 To determine the domain, remember the original denominator 0. 0 0 So,, 0 This means the domain is all real numbers ecept - and 0 or D:,,0. 0 EX. Determine the zeros and the domain of f. Zeros: the values that will make a function zero. Begin by factoring the numerator and dominator completely. 0 To find the zeros, set f() = 0 and solve. 0 0 To determine the domain, remember the original denominator 0. ( )( ) 0 0, The zeros are - and. The domain is all real numbers ecept - or D :,
7 Products and Quotients of Rational Epressions EX. Simplify. Remember, division is multiplication by the reciprocal. So begin by re-writing the given problem as multiplication. Then, factor each numerator and denominator completely. ) )( ( ) ( Net, cancel out identical factors, as shown. ) )( ( ) ( This makes your final answer. Adding or Subtracting Rational Epressions EX. Simplify. Begin by determining the LCD. To do this, first factor each denominator. So, the LCD is. The first fraction will need to be multiplied by and the second fraction will need to be multiplied by. This gives ) (. So, we have 9 ) ( Don t forget to distribute the negative.
8 Simplifying Comple Fractions EX. Simplify. Begin by determining the LCD for all of the fractions. In this case, the LCD is. Multiply the numerator and denominator by this LCD. Solving Fractional Equations EX. Solve 4. Begin by determining the LCD. To do this, first factor each denominator. Net multiply every term by the LCD. This will clear out the fractions. 4 0 Solve this by factoring. 0 So, = -,. HOWEVER, because this value would make the denominator zero. So, it must be ecluded from the domain. = is called an etraneous solution. Although we used correct algebra to determine it, it does not create a true statement when it is plugged back into the original problem. The final answer is = - Don t forget to distribute the negative.
9 EX. Solve. Begin by determining the LCD. To do this, first factor each denominator. The LCD =. Net multiply every term by the LCD. This will eliminate the fractions. 4 8 7 Solving Fractional Inequalities EX. Solve 0. Begin by determining the LCD. The LCD of,, and 0 is 0. Multiply every term in your inequality by the LCD. This will eliminate the fractions. 0 0 0 ) ( () ) ( 8 Don t forget to flip the inequality when you divide by a negative number. Remember, be sure = 4 is in the domain of the original function before you bo it out as your final answer.
Roots of Real Numbers Every positive real number has two even roots, one positive and one negative. 4 8 4 The positive root is called the principal root. When taking an even root: means giveboth positive and negative values e. means giveonly the negative value e. means giveonly the positive value or the PRINCIPAL VALUE e. Every real number has one odd root, either positive or negative, taking the number s sign. 7 7 *Remember, negative eponents mean reciprocal. n b is called a radical epression: inde n b Radical sign 0 0 0 9 9 *Remember, negative eponents mean reciprocal radicand When the inde (n) is even, the radicand(b) must be positive to get two real answers (±). When the inde is odd, the radicand can be positive or negative to get one real answer(+or -). Find the real roots of the equation. When solving equations, you must give all possible solutions. 4 0 4 4 0 No Re al Roots Since both and - make the equation true, there are two solutions. Since square roots cannot be negative, there are no real roots. 0
Properties of Radicals Product Property n y n n y; 0, y 0 for even integers n Quotient Property n y n n ; 0, y 0 for even integers n y Theorems for indees EX. Simplify. n m nm n m n m 0 Since neither radicand is a perfect cube, factor under each radical. Since neither radical can be simplified, combine factors under one cube root. Simplify. 8 8 0 0 8 8 Since the fraction under the radical cannot be simplified, use the quotient property to split into two separate radicals one in the numerator and one in the denominator. Simplify. Since 0 is divisible by, use the quotient property to make a fraction under one radical. Divide. Simplify. EX. Simplify with Variables. 7a 4 4b 7a 4b 4 a b 4 a b b b b b ab b ab b Since the fraction under the radical cannot be simplified, use the quotient property to split into two separate radicals one in the numerator and one in the denominator. Factor and Simplify. Rationalize the denominator multiply numerator and denominator by b to create a perfect cube in the denominator. Simplify. 9 Factor notice the radicand is a perfect square trinomial. Simplify.
Sums/Differences of Radicals To add or subtract radicals, first simplify each radical term, then combine like radicals. Like radicals are radicals with the same inde(n) and the same radicant(b). Like radicals are combined by adding/subtracting the coefficients. EX. Simplify. 8 98 7 9 Simplify each radical Add like radicals 8 4 Use quotient property/simplify Subtract like radicals Rationalize denominator 8 7 Distribute/Use product property Simplify each radical Unlike radicals cannot be combined Divide/Use quotient property Simplify under each radical Unlike radicals cannot be combined Simplify each radical. Combine like radicals. Binomials Containing Radicals You can FOIL binomials containing radicals just the way you would multiply any binomial. EX. Simplify. 4 7 7 8 7 7 (7) 7 4 7 FOIL. Remember 7 7 49 7. Combine like radicals and constants. 4 4 8 8 Binomial Squared: Square the first, multiple the terms and double, square the last. Combine like radicals and constants.
9 4 7 9 4 To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator. Combine like terms. Simplify by a factor or. The Imaginary Number i In the domain of real numbers, the radicant of an even inde radical must be positive. When the radicant is negative, and the inde is even (square rt, fourth rt, sith rt, etc), the epression is not real, but imaginary. The imaginary number i. When simplifying, substitute i for. EX. Simplify. 0 0 i 0 i EX. Addition/Subtractraction. i i 4i 49 4i 7i i EX. Multiplication Since i, i. Whenever i appears, replace it with -. 4 i i 0i 0 0 i i i EX. Division You may not leave an imaginary number in the denominator. To rationalize the denominator, multiply the numerator and denominator by i. Of course, you still need to rationalize radical denominators, too. i i i i i i i i i i i i i i EX. Simplify with Variables 9 i i 4i y y i y i y i y y
EX. Solving Equations with Non-Real Solutions. 8 i 9 8 Don' t forget the Comple Numbers A comple number is a number in the form a + bi. If b = 0, the number is real. E. +0i usually written If b 0, the number is imaginary. E. i If b 0 and a = 0, the number is pure imaginary. E 0+ i usually written i EX. Addition/Subtractraction. To add/subtract comple numbers, combine the real parts and combine the imaginary parts. ( + i) (4 i) + i 4 + i - + 8i EX. Multiplication FOIL. Whenever i appears, replace it with -. Simplify. ( + 4i) ( + i) +i +0i +8i + i + 8(-) + i 8 7 + i EX. Division You may not leave a binomial imaginary number in the denominator. To rationalize the denominator, multiply the numerator and denominator by the comple conjugate of the denominator. Reminder: a + bi and a bi are comple conjugates i Simplify: i i i i i 0 i i i 4 9i 0 7i 4 9 0 7i 7 7i 4 9 7 7 i Multiply the numerator and denominator by the comple conjugate of the denominator. Simplify Separate final answer into a + bi form. 4
The Quadratic Formula A quadratic equation is an equation with degree of. For eample, + = The quadratic formula is used to find the solutions or roots to quadratic equations in one variable. To use the quadratic formula, the quadratic equation must be in the form: a + b + c = 0 where a 0 The quadratic formula is: b b 4ac a Substitute the values of a, b, and c from the equation into the formula. Simplify to find the solutions/roots of the equation. As an analysis student, you must have the Quadratic Formula memorized. E. Find the roots of + = using the quadratic formula. + = + = 0 Rewrite the equation in the form a + b + c = 0 b b 4ac Substitute a =, b =, c = - into the quadratic formula a 4()( ) Simplify to find the two roots. () The roots are:, Incorporating the two roots using ± is acceptable. This equation has two real irrational roots. E. Find the roots of a = a using the quadratic formula. a + = a Rewrite the equation in the form a + b + c = 0 a a + = 0 b b 4ac Substitute a =, b = -, c = into the quadratic formula a ( ) 4()() Simplify to find the two roots. ()
0 0 4 0 i 0 i The roots are: i, i **Imaginary roots are usually written in the form a + bi, so you should split each root into two separate fractions: i, i. Incorporating the two roots using ± is acceptable. i This equation has two imaginary roots. EX. The quadratic formula can be used when finding the zeros of an equation. f() = 4 0 = 4 Substitute zero for f() ( + 4) ( 4) = 0 No GCF eists/factor difference of two squares ( + 4) ( + ) ( ) = 0 Factor difference of two squares again + 4 = 0 or + = 0 or = 0 Set each factor equal to zero = - = Solve Solve using quadratic formula or square roots The roots are: = ±, ±i Notice that is to the fourth power. Therefore, there are 4 roots. b 0 b 4ac a 0 4i i () 4()(4) 4 0 4 4
The Discriminant The discriminant is the portion of the quadratic formula found beneath the radical: b 4ac By calculating only b 4ac, you can tell the nature of the roots. That is, you can tell what kind of roots you have without solving for the roots. Here are the possibilities: Two real rational roots Two real irrational roots One real rational double root Two imaginary conjugate roots b 4ac is called the discriminant, because it discriminates (tells the difference) among the four natures (types) of roots. If the discriminant (b 4ac) is: A positive perfect square A positive non-perfect square Zero (a neutral perfect square) Negative The nature of the roots is: Two real rational roots Two real irrational roots One real rational double root Two imaginary conjugate roots Directions will be: Find the nature of the roots. Epectation: Use b 4ac only. Show work. Give one of the four descriptions based upon the value of the discriminant. 7
Find the nature of the roots. EX. + + = 0 b 4ac 4()() 0 Since is positive real number and a perfect square, answer: Two real rational roots EX. + = 0 b 4ac 4()(-) + 8 44 Since 44 is positive real number and not a perfect square, answer: Two real irrational roots EX. + + 9 = 0 b 4ac 4()(9) 0 Since 0 is a neutral real number and a perfect square, answer: One real rational double root EX 4. + + 0 = 0 b 4ac 4()(0) 40-4 Since -4 is a negative number, answer: Two imaginary conjugate roots Just to show that b 4ac works, let s solve each eample using b b 4ac a EX 4 4 4, 0, or Two real rational roots EX 44, Two real irrational roots EX 0 0, or One real rational double root EX 4 4 i i Two imaginary conjugate roots 8
Graphing Quadratic Functions A quadratic function is an equation in the form: f() = a + b + c When graphing a function, it is often helpful to replace f() with y: y = a + b + c The graph of a quadratic function is a parabola. The first step is to find the verte point (, y). b To find the -coordinate of the verte, use the formula: a. To find the y- coordinate of the verte, substitute the -coordinate into the equation and solve for y. E. Graph y = + + Step : Find the verte. Find the -coordinate. b a () Find the y-coordinate. y = + + y = (-) + (-) + y = 9 8 + y = -4 Verte point: (-, -4) Step : Make a table of values choosing -values on either side of (-coord. of the verte) Choose enough values to cross the -ais. + + y - + - 0 + 0-4 4 + - - VERTEX -4-4 + - - + 0 0 0 + 0 + Notice the symmetry of the y-coordinates Notice the zeros of the function. Zeros are the values of where y = 0. The zeros are: - and -. Step : Plot the points and connect with a smooth curve. y Notice the equation for the ais of symmetry is the vertical line = - (the -coordinate of the verte) You may recall, when a is positive, the parabola opens up and when a is negative, the parabola opens down. 9
Dividing Polynomials (Long Division) Division Algorithm: Think back to long division from rd grade. How many times does the divisor go into the dividend? Put that number on top. Multiply that number by the divisor and put the result under the dividend. Subtract and bring down the net number in the dividend. Repeat until you have used all the numbers in the dividend. EX: First, we need to write this epression in long division form. You write it the same way you do normal long division, as if they are only simple numbers. You need to use place holders if you have powers of that are missing. The dividend, the thing on the inside being divided, needs to have all powers of represented in decreasing order. The first term must be of the highest power of that is in the epression. The second term must be one lower power of. If there is none of that power, you put a zero. In this first eample, there is no ², so put a zero in to hold the place. 0 Look at the divisor, the term outside of the division symbol, (in this case, ). Only at the highest power of (in this eample,.) The coefficient is just. What times is equal to the first term of the dividend? In other words, what times equals ³? The answer is ², and this is the first term of the quotient. Write this on top of the division bar, eactly above the first term of the dividend.... 0 0
Now, like in normal long division, we multiply this first part of our answer by the divisor, the epression on the right side. Then, you subtract that from the dividend! After this subtraction, you bring down the net term in the dividend, and repeat the process. Subtract... (... 0 ) Bring down the net term Now, repeat the process until the first term of the divisor can no longer go into the term in the dividend. Look at the first term of the divisor, the. Of this new epression, ², look at the first term. What times equals ²? Then continue doing what you just did....... 0 ( )... ( 4) Bring down the net term Now, one more time, what times equals?...... 0 ( )... ( 4) ( ) 0 There is no remainder. The quotient is.
Another eample: 4 0 4 (Pay attention to lining up like terms as you multiply.) 4 4 0 (4 8) ( 0 4) 4 Since does not go into -, this is the remainder. The answer is written: 4 4 Synthetic Division To use synthetic division: There must be a coefficient for every possible power of the variable. The divisor must have a leading coefficient of. For an eample of synthetic division, consider + divided by. First, if a power of is missing from the polynomial, a term with that power and a zero coefficient must be inserted into the correct position in the polynomial. In this case the term is missing, so we must add 0 between the cubic and linear terms: + 0 + Net, all the variables and their eponents (,, ) are removed, leaving only a list of the coefficients:, 0, -,. These numbers form the dividend. We form the divisor for the synthetic division using only the constant term () of the linear factor -. Note: If the divisor were + we would put it in the format (-), resulting in a divisor of -. The numbers representing the divisor and the dividend are placed into a division-like configuration:
The first number in the dividend () is put into the first position of the result area (below the horizontal line). This number is the coefficient of the term in the original polynomial: Bring Down the first coefficient Then this first entry in the result () is multiplied by the divisor () and the product is placed under the net term in the dividend (0): Multiply Net the number from the dividend and the result of the multiplication are added together and the sum is put in the net position on the result line: Add This process is continued for the remainder of the numbers in the dividend: Multiply Add Multiply Add The result is the list,,, 4. All numbers ecept the last become the coefficients of the quotient polynomial. Since we started with a cubic polynomial and divided it by a linear term, the quotient is a nd degree polynomial: The last entry in the result list (4) is the remainder. The quotient and remainder can be combined into one epression: (Note that no division operations were performed to compute the answer to this division problem.)
Rational Eponents The word rational refers to any number that can be written as a fraction. Therefore, rational eponents are fractional eponents (or eponents that can be written as fractions). m b n n b m or The denominator of the eponent is the inde of the root. The numerator of the eponent is: the eponent on base b in the radicand OR the eponent on the entire radical epression. n b m 4 4 4 4 4 EX. 409 8 8 4 OR 4 8 Choose the arrangement that makes the work easier In this case, the second option 4 is easier. Negative eponents mean reciprocal. EX. Many decimal eponents can be written as fractions. EX. 9 9 9 4 Write in eponential form. a b a b EX. a bc c c Multiply. EX. 8 4 4 9 4 To multiply radicals with different indees, work in eponential form. Factor the radicands Write in eponential form Get common denominators to add eponents Multiply by adding eponents on like bases Change to radical form. Simplify. 4
To solve equations in eponential form, raise each side to the reciprocal power. EX. Solve 0 0 4 4 4 4 Composition of Functions EX. Solve 8 8 4 8 The functions f() and g() can be combined to form a composition function. f(g()), read f of g of, is one composition. g(f()) read g of f of, is the other composition depending on the order in which you work. They are two different compositions generally yielding two different values. EX. Find both compositions: f(g()) and g(f()), given f() = and g() =. To find f(g()), start from the inside and work out. In this eample, =. To find f(g()), substitute = into function g: g() = () =. Now, substitute the value of g() into function f: f(g()) = f() = =. So, the composition f(g()) equals. Now, find g(f()). Here, is still. But, the order of the functions has changed. Since we always work from the inside out, substitute = into function f first: f() = = 9. Substitute the value of f() into function g: g(f()) = g(9) = (9) = 8. So, the composition g(f()) equals 8. EX. Find both compositions: f(g()) and g(f()), given f() = and g() =. To find f(g()), start from the inside and work out. In this eample, =. So, g() =. Now, substitute the value of g() into function f: f(g()) = f() = () = 4. So, the composition f(g()) equals 4. Now, find g(f()). Here, is still. But, the order of the functions has changed. Since we always work from the inside out, f() =. Substitute the value of f() into function g: g(f()) = g( ) = ( ) =. So, the composition g(f()) equals.
EX. Find both compositions: f(g()) and g(f()). Given f() = and g() = +, find both compositions. f(g()) = f(+)= g(f()) = f ( ) Inverses of Functions Inverse functions are two functions, f and f - (read f inverse ) that have the following characteristics: NOTE: f - is not an eponent it is a notation or way of writing inverse of function f. Every point (a, b) on function f corresponds to a point (b, a) on the inverse function f -. In other words, all of the coordinate pairs are reversed.. The domain of function f is the range of function f -. And, the range of function f is the domain of function f -. (That should make sense as the coordinate pairs are all reversed, so the domain and range are reversed.). To find the inverse equation of a function, reverse the and the y in the equation, then solve for y. 4. Because of this reversal, the graphs of a function and its inverse are reflections over the line y =.. When you do both compositions, f(f - ()) and f - (f()), you get for both. This is a good algebraic way to see if two equations are inverses just do both compositions and see if you get for both. If you do, the equations are inverses. If you do not, the equations are not inverses. EX. Find the inverse of f() = 4. To find the inverse function, switch places with and y. It is helpful to change f()= to y=. Function: f() = 4 or y = 4 Inverse function: = y 4 Solve for y: + 4 = y y = ½ + Use function Notation: f - () = ½ + Therefore, f() = 4 and f - () = ½ + are inverse functions.
EX. Graph f() and f - () on a coordinate plane. f() = 4 f - () = ½ + y (, ) (4, 4) (-, ) (-4, 0) f - () (0, ) (, 0) (, ) (, -) y = f() (0, -4) Notice that all of the coordinate pairs for f() are reversed for f - (). Notice that the graphs of f() and f - () are a reflection over the line y =. Since f() and f - () are inverse functions, both compositions should equal. Let s verify that the functions are indeed inverses by showing that f(f - ()) and f - (f()) both equal. f() = 4 f - () = ½ + f(f - ()) = f(½ + ) = (½ + ) 4 f - (f()) = f - ( 4) = ½ ( 4) + = + 4 4 = + = = Conclusion: f(f - ()) = and f - (f()) =, therefore f() and f - () are indeed inverses. 7