Ch 5 Alg 2 L2 Note Sheet Key Do Activity 1 on your Ch 5 Activity Sheet.

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1 Ch Alg L Note Sheet Ke Do Activit 1 on our Ch Activit Sheet. Chapter : Quadratic Equations and Functions.1 Modeling Data With Quadratic Functions You had three forms for linear equations, ou will have three forms for quadratics too. Standard form is the common form for quadratics. The others are verte form and factored form. You will use algebra to get from one form to the other, just like ou did with lines. So that is wh we reviewed some Algebra 1 Skills ou will need to change from factored form to standard form. Quadratic Functions and Their Graphs Definition: Standard Form of a Quadratic Function The equation f ( ) = a + b+ c is a quadratic function if a 0. a is called the quadratic term b is called the linear term c is called the constant term The domain of a quadratic function is all real numbers. Note: If a = 0, it would be a linear equation, f ( ) = b+ c. Eample 1 Classifing Functions Determine whether each function is linear or quadratic. Identif the quadratic, linear, and constant terms. a. f( ) = ( + 3) ( ) = ( + 3) ( + 3) distribute = distribute again f( ) = 1 combine like terms compare to f ( ) = a + b+ c Is quadratic, where a =, b = & c = 1 Quadratic term: Linear term: Constant term: 1 b. f( ) = 3( ) 3( ) = distribute f ( ) = + combine like terms compare to f ( ) = a + b+ c so a = 0 so it s not quadratic! Look at the form, = m+ b. f ( ) = + The function is actuall linear with a slope of and -intercept of. Do Activit on our Ch Activit Sheet. Look at Eample on the net page, then Do.1 Part A Eercises: Page 1 #1 9 On calculator #1 1 S. Stirling Page 1 of

2 Ch Alg L Note Sheet Ke Calculator: If ou want the calculator to fit our points [STAT] EDIT [STAT PLOT] [STAT] CALC : QuadReg It gives ou the a, b & c!! Put in [Y=] and graph! Eample Application The table shows the height of a column of water as it drains from its container. Model the data with a quadratic function. Graph the data and the function. Use the model to estimate the water level at 3 seconds. = + [TABLE] when X = 3 Or [CALC] 1: value tpe X = 3 Gives Y1 = mm Vocabular for Quadratic Functions: The graph of a quadratic function is a parabola. The ais of smmetr is the line that divides a parabola into two parts that are mirror images. It is alwas a vertical line defined b the -coordinate of the verte. Points on the parabola have corresponding points on its mirror image. The two corresponding points are the same distance from the ais of smmetr. The verte of a parabola is the point at which the parabola intersects the ais of smmetr. It is where the curve turns from decreasing (downhill) to increasing (uphill) or visa versa. The -value of the verte of a parabola represents the maimum or minimum value of the function. Eample 1 Points on a Parabola Use the graph of f ( ) = +. Identif the verte, ais of smmetr, points P and Q corresponding to P and Q, and the range of f ( ). Verte: (,0 ) Ais of Smmetr: = (a vertical line) P ( 1, ) ( 3, ) Q ( 0,) (,) P both are 1 unit from =. Q both are units from =. Range: All real numbers where 0 S. Stirling Page of

3 Ch Alg L Note Sheet Ke Eample a Use the graph to identif the verte, ais of smmetr, points P and Q corresponding to P and Q, and the range of f ( ). Verte: ( 1, 1) Ais of Smmetr: = 1 (a vertical line) ( 1, 3) (3,3 Q (,0) Q ( 0,0) P P ) both are units from =1. both are 1 unit from =1. P Q Range: All real numbers where 1 Eample b Use the graph to identif the verte, ais of smmetr, points P and Q corresponding to P and Q, and the range of f ( ). Verte: ( 1, ) Ais of Smmetr: P(,1) P ( 0,1) Q( 1, ) Q ( 3, ) = 1 (a vertical line) both are 1 unit from = 1. both are units from = 1. P Q Range: All real numbers where Do.1 Part B Eercises: Page 1 #10 1, #7 9 Homework Worksheet #, page S. Stirling Page 3 of

4 Ch Alg L Note Sheet Ke. Properties of Parabolas Properties of Parabolas The graph of f ( ) = a + b+ c is a parabola when a 0. When a > 0, the parabola opens up. When a < 0, the parabola opens down. b The ais of smmetr is the line =. a b The -coordinate of the verte is. a The -coordinate of the verte is the value of the function b b when = or = f a a. The -intercept is ( 0,c ). Eample 1. Graphing Quadratics Graph and label = f( ) = Identif a, b and c in f ( ) = a + b+ c a =, b = 0 and c = Open up or down? a =, so a < 0, the parabola opens down 3. Find ais of smmetr. b 0 = = = 0 a ( ) ais of smmetr: = Find the verte. So the -coordinate of the verte is 0 and the -coordinate of the verte is The verte is at ( 0,9 ).. Find the -intercept. The -intercept is ( 0,c ), so ( 0,9 ). since f (0) = (0) + 9 = 9 7. Make a curve to go through the points. f (0) = (0) + 9 = 9. Find more points on the graph. 1 (1) + 9 = 1 ( 1) + 9 = () + 9 = 7 ( ) + 9 = 7 Notice the smmetr!! S. Stirling Page of

5 Ch Alg L Note Sheet Ke Eample Graphing Quadratics Graph and label = 3 1. Identif a, b and c in f ( ) = a + b+ c a = 1, b = and c = 3.. Open up or down? a = 1, so a > 0, the parabola opens up Find ais of smmetr. b Ais of smmetr is = = = 1. a (1). Find the verte. So the -coordinate of the verte is 1 and the -coordinate of the verte is The verte is ( 1, ). Find the -intercept. The -intercept is ( 0,c ), so (0) (0) (0) 3 3, f = = ( 0, 3). 7. Make a curve to go through the points. f (1) = (1) (1) 3 =. Find more points on the graph. () () 3 = 3 3 (3) (3) 3 = 0 () () 3 = Find points using smmetr Finding Maimum and Minimum Values Eample 3 Finding a Minimum Value What is the minimum value of the function? f ( ) = a = 3 b = 1 c = Identif, and. a = 3, so a > 0, the parabola opens up (1) Ais & the -coordinate of verte is = =. (3) The -coordinate is f ( ) = 3( ) + 1( ) + = So the minimum value is, when =. Do. Part A Eercises: Homework Worksheet #, page 7 Page # 7 S. Stirling Page of

6 Ch Alg L Note Sheet Ke Eample Application A compan knows that.p + 00 models the number it sells per month of a certain make of uniccle, where the price p can be set as low as $70 or as high as $10. Revenue from sales is the product of the price and the number sold. What price will maimize the revenue? What is the maimum revenue? Independent variable: p = price, Dependent variable: R(p) = revenue From problem: Revenue = price * number sold Substitute: Rp ( ) = p.p+ 00 ( ) Standard Form: (00) Find the verte: p = = 100, (.) Rp ( ) =.p + 00 p R (100) = 100( ) = =, A price of $100 will maimize the revenue of $,000. Eample Application Part The number of widgets the Woodget Compan sells can be modeled b p + 100, where p is the price of a widget. What price will maimize revenue? What is the maimum revenue? Independent variable: p = price per widget, Dependent variable: R(p) = revenue Now Know: Revenue = price * number sold Substitute: Rp ( ) = p( p+ 100) Standard Form: Rp ( ) = p p Find the verte: (100) p = = 10, ( ) R (10) = 10( ) = 10( 0) = 00 A price of $10 will maimize the revenue of $00. Do. Part B Eercises: Page # 30 Activit Sheet Page : Reviewing Transformations S. Stirling Page of

7 Ch Alg L Note Sheet Ke.3 Transforming Parabolas In Chapter, ou learned to graph absolute value functions as transformations of their parent function, =. Similarl, ou can graph a quadratic function as a transformation of the parent function =. Summar Parent Function: = = a Vertical Stretch a > 1 Stretch awa from -ais b a factor of a. Vertical Shrink (fraction of) 0 < a < 1 Shrink toward -ais b a factor of a. Reflection in -ais (negative) a < 0 Reflects over the -ais and stretches or shrinks. Summar Translations Shifts Vertical Translations Translate up k units Translate down k units = + k = k Horizontal Translations (counter intuitive) Translate right h units = ( h) Translate left h units = ( + h) 1 Eample A: Graph = Parent Function: 1 =, with vertical shrink (factor ½ ) and reflect over -ais. 1 1 Shrink: = 1 Shrink & Reflect: 3 = 1 Parent Shrink Flip ½ ½ 1 1 ½ ½ ½ ½ 3 9 ½ ½ 10 Eample B: Graph ( ) = + 3 = +. For, shift parent up. For, shift parent to the left and = + 3 Eample C: Graph ( ) 7 = + Eample D: Graph ( ) The blue is the stretch b a factor of. The pink shifts the blue graph left 3 and down. 10 The blue is the reflected over the -ais. The pink shifts the blue graph right and up. 10 S. Stirling Page 7 of

8 Ch Alg L Note Sheet Ke To transform the graph of a quadratic function, ou can use the verte form of a quadratic function, = a( h) + k. Verte Form of a Quadratic Function The equation = a( h) + k is a quadratic function if a 0. The graph (and verte) of = a shifts h units horizontall and k units verticall. The verte is ( hk, ) and the ais of smmetr is the line = h. Eample 1 (Method 1 Transforming): Graph = 1 ( ) Graph 1 = blue Shift obvious points to the right and up 3, pink 10 = Eample 1 (Method Verte Form): Graph ( ) Use the form to = a( h) + k Graph the verte and ais of smmetr. Verte ( hk, ) is (,3 ) Ais is =. Find some points and graph them and their reflections over the ais of smmetr. 0 1 ( 0 ) + 3= 1-1 ( ) + 3= 10.3 Part A Eercises: Homework Worksheet #7, page 9 10 S. Stirling Page of

9 Ch Alg L Note Sheet Ke.3 Transforming Parabolas: Going from the Graph to the Equation E : Writing an Equation for a Parabola Write the equation for the parabola. Use the verte form: = a h + k. ( ) The verte is ( 3, ) or ( hk, ) = a( 3) + so far Your missing a, but ou know a point on the parabola (, ) is (, ) Substitute and a. ( ) = a () 3 + ( ) ( ) = a + = a + = a = a ( ) The equation is = 3 + Connections: Both the verte form and the standard form give useful information about a parabola. The standard form makes it eas to identif the - intercept. The verte form makes it eas to identif the verte and the ais of smmetr, and to graph the parabola as a transformation of the parent function. The graph shows the relationship between the two forms. E 3: Writing an Equation for a Parabola Write = + in verte form. Find a: a =1 from the standard form. b ( ) Find the -coordinate of the verte: = = = a (1) Find the -coordinate of the verte: ( ) ( ) The verte is (, ) or (, ) hk. = + = + = Now just substitute into verte form: = a( h) + k ( ) = 1 () + () Then simplif ( ) = +.3 Part B Eercises: Homework Worksheet #, page 11 1 S. Stirling Page 9 of

10 Ch Alg L Note Sheet Ke. Factoring Quadratic Epressions Do Computation Worksheet #9 on our Ch Homework Worksheet, pages Finding Common and Binomial Factors Factoring is rewriting an epression as the product of its factors. The greatest common factor (GCF) of an epression is a common factor of the terms of the epression. It s the common factor with the greatest coefficient and the greatest eponent. You can factor an epression that has a GCF not equal to 1. E 1 Plus: Finding Common Factors Completel factor each epression. Strateg: Find the greatest common factor and un-distribute! a = ( ) = i i i 0 = iii 1 = 1iii3 b. 9n n 3n 3n = ( ) 9n = 3 3 iii n n n= 13 i i i i in A quadratic trinomial is an epression in the form a + b + c. You can factor man quadratic trinomials into two binomial factors. E a: Factoring Factor Step 1: Find two factors with a product of ac and a sum b. Step : Rewrite the b term with the factors ou found Step 3: Group and factor out the GCF from each grouping terms ( ) ( ) ( + 1) 7( 1) + + onl terms Notice the binomial factors ( + 1) in both terms product ac 1 7= 7 1 7= 7 sum b 1+ 7= Step : Factor out the common binomial factor ( ) ( ) ( + 7)( + 1) E b: Factor ( ) ( ) ( + )( + ) ac 1 = = b + = E c: Factor ( + ) + ( + ) ( + )( + ) ac 1 3= 3 = 3 b 1 + = 1 S. Stirling Page 10 of

11 Ch Alg L Note Sheet Ke E 3a: Factoring (The factors do not need to be positive!!) Factor Step 1: Find two factors with a product of ac and a sum b. Use prime factors: 7 ac 1 7= 7 9 = 7 3 b = Step : Rewrite the b term as a sum with the factors ou found Step : Factor out the common Step 3: Group and factor out the GCF from each grouping binomial factor ( 9) + ( + 7 ) ( 9) + ( 9) ( 9) + ( 9) 9 ( )( ) E 3b: Factor + E 3c: Factor ( ) ( ) ( )( ) ac 1 = = ac 1 = 3 = b + = b = ( 3) + ( 3) ( )( 3) Use prime factors: 1 3 Do. Part A Eercises: Page 3 #1 1 S. Stirling Page 11 of

12 Ch Alg L Note Sheet Ke E a: More Factoring (Same thing different numbers.) Factor 1 E b: a. Factor 1 3 Step 1: Step : Step 3: ( + 3) + ( + 3) Step : ( )( + 3) ac 1 1= 1 3 = 1 b 1 3+ = 1 ac 1 3= 3 1= 3 b 1 + 1= ( + ) + 1( + ) ( 1)( + ) Use prime factors: 3 E c: b. Factor ( ) ( ) ( )( ) ac 1 10= 10 = 10 b 3 + = 3 E d: c. Factor ( ) ( ) ( + )( 1) ac 1 = 1 = b 1+ = E a: Factoring (Same stuff, but a bit harder.) Factor Step 1: Find two factors with a product of ac and a sum b. ac 3 = 1 1 1= 1 b = 1 Step : Rewrite the b term as a sum with the factors ou found Step : Factor out the common Step 3: Group and factor out the GCF from each grouping binomial factor ( ) ( ) ( ) + 1( ) 3 ( ) ( ) ( 3 1)( ) S. Stirling Page 1 of

13 Ch Alg L Note Sheet Ke E b: a. Factor ac 1= 3= b = 11 E c: c. Factor 7 + ac = 1 3= 1 b 7 + 3= ( ) ( ) ( + )( + 3) Use prime factors: ( ) ( ) ( 3)( ) Use prime factors: 1 3 E a: Factoring (Same stuff, but a bit harder.) Factor 1 Use prime factors: Step 1: Find two factors with a product of ac and a sum b ac 1= 0 10 = 0 3 b 10 + = Step : Rewrite the b term as a sum with the factors ou found Step : Factor out the common binomial factor Step 3: Group and factor out the GCF from each grouping ( ) + 3( ) ( 10) + ( 1) ( + 3)( ) + 3 ( ) ( ) E b: a. Factor ( ) + ( 1 ) ( + 9)( 1) 1 9 ac 9= 1 9= 1 b 7 + 9= 7 E c: b. Factor ( ) + ( ) 3+ ( )( ) ac 3 1= 3 1 = 3 b = 1 Use prime factors: Do. Part B Eercises: Page 3 #19-3 Do Computation Worksheet #10 on our Ch Homework Worksheet, pages 1 1. S. Stirling Page 13 of

14 Ch Alg L Note Sheet Ke A perfect square trinomial is the product ou obtain when ou square a binomial. An eample is + 10+, which can be written as ( +. The first term and the third term of the trinomial are alwas positive, as the represent the squares of the two terms of the binomial. The middle term of the trinomial is two times the product of the terms of the binomial. ) Perfect Square Trinomials ( ) u+ v = u + uv+ v ( ) u v = u uv+ v Pre-Eample 7a: Multipl ( + 3) means ( + 3)( + 3) formula u + uv+ v with u = and v = = ( ) ( )( ) ( ) Pre-Eample 7b: Multipl ( ) means ( )( ) formula u uv+ v with u = and v = + = 0+ ( ) ( )( ) ( ) Now tr epanding, multipling, the perfect square trinomials simpl b using the formulas. Pre-Eample 7c: Multipl ( 3 + ) formula u + uv+ v with u = 3 and v = Pre-Eample 7d: Multipl ( 7) u uv+ v formula with u = 1 and v = ( ) + ( )( ) + ( ) ( ) ( )( ) ( ) To factor, using these formulas, all ou need to do is identif u and v utilizing the formulas, then work it in reverse. Eample 7a: Factor Eample 7b: Factor = 9, u= 3 =, u= since ( ) since ( ), since ( ) since ( ) 7 = 9 v = 7 3 = 9, v = 3 check u u v+ v check u + uv+ v ( 3) ( 3)( 7) + ( 7) = ( ) + ( )( 3) + ( 3) factored is ( ) = ( ) u v 3 7 factored is ( u+ v) = ( + 3) Eample 7c: Factor since, u Eample 7d: Factor = since =, u= ( ) = ( ) () 1 = 1 7 since ( 9) =, since, v = 1 v = 9 check u uv+ v check u + uv+ v ( ) ( )( 1) + ( 1) = ( ) + ( )( 9) + ( 9) ( ) ( 1) factored is u v = factored is u v = = = ( + ) ( + 9) S. Stirling Page 1 of

15 Ch Alg L Note Sheet Ke Difference of Two Squares An epression of the form a b is defined as the difference of two squares. It also follows a pattern that makes it eas to factor. ( u+ v)( u v) = u v Pre-Eample a: Multipl ( long wa Now use the formula in reverse. Pre-Eample b: Multipl ( + 3)( 3) + )( ) long wa u+ v u v u+ v u v with u = and v = formula ( )( ) with u = and v = 3 ( ) ( ) 3 = 9 Eample a: Factor u= and since ( ) =, v = check u v = ( ) ( ) = factored is ( u+ v)( u v) = ( + )( ) formula ( )( ) ( ) ( ) = Eample b: Factor 9 =, u= since ( ) since ( ) check 7 = 9, v = 7 u factored is ( ) ( ) 9 ( + 7)( 7) v = 7 = ( u+ v)( u v) = Eample c: The photo shows the thin ring that is the cross-section of the pipe. Find an epression, in factored form, that gives the area of the cross-section in completel factored form. Area of a washer = π R Outer radius R = 3 Inner radius r π r ( ) 3 π r A = π = 9π π r Factor out GCF: π ( 9 r ) Diff. Squares: ( r)( π 3+ 3 r) S. Stirling Page 1 of

16 Ch Alg L Note Sheet Ke Combined factoring: Factor out a GCF ( ) 1. Factor out a GCF. Make highest power term positive.. Test for special case: Are the first and last terms perfect squares? Epanded Form = Factored Form u + uv+ v = u+ v u uv+ v = u v ( ) ( ) ( u v)( u v) u v = + 3. Factor b ac and b method. Factor b ac and b method. = a + b + c Epanded: u + uv+ v ( ) + ( )( ) + ( ) Fits the form! Factor ( u+ v) ( + ) Epanded: + 9 u uv+ v ( ) ( )( 7) ( 7) Factor ( u v) ( + Fits the form! Epanded: 9 u v 7) ( ) ( 3) Fits the form! Factor ( u+ v)( u v) ( + 3)( 3) ac 30 1 b ( 10 1) + ( + 3) ( 3) + 1( 3) ( 1)( 3) Do Do. Part C Eercises: Page 3 #37 #1 Do Computation Worksheet #11 on our Ch Homework Worksheet, pages S. Stirling Page 1 of

17 Ch Alg L Note Sheet Ke Algebra 1 Review: Square Roots and Radicals A radical smbol indicates a square root. The epression 1 means the principal, or positive, square root of 1 or. The epression 1 means the negative square root of 1 or. In general, ± for all real numbers. Square Root Properties When simplifing radicals, three rules appl 1) You ma not have factors that are perfect squares under the Multiplication Propert of Square Roots radical sign. For an numbers a 0 and b 0, The cure? Factor the radicand (the numbers under the radical) and use the multiplication propert. ab = a b Simplif 1 = 3= 3 = 3 7 = 3 = 3 = 3 Division Propert of Square Roots For an numbers a 0 and b > 0, a a = b b 7 = 9 = 9 = 9 = 3 = Know our Perfect Squares: ) You ma not have fractions under the radical sign. The cure? Use the division propert and simplif further, if necessar. Simplif 9 = = i = = = 7 3) You ma not have radicals in the denominator of a fraction. The cure? Multipl the numerator and denominator of the fraction b the radical in the denominator (to create a perfect square). Simplif further if needed Simplif = = = = = = = = = Do Page #1 1 S. Stirling Page 17 of

18 Ch Alg L Note Sheet Ke. Solving Quadratic Equations Solving an equation means to the values that make the original sentence true. With linear equations ou usuall onl have one solution, with quadratics ou usuall have two. Zero Product Propert If p q= 0, then either p = 0 or q = 0 or both = You will use this propert to help ou solve quadratics. Eample 1a: Solving b Factoring Eample 1b: Solving b Factoring Solve 11= 1 Solve 1 = 11= 1 1 = 11+ 1= 0 get equation = 0 1 = 0 get equation = 0 ( ) + ( + 1) =0 factor ( 1) = 0 factor & use Zero Product Prop. ( 3) + ( 3) = 0 = 0 or 1= 0 so solve for s = 0 or = 1 ( )( 3) = 0 now use Zero Product Prop. =0 or 3= 0 so solve for Caution!! Don t tr alternate procedures!! = or = 3 1 = 1 = divide both sides b = 1 Oops! You lost a root! NO GOOD Eample 1c: More Practice Solve b Factoring a. Solve = 0 Eample 1d: More Practice Solve b Factoring b. Solve + 7= = = 0 ( ) ( ) ( + )( + 3) = = = = 0 ( ) ( ) ( )( + 9) = 0 + = 0 or + 3= 0 = or = 3 = 0 or + 9= 0 = or = 9 Eample 1e: More Practice Solve b Factoring c. Solve = 0= + ( )( 3) 0= 1 + 0= ( + 3) factor out GCF and divide! 1= 0 or + 3= 0 0 = = 1 or = 3 0= ( ) ( ) Do Page 70 #1, 13, 1, 17 S. Stirling Page 1 of

19 Ch Alg L Note Sheet Ke. Solving Quadratic Equations (cont.) Look at Eample below. How is it different from the previous two Eamples? VIP! You ed it!? There is no linear term. So, ou can solve b taking the square root.! Careful!! Don t lose a root! Solve =. B substitution, ou know the solution is + or, because = and ( ) = Solve. If ou solve this b taking the square root of both sides, ou need to remember that ou get a positive and a negative root. = Write ± for plus or minus = = ± =± Definition: or ± for all real numbers. Eample a: Solving b Square Roots Hint: Work in the reverse order of operations because ou re solving! Solve 10 = 0 10 = 0 Would be hard to factor, no? = 10 get the quadratic term alone! 10 = isolate the completel = 3 now take the square root, both sides =± 3 need the ±!! =± simplif Eample b: Solving b Square Roots Solve 3 = 3 = 3 = 3 3 isolate the completel = now take the square root, both sides = ± need the ± =± i =± simplif (take perfect squares out) Eample c More Practice Solve b Square Root: Alternate Method, Factoring: a. Solve =0 a. Solve = 0 = 0 Could factor, but = get the quadratic term alone! = isolate the completel = now take the square root, both sides =± need the ±!! =± simplif = 0 Tr factoring! Wh not? ( + )( ) = 0 difference of two squares + = 0 or = 0 = or = = or = = ± Do Page 70 #7 1, 1, 1, 1 S. Stirling Page 19 of

20 Ch Alg L Note Sheet Ke Not ever quadratic equation can be solved b factoring or b ing square roots. You can solve a + b + c b graphing = a + b+ c, its related quadratic function. The value of is 0 where the graph intersects the -ais. Each -intercept is a zero of the function and a root of the equation. Solving b Tables and/or b Graphing B the wa, ou can use these methods to solve an tpes of equations! But, ou need a calculator. Enter the left hand part of the equation in Y1 and enter the right hand part of the equation in Y, then ou re just looking for when the -values are the same! (The -values that make the sentence true!) Calculator Solutions [TABLE]: Calculator Solutions [GRAPH]: [Y=] enter left hand part Y1 [Y=] enter left hand part Y1 enter left hand part Y enter left hand part Y [TBLSET] start at 0 with ΔTbl = 1 [GRAPH] [ZOOM] choose a window if necessar. [TABLE] Look for when Y1 = Y. Look for both intersections if possible You can tweek the ΔTbl to a smaller nd [CALC] : intersect number until the Y values get reall close. Answer calculator s questions. Make sure ou both answers if there are two points of intersections. Eample a: Solving b Tables Solve 11= 1 Eample : Solving b Graphing Y1 = 11 Solve = Y = 1 Y1 = [TBLSET] TblStart = 0 with ΔTbl = 1 Y = Look for when Y1 = Y. [GRAPH] Tr [ZOOM] : Happens at = 3, but ou know there is probabl Look for when Y1 = Y, the intersection points. another solution, so change ΔTbl = 0.1. nd [CALC] : intersect Guess at the left intersection. For the left intersection: X = 1.3 Y = Look again nd [CALC] : intersect Guess at the right intersection. At =., there is another solution. For the right intersection: X = 3.3 Y = With quadratics, there is a maimum of solutions. With quadratics, there is a maimum of solutions. Eample b: Solving b Tables Solve + =0 Y1 = + Y = 0 [TBLSET] TblStart = 0 with ΔTbl = 1 Look for when Y1 = Y. Notice that the sign changes between = 0 and = 1, then again between = and =, so must be near zero between those values. So change ΔTbl = Look again near the -values ou found before At 0.1 and. are the solutions. Do Page 70 #0 31 #3 (even) S. Stirling Page 0 of

21 Ch Alg L Note Sheet Ke. Comple Numbers Introduction: The imaginar number i is defined as the number whose Solve =. square is 1. So i = 1 and i = 1. Tr to solve =±, but ou An imaginar number is an number of the form a + bi, can t take the square root of a where a and b are real numbers, and b 0. negative well not if the answer needs to be a real number. So to simplif comple numbers, ou will rewrite Square Root of a Negative Real Number an a as i a, then simplif as ou would do For an positive real number a, a = i a. normall. Eample: = 1 = i = i Eample 1: Simplifing Numbers Using i = i = i = 1= Simplif b using the imaginar number i. Note that ( ) ( ) NOT( ) 1 = not correct! So, ou must write it as an imaginar number first, before ou simplif. 1 i i Factor. i Take out the factor of 1 and. Eample 1b Eample 1c: Eample 1d: a. Simplif 7. b. Simplif. c. Simplif. 1 3 i 17 i 13 i i 13 i 7 i 3 i Imaginar numbers and real numbers together make up the set of comple numbers. Now we can epand our number tpes Eample a: Simplifing Imaginar Numbers Write the comple number 9+ in the form a + bi. 9+ 3i + Simplif the radical epression. + 3i Write in the form a ± bi. Eample 3: Write the comple number the form a + bi in 19 ii + 7 3i + 7 Do Page 7 #1 1 S. Stirling Page 1 of

22 Ch Alg L Note Sheet Ke You can appl the operations of real numbers to comple numbers. Simpl pretend that the i is an. Eample a: Additive Inverse of a Comple Number Find the additive inverse of + i ( + i) Find the opposite. i Simplif. Eample b: Eample c: Eample d: Find the additive Find the additive Find the additive inverse of inverse of inverse of i 3i a+ bi 3i a+ bi ( i) ( ) ( ) i 3i a bi To add or subtract comple numbers, combine the real parts and the imaginar parts separatel. Add or subtract like things. Eample a: Adding Comple Numbers Simplif the epression + i ( + 7i) + ( + i) + 7i+ i 3+ 13i Eample b: Simplif ( + 3i) ( + i) + 3i i + 3i i i Eample c: Eample d: Simplif Simplif ( + ) ( i) 7 3 i + 3i 7 3 i i+ 3i i 3i For two imaginar numbers bi and, Eample : Multipling Comple bi ci = bci = bci 1= bc. ci ( )( ) Eample 7a: Numbers Simplif ( 1i)( 7i) a. Find ( i)( i ) = i = Eample 7b: Eample 7c: 0i and since i = 1 Simplif Simplif 0i 1 = 0 b. Find ( + 3i)( 3+ i) ( 3+ i) + 3i( 3+ i) + 10i 9i+ 1i and since i 1+ 10i 9i 1+ i = 1 ( i)( 3i) ( 9i)( + 3i) 1i 0i+ 1i 1 1i 0i 9 3i 1 + 1i 3i 7i i 3i 3 i S. Stirling Page of

23 Ch Alg L Note Sheet Ke Some quadratic equations have solutions that are comple numbers. Solve like ou did before, just use comple numbers when simplifing. Eample a: Finding Comple Solutions Eample b: Simplif Solve = 0 Solve 3 + = = Isolate the. 3 = = 3 3 Now take the square root of both sides. = 1 =± Remember the ± (two roots). =± i = ± 1 = ± i Check: = i Show it works! ( i ) = 0 i i = = 0 Eample c: Check: = i Show it works! Simplif Solve ( i) = 0 i i = = 0 = = = 0 10 = ± 30 =± i 30 Eample d: Simplif Solve + = 0 Do Page 7 # = 1 = 1 1 = ± i =± i S. Stirling Page 3 of

24 Ch Alg L Note Sheet Ke.7 Completing the Square You can use the technique of completing the square to solve quadratic equations and to rewrite quadratics into verte form. You must know our perfect square trinomials well in order to appl this technique. We will start with just the technique of completing the square. Perfect Square Trinomials ( ) u+ v = u + uv+ v ( ) u v = u uv+ v Eample 1a: Complete the square What would ou have to add to the epression to get a perfect square? Eample 1b: Complete the square What would ou have to add to the epression to get a perfect square? + 10 u + uv+ v So u = and uv =, v=, u uv+ v So u= and uv = 10, v =, and v = v= 10, v = 10, and v = In order to make a perfect square, we would have to add v = = 1 In order to make a perfect square, we would have to add v = = If ou add 1, ou get a perfect square. 10+ If ou add 1, ou get a perfect square. ( + ) rewrite as a perfect square. ( ) rewrite as a perfect square. You can use this techniques to help ou solve equations. Remember from before if ou have an equation like = 100, ou can easil solve it b taking the square root of both sides, =± 100 =± 10. So if we can write an epression as a square, then we can use the square root to solve it. Eample a: Solving b Completing the Square Solve = 3 Since the left hand side is a perfect square trinomial, we can rewrite it as a perfect square. Eample 3: Solving b Completing the Square Solve 1+ = 0 Get the left hand side read for completing the square b subtracting from both sides = 3 1 = to complete the square, need to ( ) + = 3 rewrite it as a perfect square. add = 3 to both sides = =± 3 take the square root of both sides. + =± ( ) simplif = 31 rewrite it as a perfect square. =± subtract from both sides =± 31 take the square root of both sides. You get answers: =+ = 1 and = = 11 = ± 31 + adding to both sides You get answers: = + 31 and = 31 S. Stirling Page of

25 Ch Alg L Note Sheet Ke Eample : Solving b Completing the Square Solve + 3=0 Get the left hand side read for completing the square b subtracting 3 from both sides. = 3 to complete the square, need to add = 1 to both sides. + 1= 3+ 1 = 0 ( ) =± 0 = ± i simplif completel = ± i You get answers: = + i and = i You can also use completing the square to rewrite a standard form equation, form, = a( h) + k. = a + b + c, into verte Eample : Rewriting in Verte Form Eample : Rewriting in Verte Form Write = 10 in verte form. Write = + + in verte form. You will need to have a + to be able to complete the square. So if ou add, ou will need to subtract so that ou do not change the value. You will need to have a +9 to be able to complete the square. So if ou add 9, ou will need to subtract 9 so that ou do not change the value of the function. = 10+ = = ( 10+ ) + ( ) = ( + + 9) + ( 9) = ( ) 7 rewrite as a perfect square = ( + 3) 7 rewrite as a perfect square = a( h) + k = a( h) + k You can now see that the verte is at You can now see that the verte is at (, 7) and since a = 1 there is no stretching nor ( 3, 7) and since a = 1 there is no stretching nor shrinking. shrinking. Do Page 3 #1 3,, 7 1, 1, 1, 19,, 31 S. Stirling Page of

26 Ch Alg L Note Sheet Ke. The Quadratic Formula So far ou can solve quadratic equation b the factoring method, the square root method, completing the square and graphing methods (b graphing calculator). But what if the first two methods don t work and ou don t have a calculator? You can use the quadratic formula to the solutions, but be warned! The aren t alwas prett looking! WATCH ORDER OF OPERATIONS! The Quadratic Formula A quadratic equation written in standard form a + b + c = 0 can be solved with the quadratic formula. b b ac = ±. a a The discriminant is the value of b under the radical sign. ac. The epression Note: be sure to get the equation in standard form before identifing a, b and c. b He! Do ou recognize the part? a Yep, it is the ais of smmetr. Eample 1a: Solving b Quadratic Formula Eample a: Solving b Quadratic Formula Solve 3 = Solve = 7 3 = 3 = 0 get in standard form and get it = 0 = = 0 get it = 0 a = 3, b = & c = a =, b = & c = 7 Use the formula Use the formula = ± a a b b a c = ± a a b b a c () ()(7) ( ) ( ) (3)( ) () = ± Substitute. = ± Substitute. (3) (3) () () The discriminant The discriminant = ( ) (3)( ) = () ()(7) = + = 9, then substitute & simplif: = 3 = 0, then substitute & simplif: = ± 9 = ± 7 0 = ± 3 i 3 i = ± = ± So our two solutions are = + = = or = = = 3 You could do this b factoring! Notice ou get two real answers!! So our two solutions are 3 i 3 i = + or = Notice ou get two comple (imaginar) answers!! It has something to do with the discriminant. S. Stirling Page of

27 Ch Alg L Note Sheet Ke Eample 1b: Eample b: Solve 3 = Solve = =0 3= 0 a = 3, b = 1& c = + + 3= 0 b b ac a =, b = & c = 3 = ± a a b b ac = ± ( 1) ( 1) (3)( ) a a = ± (3) (3) () () ()(3) = ± () () The discriminant = ( 1) (3)( ) = 1+ = 9, then substitute & simplif: The discriminant = () ()(3) = 1 1 =, then substitute & simplif: = ± 9 = 1 ± 7 i i = 1± = 1± = 1± So our two solutions are i ± i Or = ± = = + = = or = = = 1 3 Do Page 93 #1,, 7, 11, 1, 1, 0 Quadratic equations can have real or comple solutions. You can determine the tpe and number of solutions b ing the discriminant, b ac. The table shows the relationships among the value of the discriminant, the solutions of a quadratic equation, and the graph of the related function. These relationships are true for real number values of a, b, and c. S. Stirling Page 7 of

28 Ch Alg L Note Sheet Ke Eample a: Using the Discriminant Eample b: Determine the tpe and number of solutions of Determine the tpe and number of solutions of + + = = 0. a = 1, b = & c = a = 1, b = & c = 9 The discriminant = () (1)() = 3 3 = The discriminant = () (1)(9) = 3 3 = 0 Since the discriminant is positive, + + = 0 has two real solutions. (And the graph of f ( ) = + + has two - intercepts.) Since the discriminant is zero, + + 9=0 has one real solution. (And the graph of f ( ) = has one -intercept.) Eample c: Determine the tpe and number of solutions of = 0. a = 1, b = & c = 10 The discriminant = () (1)(10) = 3 0 = Since the discriminant is negative, = 0 has no real solutions. It has two imaginar solutions. (And the graph of f( ) = has no -intercepts.) You are now familiar with five methods of solving quadratic equations: ing square roots, factoring, using tables, graphing, and using the Quadratic Formula. For a = c, ing square roots works best. The discriminant can help ou decide how to solve equations that have an -term. Do Page 93 #,,, 3 39 Summar: 1. If ou can write the equation in the form a = c, use the Square Root method.. Otherwise write the equation in the form a + b + c = 0, then Discriminant Method positive perfect square Factoring, Quadratic Formula, graphing or table. positive nonsquare Quadratic Formula, graphing or table. zero Factoring, Quadratic Formula, graphing or table. negative Quadratic Formula, graphing or table. S. Stirling Page of