Chapter 5: Quadratic Equations and Functions 5.1 Modeling Data With Quadratic Functions Quadratic Functions and Their Graphs

Ch 5 Alg Note Sheet Ke Chapter 5: Quadratic Equations and Functions 5.1 Modeling Data With Quadratic Functions Quadratic Functions and Their Graphs Definition: Standard Form of a Quadratic Function The equation f ( ) = a + b+ c is a quadratic function if a. 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 =, 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 = + 3 1 distribute again f( ) = 5 1 combine like terms Is quadratic, where a =, b = 5 & c = 1 Quadratic term: Linear term: 5 Constant term: 1 b. f ( ) = 3( ) 3( ) = 3 3 + distribute f ( ) = + combine like terms a = so it s not quadratic! Look at the form, = m+ b. The function is actuall linear with a slope of and -intercept of. 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. Also, 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 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: (, ) Ais of Smmetr: = (a vertical line) 1, P 3, both are 1 unit from =. P ( ) ( ) Q (,) (,) Q both are units from =. Range: All real numbers where S. Stirling Page 1 of 9

Ch 5 Alg Note Sheet Ke Using Quadratic Models Eample 3 Fitting a Quadratic Function to 3 Points Find a quadratic function that includes the values in the table. Note the (, ) in the quadratic form Point Substitute in for & Simplif,3 3 = a() + b() + c 3= a+ b+ c ( ) ( 3,13 ) (, 9 ) 13 (3) (3) = a + b + c represent the points on the parabola. = a + b + c 13 = 9a+ 3b+ c = a + b + c 9 = 1a+ b+ c 9 () () Now ou need to solve the sstem of equations for a, b & c. You do this in the same wa ou solved sstems before, ou simpl need to substitute or eliminate variables for 3 equations and 3 variables. Easiest variable to eliminate is c. Just multipl both sides of E, equation, b 1. Then add E1 to E then E to E3. Add E1 & 1*E Solve: 3= a+ b+ c 1 = 5a 1b 13 = 9a 3b c Take the Results: 1 = 7a+ 1b E1 3= a+ b+ c 1 = 5a 1b 1 = 5a 1b = a E 13 = 9a+ 3b+ c 1 = 7a+ 1b a = 3 E3 9 = 1a+ b+ c Now can ou Add E3 & 1*E a & b? 9 = 1a+ b+ c Find b: 13 = 9a 3b c 1 = 7(3) + 1b 1 = 7a+ 1b 1 = 1+ b b = 5 Check on the calculator: Plot the 3 points. [STAT] Put the equation in [Y=] Were all three points on our graph? Tr [ZOOM] 9 or [ZOOM] out! The quadratic function is: = a + b+ c = 3 5+ 1 Find c: 3= (3) + ( 5) + c 3= 1 1+ c c = 1 Calculator: If ou want the calculator to fit our points [STAT] EDIT [STAT PLOT] [STAT] CALC 5: QuadReg It gives ou the a, b & c!! Put in [Y=] and graph! Tr the Quick Check page answer: = + 3 1. 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 35 seconds. = + [TABLE] when X = 35 Or [CALC] 1: value tpe X = 35 Gives Y1 = 57.937 mm.9.13 1.3333 S. Stirling Page of 9

Ch 5 Alg Note Sheet Ke 5. Properties of Parabolas Properties of Parabolas The graph of f ( ) = a + b+ c is a parabola when a. When a >, the parabola opens up. When a <, 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 (,c ). Eample 1 Alt. Graphing Quadratics Graph and label = f( ) = + 9 1 Identif a =, b = and c = 9. a =, so a <, the parabola opens down Ais of smmetr is b = = =. a ( ) So the -coordinate of the verte is and the -coordinate of the verte is The verte is at (,9 ). The -intercept is (,c ), so (,9 ). f () = () + 9 = 9 1 1 3 More points on the graph. 1 (1) + 9 = 5 1 ( 1) + 9 = 5 () + 9 = 7 ( ) + 9 = 7 Notice the smmetr!! Tr: = S. Stirling Page 3 of 9

Ch 5 Alg Note Sheet Ke Eample Graphing Quadratics Graph and label = 3 Identif a = 1, b = and c = 3. 1 1 a = 1, so a >, the parabola opens up Ais of smmetr is b = = = 1. a (1) So the -coordinate of the verte is 1 and the -coordinate of the verte is The verte is ( 1, ), The -intercept is (,c ), so More points on the graph. 1 () () 3 = 3 3 (3) (3) 3 = () () 3 = 5 f (1) = (1) (1) 3 = f () = () () 3 = 3, (, 3). Find points from the smmetr. 3 1 5 3 1 1 3 5 Tr: = + + Finding Maimum and Minimum Values Eample 3 Finding a Minimum Value What is the minimum value of the function? Identif a = 3, b = 1 and c =. a = 3, so a >, the parabola opens up f ( ) = 3 + 1 + (1) Ais & the -coordinate of verte is = =. (3) The -coordinate is f ( ) = 3( ) + 1( ) + = So the minimum value is, when =. S. Stirling Page of 9

Ch 5 Alg Note Sheet Ke Eample Application A compan knows that.5p + 5 models the number it sells per month of a certain make of uniccle, where the price p can be set as low as $7 or as high as $1. 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 Rp ( ) = p.5p+ 5 Substitute: ( ) Standard Form: Find the verte: Rp ( ) =.5p + 5 p (5) p = = 1, (.5) R (1) = 1(.5 1 + 5) = 1 5 = 5, 1 5 5 1 15 5 A price of $1 will maimize the revenue of $5,. Eample Application Part The number of widgets the Woodget Compan sells can be modeled b 5p + 1, 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( 5p+ 1) Standard Form: Rp ( ) = 5p + 1 p Find the verte: (1) p = = 1, ( 5) R (1) = 1( 5 1 + 1) = 1( 5) = 5 A price of $1 will maimize the revenue of $5. S. Stirling Page 5 of 9

Ch 5 Alg Note Sheet Ke 5.3 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) < a < 1 Stretch awa from -ais b a factor of a. Reflection in -ais (negative) a < Reflects over the -ais and stretches or shrinks. Summar Translations Shifts Vertical Translations Translate up k units (k positive): Translate down k units (k positive): = + k = k Horizontal Translations (counter intuitive) Translate right h units (h positive): = ( h) Translate left h units (h positive): = ( + 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 1 3 3 9 ½ ½ 1 1 ½ ½ 1 1 ½ ½ 3 9 ½ ½ 1 Eample B: Graph ( ) = +. 5 3 = + and For, shift parent up. For 5, shift parent to the left 3. 1 = + 3 Eample C: Graph ( ) 7 = + 5 Eample D: Graph ( ) The blue is the stretch b a factor of. The pink shifts the blue graph left 3 and down. The blue is the reflected over the -ais. The pink shifts the blue graph right and up 5. 1 1 S. Stirling Page of 9

Ch 5 Alg 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. 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): Graph = 1 ( ) + 3 1 Graph 1 = blue Shift obvious points to the right and up 3, pink 1 = 1 + 3 Eample 1 (Method ): Graph ( ) Use the form to = a( h) + k Graph the verte and ais of smmetr. Verte ( hk, ) is (,3 ) Ais is =. Find points and graph them and their reflections. - 1 ( ) + 3= 1 1 ( ) + 3= 5 1 Eample : 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) + Your missing a, but ou know a point on the parabola ( 5, ) is (, ) Substitute and a. ( ) ( ) = a (5) 3 + ( ) = a + = a = a The equation is ( ) = 3 + Use plot pattern too! Over a = 1 a = up down ± 1 up 1 down ± up down ± 3 up 9 down 1 S. Stirling Page 7 of 9

Ch 5 Alg Note Sheet Ke 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. Eample : Writing an Equation for a Parabola Write = + 1+ 7 in verte form. Find a: a = from the standard form. b 1 5 Find the -coordinate of the verte: = = = =.5 a () 5 5 5 Find the -coordinate of the verte: f = + 1 + 7 5 = i 5 + 7 = 1.5 5 + 7 = 5.5 hk. The verte is (.5, 5.5) or (, ) Now just substitute into verte form: = a( h) + k ( ) Then simplif = ( + ) = (.5) + ( 5.5).5 5.5 Tr: Write = 3 + 1 + 5 in verte form. Answer: ( ) = 3 + 17 S. Stirling Page of 9

Ch 5 Alg Note Sheet Ke Eample 3: Civil Engineering The photo shows the Verrazano Narrows Bridge in New York, which has the longest span of an suspension bridge in the United States. A suspension cable of the bridge forms a curve that resembles a parabola. The curve can be modeled with the function f( ) =.13( 13), where and are measured in feet. The origin of the function s graph is at the base of one of the two towers that support the cable. How far apart are the towers? How high are the? Start b drawing a diagram. The function is in verte form. Since h = 13 and k =, the verte is at (13, ). The verte is halfwa between the towers, so the distance between the towers is (13 ft) = ft. To the tower s height, for =. f () =.13( 13) 5 The towers are ft apart and about 5 ft high. Eample 3 Part : Civil Engineering Suppose the towers in Eample 3 are ft apart and ft high. Write a function that could model the curve of the suspension cable. Use verte form. Since the distance between the towers is. The verte is at (, ). now = a( ) + or = a( ) Need to a, and ou know a point on the graph (, ). substitute = a( ) and solve for a. = a a =.15 The equation is =.15( ) S. Stirling Page 9 of 9

Ch 5 Alg Note Sheet Ke 5. Factoring Quadratic Epressions 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. Eample 1 Plus: Finding Common Factors Completel factor each epression. Strateg: Find the greatest common factor and un-distribute! a. + 1 = ( + 5 3) b. 9n n 3n 3n = ( ) A quadratic trinomial is an epression in the form a + b + c. You can factor man quadratic trinomials into two binomial factors. Eample : Factoring Factor + + 7 Step 1: Find two factors with a product of ac and a sum b. Step : Rewrite the b term with the factors ou found. + + 7 + 1+ 7+ 7 Step 3: Group and factor out the GCF from each grouping + 1 + 7+ 7 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 + 1 + 7 + 1 ( + 7)( + 1) Eample Quick Check: a. Factor + + + + + + + + ( + )( + ) ac 1 = = b + = Eample Quick Check: b. Factor + + + + + 3 + + + 1 3 ( + )( + ) ac 1 3= 3 = 3 b 1 + = 1 S. Stirling Page 1 of 9

Ch 5 Alg Note Sheet Ke Eample 3: Factoring (The factors do not need to be positive.) Factor 17+ 7 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 17 9+ = 17 3 3 Step : Rewrite the b term as a sum with the factors ou found. 17+ 7 9+ + 7 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 ( )( ) Eample 3 Quick Check: a. Factor + ac 1 = = b + = Eample 3 Quick Check: c. Factor + 11 ac 1 = 3 = b 11 3+ = 11 + + + ( )( ) 3 + + 3 3 + 3 ( )( 3) Use prime factors: 1 3 Eample : More Factoring (Same thing different numbers.) Factor 1 Eample Quick Check: a. Factor 1 3 Step 1: Step : 1 + 3+ 1 Step 3: ( + 3) + ( + 3) Step : ( )( + 3) ac 1 1= 1 3 = 1 b 1 3+ = 1 + + 1 3 + + 1 + ( 1)( + ) ac 1 3= 3 1= 3 b 1 + 1= 1 Use prime factors: 3 S. Stirling Page 11 of 9

Ch 5 Alg Note Sheet Ke Eample Quick Check: b. Factor + 3 1 + 5 1 + 5 ( + 5)( ) ac 1 1= 1 5= 1 b 3 + 5= 3 Eample Quick Check: c. Factor + 5 1+ 5 5 1 + 5 1 ( + 5)( 1) ac 1 5= 5 1 5= 5 b 1+ 5= Eample 5: Factoring (Same stuff, but a bit harder.) Factor 3 1 + 5 Step 1: Find two factors with a product of ac and a sum b. Step : Rewrite the b term as a sum with the factors ou found. 3 1 + 5 3 15+ 1+ 5 Step : Factor out the common Step 3: Group and factor out the GCF from each grouping binomial factor ( 3 3 15) + ( 1+ 5) ( 5) + 1( 5) ( 3 1)( 5) 3 5 + 1 5 ac 3 5= 15 1 15= 15 b 1 1+ 15= 1 Eample 5 Quick Check: a. Factor + + 11 1 ac 1= 3= b 11 + 3= 11 Eample 5 Quick Check: c. Factor + 7 ac = 1 3= 1 b 7 + 3= 7 + 3+ + 1 ( + 3) ( + 3) ( + )( + 3) + Use prime factors: 1 3 + 3+ ( ) + 3( ) ( 3)( ) Use prime factors: 1 3 S. Stirling Page 1 of 9

Ch 5 Alg Note Sheet Ke Eample : Factoring (Same stuff, but a bit harder.) Factor 15 Step 1: Find two factors with a product of ac and a sum b. Step : Rewrite the b term as a sum with the factors ou found. 15 1+ 15 Step 3: Group and factor out the GCF from each grouping 1 + 15 ( 5) ( 5) + 3 ac 15= 1 = b 1 + = Use prime factors: 1 3 5 Step : Factor out the common binomial factor ( 5) + 3( 5) + 3 5 ( )( ) Eample Quick Check: a. Factor + 7 9 ac 9= 1 9= 1 b 7 + 9= 7 Eample Quick Check: b. Factor 3 1 1 ac 3 1= 3 1 = 3 b 1 1 + = 1 + 9 9 ( 1) ( ) ( + 9)( 1) + 9 1 3 1+ 1 ( ) ( ) ( 3 + )( ) 3 + Use prime factors: 3 3 3 A perfect square trinomial is the product ou obtain when ou square a binomial. An eample is + 1+ 5, which can be written as ( + 5). 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 ( ) a+ b = a + ab+ b ( ) a b = a ab+ b Pre-Eample 7a: a. Multipl ( + 3) means ( + 3)( + 3) formula + 3 + 3 + 9 + + 9 a + ab+ b with a = and b = 3 + 3 + 3 = + + 9 Pre-Eample 7b: b. Multipl ( 5) means ( 5)( 5) 1 1+ 5 + 5 formula a ab+ b with a = and b = 5 5 + 5 = + 5 S. Stirling Page 13 of 9

Ch 5 Alg Note Sheet Ke Now tr epanding, multipling, the perfect square trinomials simpl b using the formulas. Pre-Eample 7c: c. Multipl ( 3 + ) formula a + ab+ b with a = 3 and b = 3 + 3 + 9 + + 1 Pre-Eample 7d: d. Multipl ( 7) formula a ab+ b with a = 1 and b = 7 7 + 7 1+ 9 To factor, using these formulas, all ou need to do is identif a and b utilizing the formulas, then work it in reverse. Eample 7: Factor 9 + 9 3 = 9, a= 3 since ( ) since ( 7) = 9, b = 7 check a ab+ b 3 3 7 + 7 = 9 + 9 factored is ( a b) = ( 3 7) Eample 7 Quick Check: a. Factor + 1 + 9 =, a= since ( ) since ( 3) = 9, b = 3 check a + ab+ b + 3 + 3 = factored is ( a+ b) = ( + 3) + 1 + 9 Eample 7 Quick Check: b. Factor 1 + 1 =, a= since ( ) since () 1 = 1, b = 7 check a ab+ b 1 + 1 = 1 + 1 factored is ( a b) = ( 1) Eample 7 Quick Check: c. Factor 5 + 9+ 1 5 = 5, a= 5 since ( ) since ( 9) = 1, b = 9 check a + ab+ b 5 + 5 9 + 9 = factored is ( a+ b) = ( 5 + 9) 5 + 9+ 1 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. Difference of Two Squares a+ b a b = a b ( )( ) Pre-Eample a: a. Multipl ( + 3)( 3) long wa 3+ 3 9 9 a+ b a b with a = and b = 3 formula ( )( ) ( ) ( 3) = 9 Pre-Eample b: b. Multipl ( + 5)( 5) long wa 1+ 1 5 5 a+ b a b with a = and b = 5 formula ( )( ) ( ) ( 5) = 5 S. Stirling Page 1 of 9

Ch 5 Alg Note Sheet Ke Now use the formula in reverse. Eample Quick Check: a. Factor a = and since ( ) =, b = check a b = ( ) ( ) = + factored is ( a b)( a b) + = ( )( ) Eample Quick Check: b. Factor 9 =, a= since ( ) since ( 7) = 9, b = 7 check a b = ( ) ( 7) = 9 + 7 7 factored is ( a b)( a b) + = ( )( ) Eample : 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: π ( 3+ r)( 3 r) Combined factoring: 1. Factor out a GCF. Make highest power term positive.. Test for special case: Perfect Square Trinomial: ( ) ( ) + + = + a ab b a b a ab+ b = a b Difference of two squares: a b = a+ b a b ( )( ) 3. Factor b ac and b method. Factor out a GCF. 15 5 5( 3 + ) A special case. Look for perfect squares. + 9 Is a=, b = 7? Check a ab+ b = ( a b) ( 7) ( 7) Factor ( a b) ( 7) Factor b ac and b method. = a + b + c 1 17 + 3 Set up: Find #s: ( 1 15) + ( + 3) ac = 3 15i 5 3 + 1 3 b = 17 15 + ( 5 1)( 3) + Yes it checks! S. Stirling Page 15 of 9

Ch 5 Alg 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, = or ± for all real numbers. When simplifing radicals, three rules appl 1) You ma not have factors that are perfect squares under the radical sign. The cure? Factor the radicand and use the multiplication propert. Simplif 1 = 3= 3 = 3 75 = 5 3 = 5 3 = 5 3 7 = 9 = 9 = 9 = 3 = Square Root Properties Multiplication Propert of Square Roots For an numbers a and b, ab = a b Division Propert of Square Roots For an numbers a and b >, a a = b b ) You ma not have fractions under the radical sign. The cure? Use the division propert and simplif further, if necessar. Simplif 9 = 9 = 5 3 9 = 5 5 5 9 = i 7 = 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 3 = 3 3 3 = = 5 3 5 3 15 15 = = = 3 3 9 3 5 = 7 5 7 35 35 = = 7 7 9 7 S. Stirling Page 1 of 9

Ch 5 Alg Note Sheet Ke 5.5 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 ab =, then either a = or b = or both =. You will use this propert to help ou solve quadratics. Eample 1: Solving b Factoring Eample 1 plus: Solving b Factoring Solve 11= 15 Solve 1 = 11= 15 11+ 15= get equation = + 5+ 15 = factor ( ) ( ) ( 5)( 3) 3 + 5 3 = = now use Zero Product Prop. 5= or 3= so solve for = 5 or = 3 1 = 1 = get equation = 1 = factor & use Zero Product Prop. ( ) = or 1= so solve for s = or = 1 Caution!! Don t tr alternate procedures!! 1 = 1 = divide both sides b = 1 Oops! You lost a root! NO GOOD Eample 1 More Practice Solve b Factoring: a. Solve + 7+ 1= + 3+ + 1= + 3 + + 3 = ( + )( + 3) = + = or + 3= = or = 3 Eample 1 More Practice Solve b Factoring: b. Solve + 7= 1 + 7 1= + 9+ 1= + 9 + + 9 = ( )( + 9) = = or + 9= = or = 9 Eample 1 More Practice Solve b Factoring: c. Solve = = + ( ) = + 3 factor out GCF and divide! = + 3+ 1 3 ( ) ( ) = + 3 + 1 + 3 ( )( ) = 1 + 3 1= or + 3= = 1 or = 3 S. Stirling Page 17 of 9

Ch 5 Alg Note Sheet Ke 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 = 5. B substitution, ou know the solution is +5 or 5, because 5 = 5 and ( 5) = 5 Solve = 5. 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 = 5 = ± 5 =± 5 Definition: = or ± for all real numbers. Note: 5 = 5 = 5 = 5 and 5 = ( 5) = 5 = 5 Eample : Solving b Finding Square Roots Hint: Work in the reverse order of operations because ou re solving! Solve 5 1 = 5 1 = Would be hard to factor, no? 5 = 1 get the quadratic term alone! 5 1 = 5 5 isolate the completel = 3 now take the square root, both sides =± 3 need the ±!! =± simplif Eample : Solving b Finding Square Roots Solve 3 = 3 = 3 = 3 3 isolate the completel = now take the square root, both sides = ± need the ± =± simplif (take perfect squares out) Eample More Practice Solve b Square Root: Alternate Method, Factoring: a. Solve 5= a. Solve 5= 5= Could factor, but = 5 get the quadratic term alone! 5 = isolate the completel 5 = now take the square root, both sides 5 =± need the ±!! 5 =± simplif 5 = Tr factoring! Wh not? + 5 5 = difference of two squares ( )( ) + 5= or 5= = 5 or = 5 5 5 = or = 5 = ± See wh ou need to use the ± when using the square root method? You would totall lose one of our roots! S. Stirling Page 1 of 9

Ch 5 Alg 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 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]: [Y=] enter left hand part Y1 enter left hand part Y [TBLSET] start at with ΔTbl = 1 [TABLE] Look for when Y1 = Y. You can tweek the ΔTbl to a smaller number until the Y values get reall close. Eample alt: Solving b Tables Solve 11= 15 Y1 = 11 Y = 15 [TBLSET] TblStart = with ΔTbl = 1 Look for when Y1 = Y. Happens at = 3, but ou know there is probabl another solution, so change ΔTbl =.1. Look again At =.5, there is another solution. With quadratics, there is a maimum of solutions. Eample : Solving b Tables Solve 5+ = Calculator Solutions [GRAPH]: [Y=] enter left hand part Y1 enter left hand part Y [GRAPH] [ZOOM] choose a window if necessar. Look for both intersections if possible nd [CALC] 5: intersect Answer calculator s questions. Make sure ou both answers if there are two points of intersections. Eample 5 alt: Solving b Graphing Solve = Y1 = Y = [GRAPH] Tr [ZOOM] : Look for when Y1 = Y, the intersection points. nd [CALC] 5: intersect Guess at the left intersection. For the left intersection: X = 1.3 Y = nd [CALC] 5: intersect Guess at the right intersection. For the right intersection: X = 3.3 Y = With quadratics, there is a maimum of solutions. Y1 = 5+ Y = [TBLSET] TblStart = with ΔTbl = 1 Look for when Y1 = Y. Notice that the sign changes between = and = 1, then again between = and = 5, so must be near zero between those values. So change ΔTbl =.1. Look again near the -values ou found before At.1 and.5 are the solutions. S. Stirling Page 19 of 9

Ch 5 Alg Note Sheet Ke 5. Comple Numbers Introduction: The imaginar number i is defined as the number whose Solve = 5. square is 1. So i = 1 and i = 1. Tr to solve =± 5, 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. 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 1 QC: a. Simplif 7. i 17 i 7 Eample 1 QC: b. Simplif 1. 13 i i i 3 Eample 1 QC: c. Simplif 3. 13 i i Note: Now that a is defined for a <, eplain wh a ( a) a = a and ( a). = a Imaginar numbers and real numbers together make up the set of comple numbers. Eample : Simplifing Imaginar Numbers Write the comple number 9+ in the form a + bi. 9+ Now we can epand our number tpes 3i + Simplif the radical epression. + 3i Write in the form a ± bi. Eample Quick Check: Write the comple number 1 + 7 in the form a + bi. 19 ii + 7 7+ 3i S. Stirling Page of 9

Ch 5 Alg Note Sheet Ke OPTIONAL: The absolute value of a comple number is a+ bi = a + b. Eample 3: Finding Absolute Value a. Find 5i. 5i = + 5 = 5 b. Find 3 i. ( ) 3 i = 3 + 9+ 1 = 5 = 5 Eample 3 QC: a. Find i. ( ) i = + 3 + 1 = 5 = i 13 = 13 Eample 3 QC: b. Find + 5i. ( ) + 5i = + 5 + 5 = 9 You can appl the operations of real numbers to comple numbers. Remember, if the sum of two comple numbers is, then each number is the opposite, or additive inverse, of the other. Eample : Additive Inverse of a Comple Number Find the additive inverse of + 5i + 5i Find the opposite. ( ) 5i Simplif. Eample QC: a. Find the additive inverse of ( 5i) 5i 5i Eample QC: b. Find the additive inverse of 3i ( 3i ) + 3i Eample QC: c. Find the additive inverse of a+ bi ( a bi) + a bi To add or subtract comple numbers, combine the real parts and the imaginar parts separatel. Eample 5: Adding Comple Numbers Simplif the epression + 5i 5+ 7i + + i 5 + 7i+ i 3+ 13i Eample 5 QC: a. Simplif + 3i + i + 3i i + 3i i i Eample 5 QC: b. Simplif ( ) 7 3+ i 7 3 i i Eample 5 QC: c. Simplif ( i) + 3i i+ 3i 3i For two imaginar numbers bi and ci, ( )( ) Eample : Multipling Comple Numbers 5i i a. Find ( )( ) i and since i = 1 i 1 = b. Find ( + 3i)( 3+ 5i) + 1i 9i+ 15i and since i = 1 15+ 1i 9i 1+ i bi ci = bci = bci 1= bc. Eample QC: a. Simplif ( 1 )( 7 ) Eample QC: b. Simplif ( 5i)( 3i) 1i i+ 15i 15 1i i 9 3i i i = i = Eample QC: c. Simplif 9i + 3i ( )( ) 1 + 1i 3i 7i 1 + 7 + 1i 3i 3 i S. Stirling Page 1 of 9

Ch 5 Alg Note Sheet Ke Some quadratic equations have solutions that are comple numbers. Eample 7: Finding Comple Solutions Solve + 1 = 1 = Isolate the. = 5 Now take the square root of both sides. =± 5 Remember the ± (two roots). =± 5i Check: = 5i Show it works! ( 5i ) + 1 = i 5i + 1 = 1 + 1 = Check: = 5i Show it works! ( i) 5 + 1 = i 5i + 1 = 1 + 1 = Eample 7 Quick Check: a. Simplif Solve 3 + = 3 = 3 3 = 1 = ± 1 = ± i Eample 7 Quick Check: b. Simplif Solve 5 15 = = 5 5 = 3 5 15 = ± 3 =± i 3 Eample 7 Quick Check: c. Simplif Solve + = = 1 = 1 1 = ± i =± i S. Stirling Page of 9

Ch 5 Alg Note Sheet Ke 5.7 Completing the Square Pre-Eample: Perfect Squares ( ) + 3 = + + 9 Geometricall: + 3 What would ou need to complete the square? Given: + 1 Geometricall: + 5 What would ou need to complete the square? Given: + 1 Geometricall: + 3 + 3 If ou have +, ou need the +9 to complete the square. 3 9 ( ) + + 9= + 3 5 + 5 Add 5 to complete the square. 5? ( ) + 1 + 5 = + 5 + Add 3 to complete the square.? ( ) + 1 + 3 = + The process of completing the square can be used to solve quadratic equations as well as used to rewrite equations into verte form. Look for patterns above. Take half of the coefficient of the -term, then add the square of it to create a trinomial. Eample 1: Solving a Perfect Square Trinomial Equation Solve + 1+ 5 = 3 + 1+ 5 = 3 The left side is a perfect square. ( + 5) = 3 Rewrite. + 5=± Now take the square root of both sides. Solve the resulting equations: + 5= and + 5= = 1 and = 11 Eample 1: QC Solve 1+ 9 = 1 1+ 9 = 1 ( 7) = 1 7=± 9 7= 9 and 7= 9 = 1 and = Take half of b. Square it and add it on. b b + + S. Stirling Page 3 of 9