Quadratic Functions and Parabolas


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1 MATH 11 Quadratic Functions and Parabolas A quadratic function has the form Dr. Neal, Fall 2008 f () = a 2 + b + c where a 0. The graph of the function is a parabola that opens upward if a > 0, and opens downward for a < 0. For a > 0 For a < 0 Given such a parabolic function, we will (a) Find the verte. (b) State the intercept. (c) Find the intercepts. (d) Graph the function. (e) Solve the equation f () = d. (f) Use the solution from (e) to solve the inequalities d, < d, d, or > d. As an eample throughout, we will use the function = The Verte The verte (, ) is the point at which the graph changes from decreasing to increasing. The coordinate of the verte is given b = b 2 a. We then substitute this value into the function to find the coordinate of the verte.
2 For = , the coordinate of the verte is = b 2 a = ( 12) = 12 2(3) = 2. The coordinate is then 3(2) 2 12(2) + =. So the verte is the point (2, ). The intercept The intercept of the function = a 2 + b + c is alwas given b the constant term c. That is, when = 0, then = c. For = , the intercept is. The intercepts To find the intercepts of the quadratic = a 2 + b + c, we set = 0 and solve for. That is, we must solve the quadratic equation a 2 + b + c = 0. If the quadratic does not factor, then we can use the quadratic formula = b ± b2 4ac 2a. Note: The epression under the radical b 2 4ac is called the discriminant which determines how man solutions there are to the equation a 2 + b + c = 0. (i) If b 2 4ac > 0, then there are two intercepts. (ii) If b 2 4ac = 0, then there is one intercept. (iii) If b 2 4ac < 0, then there are no solutions to a 2 + b + c = 0 and no intercepts. (iv) If b 2 4ac is a perfect square, then the quadratic a 2 + b + c will factor. b 2 4ac > 0 b 2 4ac = 0 b 2 4ac < 0 Two intercepts One intercept No intercepts
3 Consider = The discriminant is b 2 4ac = ( 1) 2 4(2)( 10) = = 81, which is positive and a perfect square. So there are two intercepts that can be found b factoring: = (2 5)( + 2) = = 0 or + 2 = 0. Thus, = 5 2 and = 2 are the intercepts. We now will find the intercepts of = b using the quadratic formula: = b ± b2 4ac 2a ( 12) ± ( 12) 2 4(3)() 2(3) 12 ± 72 = = 12 ± 3 2 = 2 ± 2 = 2 ± 2 The approimate decimal values are and Graphing Once we have found the verte, the intercept, and the intercepts, we can graph the parabola b plotting these points. For = , the verte is (2, ), the intercept is and the intercepts are 2 ± 2, which are about and = (2, ) The graph of =
4 Solving Other Equations Given the quadratic function f () = a 2 + b + c, we also can solve for the that makes f () = d. That is, we can solve the equation a 2 + b + c = d. To do so, alwas set the equation equal to 0 b subtracting the d term, and then either factor or use the quadratic formula to solve the resulting equation. Consider again the function f () = When does = 21? Here we must solve the equation = 21. B subtracting 21 we obtain = 0 If we divide b 3, we obtain = 0 which factors as ( 5)( +1) = 0. So the solutions are = 1 and = 5. ( 1, 21) (5, 21) 1 5 Solving Inequalities Using the solution from above, solve the inequalities 21, < 21, > 21, and > 21. From the graph, we see that 21 when 1 5 i.e., for in [ 1, 5] < 21 when 1 < < 5 i.e., for in ( 1, 5) > 21 when < 1 or > 5 i.e., for in (, 1) (5, ) 21 when 1 or 5 i.e., for in (, 1] [5, )
5 Eercises 1. Let = (a) Find the verte. (b) Find the intercept. (c) Find the intercepts. (d) Graph. Show the intercepts and verte. (e) Solve the equation = 2 b factoring. (f) For which is 2? For which is > 2? 2. Let = (a) Find the verte. (b) Find the intercept. (c) Find the intercepts. (d) Graph. Show the intercepts and verte. (e) Solve the equation = b factoring. (f) For which is? For which is <?
6 Solutions Dr. Neal, Fall (a) The coordinate of the verte is = b 2 a = ( 4) 2 3 = 4 = 2. The coordinate is 3 3(2 / 3) 2 4(2 / 3) = So the verte is 2 3, (b) When = 0, the intercept is =. (c) To find the intercepts, we will solve = 0 with the quadratic formula. For our equation, a = 3, b = 4, and c = ; hence, the solutions are = b ± b2 4ac 2a = ( 4) ± ( 4)2 4(3)( ) 2(3) = 4 ± 88 = 4 ± 2 22 = 4 ± 2 22 = 2 3 ± 22 3 The approimate numerical solutions are 2.23 and (2/3, 22/3) (e) Now solve = 2, which is equivalent to = 0. We can factor as (3 + 2)( 2) = 0. So (3 + 2) = 0 or ( 2) = 0. The solutions to = 2 are then = 2 /3 and = 2. (f) When 2/3 2, then 2. When < 2/3 or > 2, then > 2. 2/3 = 2
7 2. = (a) The coordinate of the verte is = b 2a = 12 2( 2) Dr. Neal, Fall 2008 = 3. The coordinate is 2(3) 2 +12(3) 10 = 8 So the verte is (3, 8). (b) When = 0, we obtain = 10 for the intercept. (c) To find the intercepts, we must solve = 0. Dividing b 2, we instead can solve = 0, which factors as ( 5)( 1) = 0, so = 1 and = 5 are the intercepts. Or, using the quadratic formula on the original equation = 0, we obtain = 12 ± = 12 ± ( 2)( 10) 2( 2) = 12 ± 8 4 = 3 ± 2 (3, 8) = 1 5 (d) Now we must solve =, which is equivalent to = 0. Dividing b 2, we have = 0, which factors as ( 4)( 2) = 0. Thus, = 4 and = 2 are the solutions to =. (e) We have when 2 4. And < when < 2 or > 4.
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