Chapter 2 Functions and GraphsQuadratic FunctionsVertex Form of the Quadratic FunctionCompleting the Square to Find the Vertex of a Quadratic FunctionSlide 5Completing the square (continued)Intercepts of a Quadratic FunctionSlide 8Slide 9GeneralizationSlide 11Solving Quadratic InequalitiesSlide 13Application of Quadratic FunctionsSolutionSolution (continued)Slide 17Slide 18Break-Even AnalysisSolution to Break-Even ProblemSolution to Break-Even Problem (continued)Chapter 2Functions and GraphsSection 3Quadratic Functions2Quadratic FunctionsIf a, b, c are real numbers with a not equal to zero, then the function is a quadratic function and its graph is a parabola. 2( )f x ax bx c= + +3Vertex Form of the Quadratic Function It is convenient to convert the general form of a quadratic equation to what is known as the vertex form:2( )f x ax bx c= + +2( ) ( )f x a x h k= - +4Completing the Square to Find the Vertex of a Quadratic FunctionThe example below illustrates the procedure: ConsiderComplete the square to find the vertex.2( ) 3 6 1f x x x=- + -5Completing the Square to Find the Vertex of a Quadratic FunctionThe example below illustrates the procedure: ConsiderComplete the square to find the vertex.2( ) 3 6 1f x x x=- + -Solution: Factor the coefficient of x2 out of the first two terms: f (x) = –3(x2 – 2x) –16Completing the square(continued)The vertex is (1, 2) The quadratic function opens down since the coefficient of the x2 term is –3, which is negative. Add 1 to complete the square inside the parentheses. Because of the –3 outside the parentheses, we have actually added –3, so we must add +3 to the outside.f (x) = –3(x2 – 2x +1) –1+3f (x) = –3(x – 1)2 + 27Intercepts of a Quadratic FunctionFind the x and y intercepts of2( ) 3 6 1f x x x=- + -8Intercepts of a Quadratic FunctionFind the x and y intercepts of x intercepts: Set f (x) = 0: Use the quadratic formula: x = =2( ) 3 6 1f x x x=- + -20 3 6 1x x=- + -242b b aca- � -26 6 4( 3)( 1)6 240.184,1.8162( 3) 6- � - - -- �= �- -9Intercepts of a Quadratic Functiony intercept: Let x = 0. If x = 0, then y = –1, so (0, –1) is the y intercept. 2( ) 3 6 1f x x x=- + -1)0(6)0(3)0(2f10GeneralizationIf a  0, then the graph of f is a parabola.•If a > 0, the graph opens upward.•If a < 0, the graph opens downward. Vertex is (h , k) Axis of symmetry: x = h f (h) = k is the minimum if a > 0, otherwise the maximum Domain = set of all real numbers Range: if a < 0. If a > 0, the range is { }y y k�2( ) ( )f x a x h k= - +For{ }y y k�11Generalization12Solving Quadratic InequalitiesSolve the quadratic inequality –x2 + 5x + 3 > 0 .13Solving Quadratic InequalitiesSolve the quadratic inequality –x2 + 5x + 3 > 0 .Answer: This inequality holds for those values of x for which the graph of f (x) is at or above the x axis. Thishappens for x between the two x intercepts, including the intercepts. Thus, the solution set for the quadratic inequality is – 0.5414 < x < 5.5414 or[– 0.5414, 5.5414 ] .14Application of Quadratic FunctionsA Macon, Georgia, peach orchard farmer now has 20 trees per acre. Each tree produces, on the average, 300 peaches. For each additional tree that the farmer plants, the number of peaches per tree is reduced by 10. How many more trees should the farmer plant to achieve the maximum yield of peaches? What is the maximum yield?15SolutionSolution: Yield = (number of peaches per tree)  (number of trees)Yield = 300  20 = 6000 (currently) Plant one more tree: Yield = ( 300 – 1(10))  ( 20 + 1) = 290  21 = 6090 peaches. Plant two more trees: Yield = ( 300 – 2(10)  ( 20 + 2) = 280 x 22 = 616016Solution(continued)Let x represent the number of additional trees. Then Yield =( 300 – 10x) (20 + x)= To find the maximum yield, note that the Y (x) function is a quadratic function opening downward. Hence, the vertex of the function will be the maximum value of the yield. Graph is below, with the y value in thousands.210 100 6000x x- + +17Solution(continued)Complete the square to find the vertex of the parabola: Y (x) = We have to add 250 on the outside since we multiplied –10 by 25 = –250.210( 10 25) 6000 250x x- - + + +18Solution(continued)Y (x) = Thus, the vertex of the quadratic function is (5 , 6250) . So, the farmer should plant 5 additional trees and obtain a yield of 6250 peaches. We know this yield is the maximum of the quadratic function since the the value of a is –10. The function opens downward, so the vertex must be the maximum. 210( 5) 6250x- - +19Break-Even AnalysisThe financial department of a company that produces digital cameras has the revenue and cost functions for x million cameras are as follows:R(x) = x(94.8 – 5x)C(x) = 156 + 19.7x. Both have domain 1 < x < 15Break-even points are the production levels at which R(x) = C(x). Find the break-even points algebraically to the nearest thousand cameras.20Solution to Break-Even ProblemSet R(x) equal to C(x):x(94.8 – 5x) = 156 + 19.7x–5x2 + 75.1x – 156 = 0275.1 75.1 4( 6)( 156)2( 5)x- � - - -=-x = 2.490 or 12.530The company breaks even atx = 2.490 and 12.530 million cameras.21Solution to Break-Even Problem(continued)If we graph the cost and revenue functions on a graphing utility, we obtain the following graphs, showing the two intersection