Chapter 2Functions and GraphsFunctions and GraphsSection 3Quadratic FunctionsQQuadratic FunctionsIf a, b, c are real numbers with a not equal to zero, then ,,q,the functionis a quadratic function and its graph is a parabola.2()fxaxbxc=++2Vertex Form of the Quadratic FunctionIt is convenient to convert the general form of a dti tiquadratic equation to what is known as the vertex form:2()fxaxbxc=++2() ( )fxaxh k=−+3Completing the Square to Find the Vertex of a Quadratic FunctionThe example below illustrates the procedure: ppConsiderComplete the square to find the vertex.2() 3 6 1fxxx=−+−4Completing the Square to Find the Vertex of a Quadratic FunctionThe example below illustrates the procedure: ppConsiderComplete the square to find the vertex.2() 3 6 1fxxx=− + −Solution: Factor the coefficient of x2out of the first two terms:5f (x) = –3(x2–2x) –1Completing the square(continued) Add 1 to complete the square inside the parentheses. Because fth3 t id th th h t ll dd d3Th i (1 2)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+ 26The vertex is (1, 2)  The quadratic function opens down since the coefficient of the x2term is –3, which is negative. Intercepts of a Quadratic FunctionFind thexandyintercepts ofFind the xand yintercepts of2() 3 6 1fxxx=− + −7Intercepts of a Quadratic FunctionFind thexandyintercepts ofFind the xand yintercepts of x intercepts: Set f (x) = 0: Use the quadratic formula:2() 3 6 1fxxx=−+−203 61xx=−+−8qx = =242bb aca−± −2664(3)(1)6240.184,1.8162( 3) 6−± − − −−±=≈−−Intercepts of a Quadratic Functionyintercept:Letx=0Ifx=0 theny=–1so(0–1)yintercept:Let x 0. If x 0, then y 1, so (0, 1) is the y intercept.2() 3 6 1fxxx=− + −1)0(6)0(3)0(2−+−=f9)()()(fGeneralization2() ( )fxaxh k=−+For 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)=kis the minimum ifa> 0 otherwise the maximum10f (h) kis the minimum if a> 0, otherwise the maximum  Domain = set of all real numbers  Range: if a < 0. If a > 0, the range is {}yy k≤{}yy k≥Generalization11Solving Quadratic InequalitiesSolve the quadratic inequality–x2+5x+3>0Solve the quadratic inequality x+ 5x+ 3 >0 .12Solving Quadratic InequalitiesSolve the quadratic inequality–x2+5x+3>0Solve the quadratic inequality x+ 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 the13solution set for the quadratic inequality is – 0.5414 <x < 5.5414 or[– 0.5414, 5.5414 ] .Application of Quadratic FunctionsAGihhdf h20A 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?14maximum yield?SolutionSolution:Solution: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: 15 Yield = ( 300 –2(10) × ( 20 + 2) = 280 x 22 = 6160Solution(continued)Letxrepresent the number of additionaltreesLet xrepresent the number of additionaltrees. 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 6000xx−++16Solution(continued)Complete the square to find the vertex of the parabola: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 250xx−−+++17Solution(continued)Y(x)=210(5)6250x−−+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 fth dtiftiiththlfi10()18of the quadratic function since the the value of ais –10. The function opens downward, so the vertex must be the maximum. Break-Even AnalysisThe financial department of a company that produces digital cameras has the revenue and cost functions for xmillion 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 19ppR(x) = C(x). Find the break-even points algebraically to the nearest thousand cameras. Solution 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 20breaks even atx = 2.490 and 12.530 million cameras.Solution to Break-Even Problem(continued)If we graph the cost and revenue functions on a gpgraphing utility, we obtain the following graphs, showing the two intersection