 # Chapter 10 Quadratic Equations and Functions Section 5 Graphing Quadratic Functions Using Properties.

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Chapter 10 Quadratic Equations and Functions Section 5 Graphing Quadratic Functions Using Properties

Sullivan, III & Struve, Elementary and Intermediate Algebra10.5 - 2 Copyright © 2010 Pearson Education, Inc. Section 10.5 Objectives 1Graph Quadratic Functions of the Form f (x) = ax 2 + bx + c 2Find the Maximum or Minimum Value of a Quadratic Function 3Model and Solve Optimization Problems Involving Quadratic Functions

Sullivan, III & Struve, Elementary and Intermediate Algebra10.5 - 3 Copyright © 2010 Pearson Education, Inc. The Vertex of a Parabola Any quadratic function f(x) = ax 2 + bx + c, a  0, will have vertex The x-intercepts, if there are any, are found by solving the quadratic equation f(x) = ax 2 + bx + c = 0.

Sullivan, III & Struve, Elementary and Intermediate Algebra10.5 - 4 Copyright © 2010 Pearson Education, Inc. The x-Intercepts of a Parabola The x-Intercepts of the Graph of a Quadratic Function 1.If the discriminant b 2 – 4ac > 0, the graph of f(x) = ax 2 + bx + c has two different x-intercepts. The graph will cross the x-axis at the solutions to the equation ax 2 + bx + c = 0. 2.If the discriminant b 2 – 4ac = 0, the graph of f(x) = ax 2 + bx + c has one x-intercept. The graph will touch the x-axis at the solution to the equation ax 2 + bx + c = 0. 3.If the discriminant b 2 – 4ac < 0, the graph of f(x) = ax 2 + bx + c has no x-intercepts. The graph will not cross or touch the x-axis.

Sullivan, III & Struve, Elementary and Intermediate Algebra10.5 - 5 Copyright © 2010 Pearson Education, Inc. Graphing Using Properties Graphing a Quadratic Function Using Its Properties Step 1: Determine whether the parabola opens up or down. Step 2: Determine the vertex and axis of symmetry. Step 3: Determine the y-intercept, f(0). Step 4: Determine the discriminant, b 2 – 4ac. If b 2 – 4ac > 0, then the parabola has two x-intercepts, which are found by solving f(x) = 0. If b 2 – 4ac = 0, the vertex is the x-intercept. If b 2 – 4ac < 0, there are no x-intercepts. Step 5: Plot the points. Use the axis of symmetry to find an additional point. Draw the graph of the quadratic function.

Sullivan, III & Struve, Elementary and Intermediate Algebra10.5 - 6 Copyright © 2010 Pearson Education, Inc. Graphing Using Properties Example: Graph f(x) = –2x 2 – 8x + 4 using its properties. Continued. f(x) = –2x 2 – 8x + 4 abc The x-coordinate of the vertex is The y-coordinate of the vertex is

Sullivan, III & Struve, Elementary and Intermediate Algebra10.5 - 7 Copyright © 2010 Pearson Education, Inc. Graphing Using Properties Example continued: The x-intercepts occur where f(x) = 0. –2x 2 – 8x + 4 = 0 Use the quadratic formula to determine the x-intercepts. x   4.4 x  0.4 x y 22 44 6 6 8 8 24 6 8 88  12  16 4 8 12 16 f(x) = –2x 2 – 8x + 4 The y-intercept is f(0) = –2(0) 2 – 8(0) + 4 = 4 The vertex is (–2, 12). y-intercept (0, 4) x-intercepts vertex (– 2, 12) The axis of symmetry is x = – 2.

Sullivan, III & Struve, Elementary and Intermediate Algebra10.5 - 8 Copyright © 2010 Pearson Education, Inc. Maximum and Minimum Values The graph of a quadratic function has a vertex at Maximum Minimum Opens up a > 0 Opens down a < 0 The vertex will be the highest point on the graph if a < 0 and will be the maximum value of f. The vertex will be the lowest point on the graph if a > 0 and will be the minimum value of f.

Sullivan, III & Struve, Elementary and Intermediate Algebra10.5 - 9 Copyright © 2010 Pearson Education, Inc. Maximum and Minimum Values Example: Determine whether the quadratic function f(x) = –3x 2 + 12x – 1 has a maximum or minimum value. Find the value. Because a < 0, the graph will open down and will have a maximum. f(x) = –3x 2 + 12x – 1 abc The maximum of f is 11 and occurs at x = 2.

Sullivan, III & Struve, Elementary and Intermediate Algebra10.5 - 10 Copyright © 2010 Pearson Education, Inc. Applications Involving Maximization Example: The revenue received by a ski resort selling x daily ski lift passes is given by the function R(x) = – 0.02x 2 + 24x. How many passes must be sold to maximize the daily revenue? Step 1: Identify We are trying to determine the number of passes that must be sold to maximize the daily revenue. Continued. Step 2: Name We are told that x represents the number of daily lift passes. Step 3: Translate We need to find the maximum of R(x) = – 0.02x 2 + 24x. ab

Sullivan, III & Struve, Elementary and Intermediate Algebra10.5 - 11 Copyright © 2010 Pearson Education, Inc. Continued. Step 4: Solve Example continued: Applications Involving Maximization R(x) = – 0.02x 2 + 24x ab The maximum revenue is

Sullivan, III & Struve, Elementary and Intermediate Algebra10.5 - 12 Copyright © 2010 Pearson Education, Inc. Example continued: Step 5: Check Applications Involving Maximization R(x) = – 0.02x 2 + 24x R(600) = – 0.02(600) 2 + 24(600) = – 0.02(360000) + 144000 = – 7200 + 14400 = 7200 The ski resort needs to sell 600 daily lift tickets to earn a maximum revenue of \$7200 per day. Step 6: Answer