Chapter 10 Quadratic Equations and Functions

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

2. Section 10.5 Objectives 1 Graph Quadratic Functions of the Formf (x) = ax2 + bx + c 2 Find the Maximum or Minimum Value of a Quadratic Function 3 Model and Solve Optimization Problems Involving Quadratic Functions

3. The Vertex of a Parabola The Vertex of a Parabola Any quadratic function f(x) = ax2 + bx + c, a  0, will have vertex The x-intercepts, if there are any, are found by solving the quadratic equation f(x) = ax2 + bx + c = 0.

4. The x-Intercepts of a Parabola • The x-Intercepts of the Graph of a Quadratic Function • If the discriminant b2 – 4ac > 0, the graph of f(x) = ax2 + bx + c has two different x-intercepts. The graph will cross the x-axis at the solutions to the equation ax2 + bx + c = 0. • If the discriminant b2 – 4ac = 0, the graph of f(x) = ax2 + bx + c has one x-intercept. The graph will touch the x-axis at the solution to the equation ax2 + bx + c = 0. • If the discriminant b2 – 4ac < 0, the graph of f(x) = ax2 + bx + c has no x-intercepts. The graph will not cross or touch the x-axis.

5. 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, b2 – 4ac. • If b2 – 4ac > 0, then the parabola has two x-intercepts, • which are found by solving f(x) = 0. • If b2 – 4ac = 0, the vertex is the x-intercept. • If b2 – 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.

6. b a c Graphing Using Properties Example: Graph f(x) = –2x2 – 8x + 4 using its properties. f(x) = –2x2 – 8x + 4 The x-coordinate of the vertex is The y-coordinate of the vertex is Continued.

7. y vertex (– 2, 12) 16 y-intercept (0, 4) 12 8 4 x 8 8 6 4 2 2 4 6 8 12 x-intercepts 16 Graphing Using Properties Example continued: f(x) = –2x2 – 8x + 4 The vertex is (–2, 12). The axis of symmetry is x = – 2. The y-intercept is f(0) = –2(0)2 – 8(0) + 4 = 4 The x-intercepts occur where f(x) = 0. –2x2 – 8x + 4 = 0 Use the quadratic formula to determine the x-intercepts. x 4.4 x 0.4

8. 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. Maximum and Minimum Values The graph of a quadratic function has a vertex at Opens up a > 0 Maximum Opens down a < 0 Minimum

9. b a c Maximum and Minimum Values Example: Determine whether the quadratic function f(x) = –3x2 + 12x – 1 has a maximum or minimum value. Find the value. f(x) = –3x2 + 12x – 1 Because a < 0, the graph will open down and will have a maximum. The maximum of f is 11 and occurs at x = 2.

10. b a Applications Involving Maximization Example: The revenue received by a ski resort selling x daily ski lift passes is given by the function R(x) = – 0.02x2 + 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. 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) = – 0.02x2 + 24x. Continued.

11. b a Applications Involving Maximization Example continued: R(x) = – 0.02x2 + 24x Step 4: Solve The maximum revenue is Continued.

12. Applications Involving Maximization Example continued: Step 5: Check R(x) = – 0.02x2 + 24x R(600) = – 0.02(600)2 + 24(600) = – 0.02(360000) + 144000 = – 7200 + 14400 = 7200  Step 6: Answer The ski resort needs to sell 600 daily lift tickets to earn a maximum revenue of $7200 per day.