Chapter 10 Quadratic Equations and Functions

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# Chapter 10 Quadratic Equations and Functions - PowerPoint PPT Presentation

Chapter 10 Quadratic Equations and Functions. Section 5 Graphing Quadratic Functions Using Properties. Section 10.5 Objectives. 1 Graph Quadratic Functions of the Form f ( x ) = ax 2 + bx + c 2 Find the Maximum or Minimum Value of a Quadratic Function

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Presentation Transcript
Chapter 10

Section 5

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

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.

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.
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.

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.

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

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

b

a

c

Maximum and Minimum Values

Example:

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.

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.

b

a

Applications Involving Maximization

Example continued:

R(x) = – 0.02x2 + 24x

Step 4: Solve

The maximum revenue is

Continued.

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