Chapter 10 quadratic equations and functions
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Chapter 10 Quadratic Equations and Functions. Section 5 Graphing Quadratic Functions Using Properties. 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

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Chapter 10 Quadratic Equations and Functions

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Chapter 10 quadratic equations and functions

Chapter 10

Quadratic Equations and Functions

Section 5

Graphing Quadratic Functions Using Properties


Section 10 5 objectives

Section 10.5 Objectives

1Graph Quadratic Functions of the Formf (x) = ax2 + bx + c

2Find the Maximum or Minimum Value of a Quadratic Function

3Model and Solve Optimization Problems Involving Quadratic Functions


The vertex of a parabola

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 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 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.


  • Graphing using properties1

    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.


    Graphing using properties2

    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


    Maximum and minimum values

    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


    Maximum and minimum values1

    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.


    Applications involving maximization

    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.


    Applications involving maximization1

    b

    a

    Applications Involving Maximization

    Example continued:

    R(x) = – 0.02x2 + 24x

    Step 4: Solve

    The maximum revenue is

    Continued.


    Applications involving maximization2

    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.


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