- By
**nira** - Follow User

- 150 Views
- Uploaded on

Quadratic Functions. Graphs of Quadratic Functions.

Download Presentation
## Quadratic Functions

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

The graph of any quadratic function is called a parabola. Parabolas are shaped like cups, as shown in the graph below. If the coefficient of x2 is positive, the parabola opens upward; otherwise, the parabola opens downward. The vertex (or turning point) is the minimum or maximum point.

The Standard Form of a Quadratic Function

The quadratic function

f (x)=a(x - h)2+ k, a 0

is in standard form. The graph of f isa parabola whose vertex is the point (h, k). The parabola is symmetric to the line x = h. If a >0, the parabola opens upward; if a< 0, the parabola opens downward.

Graphing Parabolas With Equations in Standard Form

To graph f (x)=a(x - h)2+ k:

- Determine whether the parabola opens upward or downward. If a >0, it opens upward. If a < 0, it opens downward.
- Determine the vertex of the parabola. The vertex is (h, k).
- Find any x-intercepts by replacing f (x) with 0. Solve the resulting quadratic equation for x.
- Find the y-intercept by replacing x with zero.
- Plot the intercepts and vertex. Connect these points with a smooth curve that is shaped like a cup.

SolutionWe can graph this function by following the steps in the preceding box. We begin by identifying values for a, h, and k.

Standard formf (x)=a(x - h)2+ k

a = -2 h=3 k=8

Given equationf (x)=-2(x- 3)2+ 8

Text ExampleGraph the quadratic function f (x)=-2(x- 3)2+ 8.

Step 1 Determine how the parabola opens. Note that a, the coefficient of x2, is -2. Thus, a < 0; this negative value tells us that the parabola opens downward.

Solve for x. Add 2(x- 3)2 to both sides of the equation.

Text Example cont.

Step 2 Find the vertex. The vertex of the parabola is at (h, k). Because h =3and k =8, the parabola has its vertex at (3, 8).

Step 3 Find the x-intercepts.Replace f(x) with 0 in f(x)=-2(x- 3)2+ 8.

0 =-2(x- 3)2+ 8Find x-intercepts, setting f(x) equal to zero.

(x- 3)2= 4 Divide both sides by 2.

(x- 3) =±2 Apply the square root method.

x- 3= -2 or x- 3= 2 Express as two separate equations.

x= 1 or x= 5Add 3 to both sides in each equation.

The x-interceptsare 1 and 5. The parabola passes through (1, 0) and (5, 0).

Step 4 Find the y-intercept. Replace xwith 0 in f(x)=-2(x- 3)2+ 8.

f(0) =-2(0 - 3)2+ 8=-2(-3)2+ 8 = -2(9) + 8=-10

The y-intercept is –10.The parabola passes through (0, -10).

Text Example cont.

Step 5 Graph the parabola.With a vertex at (3, 8), x-intercepts at 1 and 5, and a y-intercept at –10, the axis of symmetry is the vertical line whose equation is x =3.

The Vertex of a Parabola Whose Equation Isf (x)=ax2+ bx + c

Consider the parabola defined by the quadratic function f (x)=ax 2+ bx + c. The parabola's vertex is at

Example

Graph the quadratic function f (x)=-x2+6x -.

Solution:

Step 1 Determine how the parabola opens. Note that a, the coefficient of x2, is -1. Thus, a < 0; this negative value tells us that the parabola opens downward.

Step 2 Find the vertex. We know the x-coordinate of the vertex is –b/2a.

We identify a, b, and c to substitute the values into the equation for the x-coordinate:

x = -b/(2a) = -6/2(-1) = 3.

The x-coordinate of the vertex is 3. We substitute 3 for x in the equation of the function to find the y-coordinate:

the parabola has its vertex at (3,7).

Example

Graph the quadratic function f (x)=-x2+6x -.

Step 3 Find the x-intercepts.Replace f(x) with 0 in f(x)=-x2+ 6x - 2. 0 = -x2+ 6x - 2

Example

Graph the quadratic function f (x)=-x2+6x -.

Step 4 Find the y-intercept. Replace xwith 0 in f(x)=-x2+ 6x - 2.

f(0) =-02+ 6 • 0 - 2=-

The y-intercept is –2.The parabola passes through (0, -2).

Step 5 Graph the parabola.

Minimum and Maximum: Quadratic Functions

- Consider f(x) = ax2 + bx +c.
- If a > 0, then f has a minimum that occurs at x = -b/(2a). This minimum value is f(-b/(2a)).
- If a < 0, the f has a maximum that occurs at x = -b/(2a). This maximum value is f(-b/(2a)).

Strategy for Solving Problems Involving Maximizing or Minimizing Quadratic Functions

- Read the problem carefully and decide which quantity is to be maximized or minimized.
- Use the conditions of the problem to express the quantity as a function in one variable.
- Rewrite the function in the form f(x) = ax2 + bx +c.
- Calculate -b/(2a). If a > 0, then f has a minimum that occurs at x = -b/(2a). This minimum value is f(-b/(2a)). If a < 0, the f has a maximum that occurs at x = -b/(2a). This maximum value is f(-b/(2a)).
- Answer the question posed in the problem.

Download Presentation

Connecting to Server..