Download presentation

Presentation is loading. Please wait.

1
Quadratic Functions

2
**Graphs of Quadratic Functions**

A Parabola.

3
**The Quadratic Function**

A quadratic function can be expressed in three formats: f(x) = ax2 + bx + c is called the standard form is called the factored form f (x) = a(x - h)2 + k, is called the vertex form

4
**The Standard Form f(x)=ax2 + bx + c If a > 0 then parabola opens up**

If a < 0 then parabola opens down (0,c) is the y-intercept Axis of symmetry is

5
**The Vertex Form f (x) = a(x - h)2 + k, a 0**

***y – k = a(x-h)2 The graph of f is a 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

6
**Graphing Parabolas-Vertex Form**

To graph f (x) = a(x - h)2 + k: Determine whether the parabola opens up or down. If a > 0, it opens up. If a < 0, it opens down. Determine the vertex of the parabola--- V(h, k). Find any x-intercepts. Replace 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.

7
**The Factored Form Factored Form of a quadratic**

r represents roots or x intercepts Axis of symmetry:

8
**Vertex form f (x) = a(x - h)2 + k **

Example Graph the quadratic function f (x) = -2(x - 3)2 + 8. Vertex form f (x) = a(x - h)2 + k a=-2 h =3 k = 8 Given equation f (x) = -2(x - 3)2 + 8 This is a parabola that opens down with vertex V(3,8)

9
**Example cont. The x-intercepts are 1 and 5.**

Find the vertex. The vertex of the parabola is at (h, k). Because h = 3 and k = 8, the parabola has its vertex at (3, 8). Find the x-intercepts. Replace f (x) with 0: f (x) = -2(x - 3)2 + 8. 0 = -2(x - 3) Find x-intercepts, setting f (x) equal to zero. 2(x - 3)2 = 8 (x - 3)2 = Divide both sides by 2. (x - 3) = ± Apply the square root method. x - 3 = or x - 3 = Express as two separate equations. x = or x = Add 3 to both sides in each equation. The x-intercepts are 1 and 5. The parabola passes through (1, 0) and (5, 0).

10
**Example cont. Graph the parabola. With a vertex at (3, 8)**

f (0) = -2(0 - 3) = -2(-3) = -2(9) + 8 = -10 Find the y-intercept. Replace x with 0 in f (x) = -2(x - 3)2 + 8. The y-intercept is –10. The parabola passes through (0, -10). Graph the parabola. With a vertex at (3, 8) x-intercepts at (1,0) and (5,0) and a y-intercept at (0,–10) the axis of symmetry is the vertical line x = 3.

11
**The Vertex of a Parabola Whose Equation Is f (x) = ax 2 + bx + c**

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

12
**Example Graph the quadratic function f (x) = -x2 + 6x -.**

Determine how the parabola opens. Note that a, the coefficient of x 2, is -1. Thus, a < 0; This negative value tells us that the parabola opens downward. Find the vertex. x = -b/(2a). 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: y=f(3) = -(3)^2+6(3)-2= =7, the parabola has its vertex at (3,7).

13
**Example Graph the quadratic function f (x) = -x2 + 6x -.**

Find the x-intercepts. Replace f (x) with 0 in f (x) = -x2 + 6x - 2. 0 = -x2 + 6x (If you cannot factor)

14
**Example Graph the quadratic function f (x) = -x2 + 6x -.**

Find the y-intercept. Replace x with 0 in f (x) = -x2 + 6x - 2. f (0) = • = - The y-intercept is –2. The parabola passes through (0, -2). Graph the parabola.

15
**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)).

16
**The Discriminant and the Kinds of Solutions to ax2 + bx +c = 0**

No x-intercepts No real solution; two complex imaginary solutions b2 – 4ac < 0 One x-intercept One real solution (a repeated solution) b2 – 4ac = 0 Two x-intercepts Two unequal real solutions b2 – 4ac > 0 Graph of y = ax2 + bx + c Kinds of solutions to ax2 + bx + c = 0 Discriminant b2 – 4ac

17
Quadratic Functions

Similar presentations

© 2020 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google