Quadratic functions

Section 1-7: Quadratic Functions - Graphs

A quadratic function is in the form f(x) = ax² + bx + c. A quadratic function graphs into a parabola, a curve shape like the McDonald's arches. Every quadratic function has a vertex and a vertical axis of symmetry. This means it is the same graph on either side of a vertical line. The axis of symmetry goes through the vertex point. The vertex point is either the maximum or minimum point for the parabola. A summary of what we talked about last year is shown below:

a>0 a<0	graph opens up graph opens down
b² - 4ac > 0 b² - 4ac = 0 b² - 4ac < 0	Graph has 2 x-intercepts Graph has 1 x-intercept Graph has no x-intercepts
Vertex Point	at x = -b/2a to find y substitute above value for x
x-intercepts	factor or use quadratic formula
y-intercept	c value in ax² + bx + c

Sample Problems Find the vertex point, axis of symmetry, x and y intercepts and graph. 1) F(x) = x² + 2x - 3 Solution: The graph opens up because a is positive. The vertex point is at x = -b/2a = -2/2 = -1 y = (-1)² + 2(-1) - 3 = -4 Vertex point is at (-1, -4) Axis of symmetry is the equation x = -1 x-intercepts found by factoring (x + 3)(x - 1) = 0 They are at x = -3, x = 1 y-intercept at (0, c) at (0, -3) The graph is as follows:

Alternate method! A second form of a quadratic can be written in the following form: f(x) = a(x - h)² + k Where (h, k) is the vertex point

Example Find the vertex point using the alternate method.

1) y = 3(x - 3)² + 4

Solution: The vertex point is (3, 4)

2) y = 2(x + 1)² + 3

Solution: The vertex point is (-1, 3)

3) y = 5(x + 2)² - 7

Solution: The vertex point is (-2, -7)

4) y = 2x² + 8x - 5

Solution: Put the equation in the correct form:

y = 2(x² + 4x + __ ) - 5

y = 2(x² +4x + 4) -5 - 8(completing the square)

y = 2(x + 2)² - 13 Therefore, the vertex point is (-2, -13)

2nd Sample Problem! Use the second method to graph the following equation!

1) y = x² - 6x -7

Solution:

Graph opens up because a is positive

x-intercepts found by factoring (x - 7)(x + 1) = 0 The x-intercepts are x = 7, and x = -1

y-intercept is at (0,c) or (0, -7)

Using method two to find the vertex point

y = (x² - 6x + ___) - 7

y = (x² - 6x + 9) - 7 - 9

y = (x - 3)² - 16

The vertex point is (3, -16)

The axis of symmetry is at x = 3

The graph looks like this:

That concludes our review of quadratics and graphing. On to quadratic modeling!!