answer:

Graph a parabola.

Find the intercepts and vertex of a parabola.

Find the vertex of a parabola by completing the square.

The Graph of a Quadratic Equation

We know that any linear equation with two variables can be written in the form Â

y

=

m

x

+

b

and that its graph is a line. In this section, we will see that any quadratic equation of the form Â

y

=

a

x

2

+

b

x

+

c

has a curved graph called a parabola.

Two points determine any line. However, since a parabola is curved, we should find more than two points. In this text, we will determine at least five points as a means to produce an acceptable sketch. To begin, we graph our first parabola by plotting points. Given a quadratic equation of the form Â

y

=

a

x

2

+

b

x

+

c

, x is the independent variable and y is the dependent variable. Choose some values for x and then determine the corresponding y-values. Then plot the points and sketch the graph.

Â

Example 1: Graph by plotting points: Â

y

=

x

2

âˆ’

2

x

âˆ’

3

.

Solution: In this example, choose the x-values {âˆ’2, âˆ’1, 0, 1, 2, 3, 4} and calculate the corresponding y-values.

Plot these points and determine the shape of the graph.

answer:

When graphing, we want to include certain special points in the graph. The y-intercept is the point where the graph intersects the y-axis. The x-intercepts are the points where the graph intersects the x-axis. The vertex is the point that defines the minimum or maximum of the graph. Lastly, the line of symmetry (also called the axis of symmetry) is the vertical line through the vertex, about which the parabola is symmetric.

For any parabola, we will find the vertex and y-intercept. In addition, if the x-intercepts exist, then we will want to determine those as well. Guessing at the x-values of these special points is not practical; therefore, we will develop techniques that will facilitate finding them. Many of these techniques will be used extensively as we progress in our study of algebra.

Given a quadratic equation of the form Â

y

=

a

x

2

+

b

x

+

c

, find the y-intercept by setting Â

x

=

0

and solving. In general, Â

y

=

a

(

0

)

2

+

b

(

0

)

+

c

=

c

, and we have

Next, recall that the x-intercepts, if they exist, can be found by setting Â

y

=

0

. Doing this, we have Â

-intercepts have this general form: