5. Draw
the graph of *y* = *f*(*x*) = *x*^{2} + 2*x* - 8.

- a) What is the
*y*-intercept? - b) What are the roots?
- c) What are the coordinates of the vertex?
- d) Write the equation in standard or vertex form.
- e) Solve for
*x*

*x*^{2} + 2*x* - 8 __>__ 0

Before we draw the graph it would help to answer questions a), b), and c).

The *y*-intercept is the point on the graph whose *x*-coordinate is 0.
So let *x* = 0 in the equation, and we get

This illustrates that the *y*-intercept is always the constant term
in a polynomial.

The roots are another name for the *x*-intercepts. They are the
points on the graph whose *y*-coordinates are 0. So let *y* = 0 in the
equation and we get

This gives us a quadratic equation in *x*. &mbsp; We are fortunate enough
to already have a 0 on one side so we are ready to factor.

Set the factors = 0.

Solve for *x*

c) What are the coordinates of the vertex?

If we use the formula for the *x*-coordinate of the vertex,

we get

or

Now that we have the *x-*coordinate of the vertex, we run this
number through the function to find the *y-*coordinate

So the vertex is at (-1, -9)

It is a good idea to get all of this information before plotting
points to draw the graph. When we plot points, since we now know that
the *x* coordinate of the vertex is -1, we will want -1 to be in the
middle of the *x*'s that we plot. We will also want the roots, *x* = -4
and 2 to be in the interval containing the *x*'s that we plot, and at
least one point on the other sides of the roots from the vertex.

When we plot these points, we get

d) Write the equation in vertex form.

The equation is given to be

Half of the linear coefficient is 1, and the square of 1 is 1, so we add and subtract 1

This simplifies to

and we see the coordinates of the vertex in the equation.

In the process of drawing the graph we found
that the roots were at -4 and 2. When we divide the real number line
into the intervals between the roots, and check out an *x* in each
interval.

Since equality is permitted, the endpoints of the intervals, the
roots, being the *x*'s which will give answers which are equal to 0
when substituted into the formula, will be solutions. So we draw
square brackets around the endpoints.

The set of solutions is the set of *x*'s which satisfy

*x* ≤ -4 or *x* ≥ 2

In interval notation, this comes out to be

(-∞, -4] ∪ [2, ∞)