# How do you identify the important parts of x^2-2x-8=0 to graph it?

Sep 23, 2015

x-intercepts: $\left(4 , 0\right)$ and $\left(- 2 , 0\right)$
vertex: $\left(1 , - 9\right)$

#### Explanation:

I think the most important parts are the x-intercepts and the vertex.

x-intercepts (The value of $x$ when $y = 0$)
Given the function $y = {x}^{2} - 2 x - 8$, set $y$ to $0$. Compute for the x-intercepts by factoring.

$y = {x}^{2} - 2 x - 8$
$0 = {x}^{2} - 2 x - 8$
$0 = \left(x - 4\right) \left(x + 2\right)$

Now that you've factored it, you can say that the x-intercepts are:

$x - 4 = 0$
$x = 4$
color(blue)((4,0)

$x + 2 = 0$
$x = - 2$
color(blue)((-2,0)

Mark the points $\left(4 , 0\right)$ and $\left(- 2 , 0\right)$ on your graph.

vertex
To solve for the vertex, you will have to convert the function into the vertex form $y = a {\left(x - h\right)}^{2} + k$, where $\left(h , k\right)$ is the vertex.

$y = {x}^{2} - 2 x - 8$
$y + 8 = {x}^{2} - 2 x$
$y + 8 + 1 = {x}^{2} - 2 x + 1$
$y + 9 = {\left(x - 1\right)}^{2}$
$y = {\left(x - 1\right)}^{2} - 9$

You can see here that $h = 1$ and $k = - 9$. Therefore, your vertex is $\left(1 , - 9\right)$.

After you have plotted the vertex and x-intercepts, just add a few more points by inserting any value to $x$. This will help to make your graph more accurate.

It should look like this:
graph{x^2-2x-8 [-8.97, 11.03, -9.32, 0.68]}