# How do you solve x^2-2x-8=0 by graphing?

Jun 24, 2018

We need to find the vertex, x-intercepts, and y-intercept to give us an idea of what the graph will look like.

#### Explanation:

Let's find the vertex first. We can find the x-coordinate by using $x = - \frac{b}{2 a}$:

$x = - \frac{2}{2 \cdot 1} = - \frac{2}{2} = 1$

The y-coordinate can be found by plugging the x-value back into the equation:

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

The vertex is $\left(1 , 9\right)$. Next, let's find the x-intercepts, which are the zeroes of the equation. Let's factor the polynomial and find them:

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

The x-intercepts are $\left(4 , 0\right)$ and $\left(- 2 , 0\right)$. Finally, the y-intercept can be found by letting $x = 0$:

${0}^{2} - 2 \left(0\right) - 8 = - 8$

The y-intercept is $\left(0 , - 8\right)$. You can plot the bolded points, which will be able to create a decent outline of the parabola, If you need them, you can plot other points on the graph, such as $\left(2 , - 8\right)$, $\left(- 1 , - 5\right)$, and $\left(3 , - 5\right)$. This is what the graph will look like:

graph{x^2-2x-8 [-9, 11, -10, 1]}