# How do you find the vertex and intercepts for y = x^2 - 8?

Nov 20, 2015

vertex = $\left(0 , - 8\right)$
y-intercept = $- 8$
x-intercepts = $2 \sqrt{2}$ and $- 2 \sqrt{2}$

#### Explanation:

The general equation for a quadratic function in vertex form is:

$y = a {\left(x - h\right)}^{2} + k$

With this equation, $y = {x}^{2} - 8$ can be rewritten as:

$y = 1 {\left(x - 0\right)}^{2} - 8$

The vertex of a quadratic equation in vertex form is $\left(h , k\right)$, or in this case $\left(0 , - 8\right)$.

To find the y-intercept, substitute $x$ as $0$, since the x-coordinate of the y-intercept is $0$:

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

To find the x-intercept, substitute $y$ as $0$, since the y-coordinate of the x-intercept is $0$:

$y = {x}^{2} - 8$
$0 = {x}^{2} - 8$
$8 = {x}^{2}$
$x = \pm \sqrt{8}$
$x = 2 \sqrt{2}$ or $- 2 \sqrt{2}$
$x = 2.83$ or $- 2.83$

Here is a graph of the equation:
graph{y= x^2-8 [-12.78, 12.53, -11.14, 2.05]}

As you can see, the graph has a y-intercept of $- 8$ and x-intercepts of $2 \sqrt{2}$ and $- 2 \sqrt{2}$.