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

Apr 14, 2018

$\left(- \sqrt{2} , 0\right) , \left(\sqrt{2} , 0\right)$ and $\left(0 , - 2\right)$ are the intercepts
$\left(0 , - 2\right)$ is also the vertex of the parabola

Explanation:

For x-intercepts, $y = 0$

$0 = {x}^{2} - 2$
$2 = {x}^{2}$
$x = \pm \sqrt{2}$ ie $\left(- \sqrt{2} , 0\right) , \left(\sqrt{2} , 0\right)$

For y-intercepts, $x = 0$

$y = 0 - 2$
$y = - 2$ ie $\left(0 , - 2\right)$

Vertex: $x = - \frac{b}{2 a}$ = $x = - \frac{0}{1}$ = $x = 0$

Therefore, the vertex is the same as the y-intercept