# How to find the x and y-intercept given y = 1/2x^2 - 2?

May 10, 2015

This parabola intersects the $y$ axis at $\left(0 , - 2\right)$ and the $x$ axis at $\left(- 2 , 0\right)$ and $\left(2 , 0\right)$.

To find these intersections, start by substituting $x = 0$ and $y = 0$ into the equation of the curve.

First the $y$ axis: Substitute $x = 0$ ...

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

Now the $x$ axis: Substitute $y = 0$ to get:

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

$\left(\frac{1}{2}\right) {x}^{2} = 2$
${x}^{2} = 4$
Hence $x = \pm 2$