# How do you find the vertex and intercepts for y=x^2 -4?

Apr 29, 2016

I found:
1) y-intercept $\left(0 , - 4\right)$;
2) x-intercepts $\left(- 2 , 0\right) \mathmr{and} \left(2 , 0\right)$;
3) vertex at $\left(0 , - 4\right)$

#### Explanation:

Your function is a Quadratic and can be represented by a Parabola:

We first find the intercepts:

1) y-axis:
set $x = 0$
to get: $y = - 4$

2) x-axis (if exists):
set $y = 0$
solve: ${x}^{2} - 4 = 0$
that gives you $x = \pm 2$
so you have 2 intercepts on the $x$ axis.

We now find the vertex:
The $x$ coordinate of the vertex can be found as:
${x}_{v} = - \frac{b}{2 a}$

where $b \mathmr{and} a$ refer to the numerical coefficients of the general Quadratic Function:
$y = a {x}^{2} + b x + c$
$a = 1$
$b = 0$
$c = - 4$
so that ${x}_{v} = - \frac{0}{2} = 0$
the $y$ coordinate of the vertex can be found substituting ${x}_{v} = 0$ into the original function that gives us:
${y}_{v} = - 4$