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

Apr 8, 2016

vertex: $\left(0 , - 4\right)$
x-intercepts: $\pm 2$
y-intercept: $\left(- 4\right)$

#### Explanation:

General vertex form
$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = \textcolor{g r e e n}{m} {\left(x - a\right)}^{2} + b$ with vertex at $\left(a , b\right)$

Writing given equation: $f \left(x\right) = {x}^{2} - 4$ in vertex form:
$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = 1 {\left(x - 0\right)}^{2} + \left(- 4\right)$ with vertex at $\left(0 , - 4\right)$

x-intercepts are the possible values of $x$ when $f \left(x\right) = 0$
$\textcolor{w h i t e}{\text{XXX}} 0 = {x}^{2} - 4$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow x = \pm 2$

y-intercept is the value of $y \left(= f \left(x\right)\right)$ when $x = 0$
$\textcolor{w h i t e}{\text{XXX}} y = f \left(x = 0\right) = {0}^{2} - 4 = - 4$

graph{x^2-4 [-5.717, 6.773, -4.885, 1.355]}