How do you find the vertex of a parabola f(x)=x^2-4?

Jun 30, 2015

The vertex of $f \left(x\right) = {x}^{2} - 4$ is at $\left(0 , - 4\right)$

Explanation:

$f \left(x\right) = {x}^{2} - 4$
is a simple variant of the vertex form:
$\textcolor{w h i t e}{\text{XXXX}}$$f \left(x\right) = {\left(x - a\right)}^{2} + b$ with a vertex at $\left(a , b\right)$

$f \left(x\right) = {x}^{2} - 4$
is equivalent to
$\textcolor{w h i t e}{\text{XXXX}}$$f \left(x\right) = {\left(x - 0\right)}^{2} + \left(- 4\right)$