# What is the vertex of  y= -x^2 + 3?

Jan 11, 2016

The vertex is at $\left(0 , 3\right)$

#### Explanation:

One way to see this is to convert the given equation into the general "vertex form" for a parabola:
$\textcolor{w h i t e}{\text{XXX}} y = \left(m\right) {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$ with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$

Since
$\textcolor{w h i t e}{\text{XXX}} y = - {x}^{2} + 3$
is equivalent to
$\textcolor{w h i t e}{\text{XXX}} y = \left(- 1\right) {\left(x - \textcolor{red}{0}\right)}^{2} + \textcolor{b l u e}{3}$
the vertex is at $\left(\textcolor{red}{0} , \textcolor{b l u e}{3}\right)$