How do you find the vertex and the intercepts for y= -x^2 + 9?

Feb 9, 2017

Vertex: $\left(0 , 9\right)$
Y-intercept: $y = 9$
X-intercepts: $x = - 3 \mathmr{and} x = + 3$

Explanation:

Remember the standard vertex form for a parabola is
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{m} {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$
with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$

Re-writing the given equation: $y = - {x}^{2} + 9$
into explicit standard vertex form:
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{- 1} {\left(x - \textcolor{red}{0}\right)}^{2} + \textcolor{b l u e}{9}$
with vertex at $\left(\textcolor{red}{0} , \textcolor{b l u e}{9}\right)$

The Y-intercept is the value of $y$ when $x = \textcolor{m a \ge n t a}{0}$
$\textcolor{w h i t e}{\text{XXX}} y = - {\textcolor{m a \ge n t a}{0}}^{2} + 9 = 9$

The X-intercepts are the values of $x$ possible when $y = \textcolor{\mathmr{and} a n \ge}{0}$
$\textcolor{w h i t e}{\text{XXX}} \textcolor{\mathmr{and} a n \ge}{0} = - {x}^{2} + 9$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow {x}^{2} = 9$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow x = \pm 3$

Here is the graph for verification purposes: