# How do you find the intercept and vertex of y= -x^2 + 9?

Jul 14, 2016

Vertex is $\left(0 , 9\right)$

Its y-intercept is $= 9$

#### Explanation:

Given -

$y = - {x}^{2} + 9$

It has no $x$ term. So we shall have it as -

$y = - {x}^{2} + 0 x + 9$

Then it is in the form $y = a {x}^{2} + b x + c$

$x$ cor-ordinate of the vertex is -

$x = \frac{- b}{2 a} = \frac{0}{2 \times \left(- 1\right)} = 0$

At x=0; y=(0^2)+9=9

Vertex is $\left(0 , 9\right)$

Its y-intercept is $= 9$