# How do you identify the important parts of g(x)=-x^2+12x-36 to graph it?

##### 1 Answer
Sep 30, 2015

Its vertex is $\left(6 , 0\right)$
Axis of symmetry is $x = 6$
Since co-efficient of x is $- 1$, the curve is concave downwards.
It has a maxima at $\left(6 , 0\right)$

#### Explanation:

g(x)=−x^2+12x−36

$x = \frac{- b}{2 a} = \frac{- 12}{2 \times \left(- 1\right)} = \frac{- 12}{- 2} = 6$

At x=6; f(x) =-(6^2)+12(6)-36
$f \left(x\right) = - 36 + 72 - 36 = - 72 + 72 = 0$

Its vertex is $\left(6 , 0\right)$
Axis of symmetry is $x = 6$
Since co-efficient of x is $- 1$, the curve is concave downwards.

It has a maxima at $\left(6 , 0\right)$