# How do you find the vertex and axis of symmetry for a quadratic equation y = -x^2 + 6x -9?

Jul 11, 2015

The vertex is at ($3 , 0$).
The axis of symmetry is $x = 3$.

#### Explanation:

The standard form for the equation of a parabola is

$y = {a}^{2} + b x + c$

$y = - {x}^{2} + 6 x - 9$

So

$a = - 1$, $b = 6$, and $c = - 9$.

Vertex

Since $a < 0$, the parabola opens downwards.

The $x$-coordinate of the vertex is at

x = –b/(2a) = -6/(2(-1)) = -6/-2 = 3.

Insert this value of $x$ back into the equation.

y = -x^2 + 6x - 9 = -3^2 + 6×3 - 9 = -9 + 18 – 9 = 0

The vertex is at ($3 , 0$).

Axis of symmetry

The axis of symmetry must pass through the vertex, so

The axis of symmetry is $x = 3$.