# How do you find the axis of symmetry and vertex point of the function: y = -x^2 - 8x + 10?

Oct 7, 2015

The axis of symmetry is $x = - 4$.
The vertex is $\left(- 4 , 26\right)$.

#### Explanation:

$y = - x - 8 x + 10$ is a quadratic equation in the form $y = a x + b x + c$, where $a = - 1 , b = - 8 , c = 10$

Axis of Symmetry
The axis of symmetry is the imaginary vertical line that divides the parabola into two equal halves.

The formula for the axis of symmetry is $x = \frac{- b}{2 a}$.

$x = \frac{- b}{2 a} = \frac{- \left(- 8\right)}{2 \left(- 1\right)} = \frac{8}{- 2} = - 4$

The axis of symmetry is $x = - 4$.

This is also the $x$ value of the vertex.

Vertex
The vertex is the maximum or minimum point of the parabola. Since $a$ is a negative number in this equation, the parabola opens downward so the vertex is the maximum point.

Since we know that $x = - 4$, we substitute it into the equation and solve for $y$.

$y = - {x}^{2} - 8 x + 10$

$y = - {\left(4\right)}^{2} - \left(8\right) \left(- 4\right) + 10 =$

$y = - 16 + 32 + 10 = 26$

The vertex is $\left(- 4 , 26\right)$

graph{y=-x^2-8x+10 [-18.16, 13.86, 14.09, 30.11]}