# How do you find the axis of symmetry and vertex point of the function: y= -x^2-4x+5?

Oct 20, 2015

The axis of symmetry is $x = - 2$.
The vertex is $\left(- 2 , 9\right)$.

#### Explanation:

$y = - {x}^{2} - 4 x + 5$ is a quadratic equation in the form $y = a x + b x + c$, where $a = - 1 , b = - 4 , \mathmr{and} c = 5$.

Axis of Symmetry
An imaginary vertical line that divides the parabola into two equal halves. The formula for determining the axis of symmetry is $x = \frac{- b}{2 a}$

$x = \frac{- \left(- 4\right)}{\left(2 \cdot - 1\right)} = \frac{4}{- 2} = - 2$

The axis of symmetry is $x = - 2$

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

The $x$ value for the vertex is the same as the axis of symmetry. To find the $y$ value, substitute $- 2$ for $x$ in the equation. Solve for $y$.

$y = - {x}^{2} - 4 x + 5$

$y = - {\left(- 2\right)}^{2} - 4 \left(- 2\right) + 5 =$

$y = - 4 + 8 + 5 = 9$

The vertex is $\left(- 2 , 9\right)$.

graph{y=-x^2-4x+5 [-14.95, 13.52, -3.76, 10.48]}