# What is the axis of symmetry and vertex for the graph y = x^2-2x-5?

Apr 11, 2018

The axis of symmetry is $x = 1$.

The vertex is $\left(1 , - 6\right)$.

#### Explanation:

Given:

$y = {x}^{2} - 2 x - 5$ is a quadratic equation in standard form:

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

where:

$a = 1$, $b = - 2$, $c = - 5$

Axis of symmetry: the vertical line that divides a parabola into two equal halves.

For a quadratic equation in standard form, the formula for determining the axis of symmetry is:

$x = \frac{- b}{2 a}$

Plug in the known values and solve.

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

$x = \frac{2}{2}$

$x = 1$

The axis of symmetry is $x = 1$.

Vertex: maximum or minimum point of the parabola. Since $a > 0$, the vertex will be the minimum point and the parabola will open upward.

Substitute $1$ for $x$ in the equation, and solve for $y$.

$y = {\left(1\right)}^{2} - 2 \left(1\right) - 5$

$y = 1 - 2 - 5$

$y = - 6$

The vertex is $\left(1 , - 6\right)$.

graph{y=x^2-2x-5 [-10.875, 11.625, -8.955, 2.295]}