# How do you find the vertex and axis of symmetry for a quadratic equation y = x^2-5?

Jul 11, 2015

The vertex is at ($0 , - 5$).
The axis of symmetry is $x = 0$.

#### Explanation:

The standard form for the equation of a parabola is

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

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

So

$a = 1$, $b = 0$, and $c = - 5$.

Vertex

Since $a > 0$, the parabola opens upwards.

The $x$-coordinate of the vertex is at

x = –b/(2a) = -0/(2×1) = -0/2 = 0.

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

$y = {x}^{2} - 5 = {0}^{2} - 5 = 0 - 5 = - 5$

The vertex is at ($0 , - 5$).

Axis of symmetry

The axis of symmetry must pass through the vertex, so

The axis of symmetry is $x = 0$.