# How do find the vertex and axis of symmetry for a quadratic equation y =x^2+2x-3?

Jun 18, 2015

Rearrange the equation into vertex form by completing the square to get:

$y = {\left(x - \left(- 1\right)\right)}^{2} + \left(- 4\right)$

Hence, the vertex is at $\left(- 1 , - 4\right)$

and the axis of symmetry is $x = - 1$

#### Explanation:

Vertex form of the equation of a parabola with vertical axis is:

$y = a {\left(x - {x}_{0}\right)}^{2} + {y}_{0}$

Starting with $y = {x}^{2} + 2 x - 3$, rearrange as follows:

$y = {x}^{2} + 2 x - 3$

$= {x}^{2} + 2 x + 1 - 1 - 3$

$= {\left(x + 1\right)}^{2} - 4$

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

This is vertex form, with $a = 1$, ${x}_{0} = - 1$ and ${y}_{0} = - 4$

So the vertex is at $\left(- 1 , - 4\right)$ and the axis of symmetry is $x = - 1$