# How do find the vertex and axis of symmetry, and intercepts for a quadratic equation ?

Jun 14, 2015

Complete the square using:

$a {x}^{2} + b x + c = a {\left(x + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$

From this the vertex, axis of symmetry, etc is easy to derive.

#### Explanation:

Given quadratic equation $a {x}^{2} + b x + c = 0$

We can complete the square using the equation:

$a {x}^{2} + b x + c = a {\left(x + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$

Then the vertex is at $\left(- \frac{b}{2 a} , c - {b}^{2} / \left(4 a\right)\right)$

Axis of symmetry: $x = - \frac{b}{2 a}$

$y$ intercept $\left(0 , c\right)$

$x$ intercepts $\left(\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} , 0\right)$