# What is the vertex form of y=x^2+7x-3?

Jan 14, 2016

$y = {\left(x + \frac{7}{2}\right)}^{2} - \frac{61}{4}$ or $4 y = {\left(2 x + 7\right)}^{2} - 61$

#### Explanation:

For a quadratic of the form $y = a {x}^{2} + b x + c$ the vertex form is $y = a \left[{\left(x + \frac{b}{2 a}\right)}^{2} - {\left(\frac{b}{2 a}\right)}^{2}\right] + c$
In this case that gives us
$y = {\left(x + \frac{7}{2}\right)}^{2} - \frac{49}{4} - 3$
$y = {\left(x + \frac{7}{2}\right)}^{2} - \frac{61}{4}$

The vertex is then $\left(- \frac{7}{2} , - \frac{61}{4}\right)$

Multiplying throughout by $4$ gives
$4 y = {\left(2 x + 7\right)}^{2} - 61$