# What is the vertex form of y=x^2+45x+31 ?

Aug 22, 2017

Vertex form of equation is $y = {\left(x + 22.5\right)}^{2} - 475.25$

#### Explanation:

$y = {x}^{2} + 45 x + 31 \mathmr{and} y = {x}^{2} + 45 x + {\left(\frac{45}{2}\right)}^{2} - {\left(\frac{45}{2}\right)}^{2} + 31$

$y = {\left(x + \frac{45}{2}\right)}^{2} - \frac{2025}{4} + 31 \mathmr{and} y = {\left(x + \frac{45}{2}\right)}^{2} - \frac{1901}{4}$ or

$y = {\left(x + 22.5\right)}^{2} - 475.25$ . Comparing with vertex form of

equation  y = a(x-h)^2+k ; (h,k) being vertex , we find here

$h = - 22.5 , k = - 475.25 \therefore$ Vertex is at $\left(- 22.5 , - 475.25\right)$

and vertex form of equation is $y = {\left(x + 22.5\right)}^{2} - 475.25$ [Ans]