# What is the vertex form of y= x^2-3x+108 ?

Dec 31, 2015

Complete The square to find the vertex

#### Explanation:

y = ${x}^{2}$ - 3x + 108

y = 1(${x}^{2}$ - 3x + - ) + 108

___= ${\left(\frac{b}{2}\right)}^{2}$

___= ${\left(\frac{3}{2}\right)}^{2}$

___= $\frac{9}{4}$

y = 1(${x}^{2}$ - 3x + 9/4 - 9/4) + 108

y = 1(x - 3/2)^2 - $\frac{9}{4}$ + 108

y = 1(x - 3/2)^2 + $\frac{423}{4}$

The vertex is at ($\frac{3}{2}$, $\frac{423}{4}$)