# What is the vertex form of 3y = 3x^2-4x+11?

Dec 9, 2015

$y = {\left(x - \frac{2}{3}\right)}^{2} + \frac{29}{9}$

#### Explanation:

Vertex form of a quadratic equation: $y = a {\left(x - h\right)}^{2} + k$

The parabola's vertex is the point $\left(h , k\right)$.

First, divide everything by $3$.

$y = {x}^{2} - \frac{4}{3} x + \frac{11}{3}$

Complete the square using only the first $2$ terms on the right. Balance the term you've added to complete the square by also subtracting it from the same side of the equation.

y=(x^2-4/3xcolor(blue)+color(blue)(4/9))+11/3color(blue)-color(blue)(4/9

$y = {\left(x - \frac{2}{3}\right)}^{2} + \frac{33}{9} - \frac{4}{9}$

$y = {\left(x - \frac{2}{3}\right)}^{2} + \frac{29}{9}$

From this, we can determine that the vertex of the parabola is at the point $\left(\frac{2}{3} , \frac{29}{9}\right)$.