# What is the vertex of y=3x^2+2x+5?

Oct 4, 2017

Vertex $\to \left(x , y\right) = \left(- \frac{1}{3} , \frac{14}{3}\right)$

#### Explanation:

Given: $y = 3 {x}^{2} + 2 x + 5$

This is part of the process of completing the square.

Write as $y = 3 \left({x}^{2} \textcolor{red}{+ \frac{2}{3}} x\right) + 5$

To complete the square you would 'do other things' to this. I am not going to do that!

${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \left(\textcolor{red}{+ \frac{2}{3}}\right) = - \frac{1}{3}$

Substitute for $x$ to determine ${y}_{\text{vertex}}$

${y}_{\text{vertex}} = 3 {\left(- \frac{1}{3}\right)}^{2} + 2 \left(- \frac{1}{3}\right) + 5$

${y}_{\text{vertex}} = + \frac{1}{3} - \frac{2}{3} + 5 = 4 \frac{2}{3} \to \frac{14}{3}$

Vertex $\to \left(x , y\right) = \left(- \frac{1}{3} , \frac{14}{3}\right)$