# What is the vertex of y=8(3x+7)^2+5?

Oct 20, 2017

$\left(- \frac{7}{3} , 5\right) = \left(- 2. \overline{3} , 5\right)$

#### Explanation:

First get this into vertex form:

$y = a {\left(b \left(x - h\right)\right)}^{2} + k$ where $\left(h , k\right)$ is the vertex

by factoring out the $3$ in the parentheses:

$y = 8 {\left(3 \left(x + \frac{7}{3}\right)\right)}^{2} + 5$

Then factor out a negative $1$:

$y = 8 {\left(3 \left(x - 1 \left(- \frac{7}{3}\right)\right)\right)}^{2} + 5$

So it's now in vertex form:

$y = 8 {\left(3 \left(x - \left(- \frac{7}{3}\right)\right)\right)}^{2} + 5$ where $h = - \frac{7}{3}$ and $k = 5$

So our vertex is $\left(- \frac{7}{3} , 5\right) = \left(- 2. \overline{3} , 5\right)$