# What is the vertex form of x = (2y +5 )^2 + 21?

Aug 21, 2017

$x = 4 {\left(y - \left(- 2.5\right)\right)}^{2} + 21$

#### Explanation:

Given: $x = {\left(2 y + 5\right)}^{2} + 21$

Note: There is a quick way to do this but it is easy to confuse yourself so I will do it the following way.

Expand the square:

$x = 4 {y}^{2} + 20 y + 25 + 21$

$x = 4 {y}^{2} + 20 y + 46 \text{ }$

This is the standard form

$x = a {y}^{2} + b y + c$

where $a = 4 , b = 20 \mathmr{and} c = 46$

The general vertex form is:

$x = a {\left(y - k\right)}^{2} + h \text{ }$

We know that $a$ in the vertex form is the same as $a$ in the standard form:

$x = 4 {\left(y - k\right)}^{2} + h \text{ [2.1]}$

To find the value of k, use the formula:

$k = - \frac{b}{2 a}$

$k = - \frac{20}{2 \left(4\right)} = - 2.5$

$x = 4 {\left(y - \left(- 2.5\right)\right)}^{2} + h \text{ [2.2]}$

To find h, evaluate equation  at $x = k = - 2.5$

$h = 4 {\left(- 2.5\right)}^{2} + 20 \left(- 2.5\right) + 46$

$h = 21$

$x = 4 {\left(y - \left(- 2.5\right)\right)}^{2} + 21 \text{ [2.3]}$