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

Dec 18, 2015

Vertex form: $x = 4 {\left(y - \frac{3}{2}\right)}^{2} + \left(- 11\right)$
Note this is a parabola with a horizontal axis of symmetry.

#### Explanation:

Vertex form (for a parabola with horizontal axis of symmetry):
$\textcolor{w h i t e}{\text{XXX}} x = m {\left(y - b\right)}^{2} + a$
with vertex at $\left(a , b\right)$

Conversion of given equation: $x = {\left(2 y - 3\right)}^{2} - 11$ into vertex form:

$\textcolor{w h i t e}{\text{XXX}} x = {\left(\left(2\right) \cdot \left(y - \frac{3}{2}\right)\right)}^{2} - 11$

$\textcolor{w h i t e}{\text{XXX}} x = {2}^{2} \cdot {\left(y - \frac{3}{2}\right)}^{2} - 11$

$\textcolor{w h i t e}{\text{XXX}} x = 4 {\left(y - \frac{3}{2}\right)}^{2} + \left(- 11\right)$
(which is the vertex form with vertex at $\left(- 11 , \frac{3}{2}\right)$).
graph{x=(2y-3)^2-11 [-11.11, 1.374, -0.83, 5.415]}