# What is the vertex form of x = (12y - 3)^2 -144x+1?

Feb 16, 2018

The vertex is at $\left(\frac{1}{145} , \frac{1}{4}\right)$ and vertex form of equation
is
$x = \frac{144}{145} {\left(y - \frac{1}{4}\right)}^{2} + \frac{1}{145}$

#### Explanation:

$x = {\left(12 y - 3\right)}^{2} - 144 x + 1 \mathmr{and} 145 x = {\left(12 y - 3\right)}^{2} + 1$ or

$145 x = 144 {\left(y - \frac{1}{4}\right)}^{2} + 1 \mathmr{and} x = \frac{144}{145} {\left(y - \frac{1}{4}\right)}^{2} + \frac{1}{145}$

The vertex form of equation is $x = a {\left(y - k\right)}^{2} + h$

If a is positive the parabola opens right , if a is negative the

parabola opens left . Vertex: (h, k); h=1/145, k =1/4,a=144/145

The vertex is at $\left(\frac{1}{145} , \frac{1}{4}\right)$ and vertex form of equation

is $x = \frac{144}{145} {\left(y - \frac{1}{4}\right)}^{2} + \frac{1}{145}$

graph{x=144/145(y-1/4)^2+1/145 [-10, 10, -5, 5]} [Ans]