# What is the vertex form of 4(y-1)=(x-2)^2?

May 2, 2017

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

#### Explanation:

Vertex form has the form $y = a {\left(x - b\right)}^{2} + c$

Divide both sides by $4$

$4 \cdot 4 \left(y - 1\right) = \frac{1}{4} {\left(x - 2\right)}^{2}$

$\cancel{4} \cdot \cancel{4} \left(y - 1\right) = \frac{1}{4} {\left(x - 2\right)}^{2}$

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

Add $1$ to both sides

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

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

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