# What is the the vertex of x = (y - 6)^2 - y+1?

Dec 28, 2017

Vertex is $\left(- 5 \frac{1}{4} , - 6 \frac{1}{2}\right)$

#### Explanation:

We can write $x = {\left(y - 6\right)}^{2} - y + 1$ as

$x = {y}^{2} - 12 y + 36 - y + 1$

= ${y}^{2} - 13 y + {\left(\frac{13}{2}\right)}^{2} - \frac{169}{4} + 37$

= ${\left(y - \frac{13}{2}\right)}^{2} - \frac{169 - 148}{4}$

= ${\left(y - \frac{13}{2}\right)}^{2} - \frac{21}{4}$

Hence vertex is $\left(- \frac{21}{4} , - \frac{13}{2}\right)$ or $\left(- 5 \frac{1}{4} , - 6 \frac{1}{2}\right)$