# How do you write x^2+42=12x-3y into vertex form?

Jun 22, 2018

$y = - \frac{1}{3} {\left(x - 6\right)}^{2} - 2$

#### Explanation:

Convert it to the form $y = a {\left(x - b\right)}^{2} + c$.

${x}^{2} + 42 = 12 x - 3 y$

Move all terms containing $y$ to the LHS.

$- 3 y = {x}^{2} - 12 x + 42$

Divide both sides by $- 3$.

$y = - \frac{1}{3} \left({x}^{2} - 12 x + 42\right)$

Because ${\left(x + p\right)}^{2} = {x}^{2} + 2 p x + {p}^{2}$, the following rule can be used:
${\left(x + \frac{1}{2} p\right)}^{2} - {p}^{2} = {x}^{2} + p x$

This gives us:

$y = - \frac{1}{3} \left({\left(x - 6\right)}^{2} - 36 + 42\right)$

Simplify to $y = a {\left(x - b\right)}^{2} + c$.

$y = - \frac{1}{3} \left({\left(x - 6\right)}^{2} + 6\right)$

$y = - \frac{1}{3} {\left(x - 6\right)}^{2} - 2$