# What is the the vertex of x = (y + 6)^2 - 3 ?

Nov 26, 2015

The vertex is $\left(- 3 , - 6\right)$.

#### Explanation:

Expand the parabola:

${\left(y + 6\right)}^{2} - 3 = {y}^{2} + 12 y + 36 - 3 = {y}^{2} + 12 y + 33$

The vertex is the minimum of a parabola, so we can derive it and set the derivative to zero:

$2 y + 12 = 0 \setminus \iff y = - 6$.

So, the vertex has $y$-coordinate $- 6$. To find the $x$-coordinate, simply compute

$f \left(- 6\right) = \left(- 6 + 6\right) - 3 = - 3$