# What is the the vertex of x = (y - 6)^2 - 11?

Jan 16, 2016

$\text{Vertex} \to \left(x , y\right) \to \left(- 11 , 6\right)$

#### Explanation:

Given: $\textcolor{w h i t e}{\ldots .} x = {\left(y - 6\right)}^{2} - 11$............................(1)

View as the same thing as the vertex form for the U shaped quadratic but instead it is expressed in terms of $y$ instead of $x$

So the instead of stating that ${x}_{\text{vertex}} = \left(- 1\right) \times \left(- 6\right)$ as in the U curve format we say ${y}_{\text{vertex}} = \left(- 1\right) \times \left(- 6\right) = 6$

${y}_{\text{vertex}} = 6$

Substitute in equation (1) gives:

So ${x}_{\text{vertex}} = {\left(6 - 6\right)}^{2} - 11 = - 11$

$\text{Vertex} \to \left(x , y\right) \to \left(- 11 , 6\right)$