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

Vertex is at $\left(2 , - 11\right)$

#### Explanation:

This is a parabola which opens upward

of the form ${\left(x - h\right)}^{2} = 4 p \left(y - k\right)$ where vertex is $\left(h , k\right)$

from the given $y = 2 {\left(x - 2\right)}^{2} - 11$

transform first to the form

$y = 2 {\left(x - 2\right)}^{2} - 11$

$y + 11 = 2 {\left(x - 2\right)}^{2}$

$\frac{y + 11}{2} = \frac{2 {\left(x - 2\right)}^{2}}{2}$

$\frac{y + 11}{2} = \frac{\cancel{2} {\left(x - 2\right)}^{2}}{\cancel{2}}$

$\frac{1}{2} \cdot \left(y + 11\right) = {\left(x - 2\right)}^{2}$

${\left(x - 2\right)}^{2} = \frac{1}{2} \cdot \left(y + 11\right)$

${\left(x - 2\right)}^{2} = \frac{1}{2} \cdot \left(y - - 11\right)$

so that $h = 2$ and $k = - 11$

vertex is at $\left(2 , - 11\right)$

Kindly see the graph
graph{y=2(x-2)^2-11[-5,40,-15,10]}

Have a nice day ! from the Philippines...