What is the vertex of  y= -x^2-4x-3-2(x-3)^2?

Mar 24, 2017

The vertex is $\left(\frac{4}{3} , - \frac{47}{3}\right)$

Explanation:

$y = - {x}^{2} - 4 x - 3 - 2 {\left(x - 3\right)}^{2}$

This is not in vertex form yet, so we need to expand and organize the quadratic, complete the square, then determine the vertex.

Expand:
$y = - {x}^{2} - 4 x - 3 - 2 \left({x}^{2} - 6 x + 9\right)$

$y = - {x}^{2} - 4 x - 3 - 2 {x}^{2} + 12 x - 18$

Organize:
$y = - 3 {x}^{2} + 8 x - 21$

Complete the square:
$y = - 3 \left[{x}^{2} - \frac{8 x}{3} + 7\right]$

$y = - 3 \left[{\left(x - \frac{4}{3}\right)}^{2} - \frac{16}{9} + 7\right]$

$y = - 3 \left[{\left(x - \frac{4}{3}\right)}^{2} + \frac{47}{9}\right]$

$y = - 3 {\left(x - \frac{4}{3}\right)}^{2} - 3 \left(\frac{47}{9}\right)$

$y = - 3 {\left(x - \frac{4}{3}\right)}^{2} - \frac{47}{3}$

Determine vertex:
Vertex form is $y = a {\left(x - \textcolor{red}{h}\right)}^{2} + \textcolor{b l u e}{k}$ where $\left(\textcolor{red}{h} , \textcolor{b l u e}{k}\right)$ is the vertex of the parabola.

The vertex is therefore at $\left(\textcolor{red}{\frac{4}{3}} , \textcolor{b l u e}{- \frac{47}{3}}\right)$.

Double check with graph:
graph{y=-x^2-4x-3-2(x-3)^2 [-30, 30, -30, 5]}