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

Mar 30, 2018

Vertex is $\left(- \frac{5}{6} , - \frac{71}{12}\right)$

#### Explanation:

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

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

= $- 3 {x}^{2} - 5 x - 8$

= $- 3 \left({x}^{2} + \frac{5}{3} x + {\left(\frac{5}{6}\right)}^{2}\right) - \left(- 3\right) {\left(\frac{5}{6}\right)}^{2} - 8$

= $- 3 {\left(x + \frac{5}{6}\right)}^{2} + \frac{25}{12} - 8$

= $- 3 {\left(x + \frac{5}{6}\right)}^{2} - \frac{71}{12}$

Now it is in vertex form $y = a {\left(x - h\right)}^{2} + k$

and vertex is $\left(- \frac{5}{6} , - \frac{71}{12}\right)$

graph{-(x+2)^2-2x^2-x-4 [-6.876, 3.124, -8.7, -3.7]}