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

Jan 30, 2017

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#### Explanation:

graph{-x^2-x-(3x+2)^2 [-1.5, 0.5, -0.5, 1.5]}

It looks like this with an $x$-intercept at around $x = - 0.5$. This will be helpful in checking our answer.

$1.$ First, we have to expand. So

$- {x}^{2} - x - 9 {x}^{2} - 4 - 12 x$

$= - 10 {x}^{2} - 13 x - 4$

Luckily, if we plug in $x = 0.5$ for $x$, we get a value of zero in accordance to our $y$ intercept, so we are on the right track.

$2.$ Then we complete the square so that the equation is in vertex form (to find the vertex). So:

$y = - 10 {x}^{2} - 13 x - 4$

$- \frac{y}{10} = {x}^{2} + 1.3 x + 0.4$

$- \frac{y}{10} + 0.0225 = {x}^{2} + 1.3 y + 0.4225$

$- \frac{y}{10} + 0.0225 = {\left(x + 0.65\right)}^{2}$

$- \frac{y}{10} = {\left(x + 0.65\right)}^{2} - 0.0225$

$y = - 10 {\left(x + 0.65\right)}^{2} + 0.225$

So the vertex would be $\left(- 0.65 , 0.225\right)$