What is the vertex form of y=-x^2-5x-19?

1 Answer
Apr 4, 2018

$y = - {\left(x + \frac{5}{2}\right)}^{2} - \frac{51}{4}$

Explanation:

$y = - {x}^{2} - 5 x - 19$

Factor out a -1.

$y = \textcolor{red}{-} \left({x}^{2} \textcolor{red}{+} 5 x \textcolor{red}{+} 19\right)$

Add and subtract ${\left(\frac{5}{2}\right)}^{2}$ within the grouping.

$y = - \left({x}^{2} + 5 x \textcolor{red}{+ \frac{25}{4}} + 19 \textcolor{red}{- \frac{25}{4}}\right)$

Factor the first 3 terms of the right hand side.

$y = - \left[\textcolor{red}{{\left(x + \frac{5}{2}\right)}^{2}} + 19 - \frac{25}{4}\right]$

Simplify.

$y = - \left[{\left(x + \frac{5}{2}\right)}^{2} + \textcolor{red}{\frac{51}{4}}\right]$

$y = \textcolor{red}{-} {\left(x + \frac{5}{2}\right)}^{2} \textcolor{red}{-} \frac{51}{4}$

graph{-x^2-5x-19 [-5, 1, -20, -10]}