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

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Answer 1 · 2018-04-04T22:44:36

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

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

Factor out a -1.

$y = - (x^{2} + 5 x + 19)$ $y=color(red)(-)(x^2color(red)+5xcolor(red)+19)$

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

$y = - (x^{2} + 5 x + \frac{25}{4} + 19 - \frac{25}{4})$ $y=-(x^2+5xcolor(red)(+25/4)+19color(red)(-25/4))$

Factor the first 3 terms of the right hand side.

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

Simplify.

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

$y = - {(x + \frac{5}{2})}^{2} - \frac{51}{4}$ $y=color(red)-(x+5/2)^2color(red)-51/4$

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

What is the vertex form of y=−x2−5x−19y=-x^2-5x-19?