What is the vertex of $y = - x^{2} - 18 x + 9$ $y=-x^2 - 18x + 9$ ?

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Answer 1 · 2017-07-06T15:36:04

$y = - x^{2} - 18 x + 9$ $y=-x^2-18x+9$

Factor out the coefficient of the highest power of $x$ $x$ ( $a$ $a$ value):
$y = - [x^{2} + 18 x - 9]$ $y=-[x^2+18x-9]$

Rewrite what's inside the brackets using vertex form
$y = - [{(x + 9)}^{2} - 81 + 9]$ $y=-[(x+9)^2-81+9]$

$y = - [{(x + 9)}^{2} - 72]$ $y=-[(x+9)^2-72]$

Finally distribute the negative sign throughout the brackets
$y = - {(x + 9)}^{2} + 72$ $y=-(x+9)^2+72$

$The vertex of the parabola is at (- 9, 72)$ $color(blue)("The vertex of the parabola is at "(-9,72))$

What is the vertex of y=−x2−18x+9y=-x^2 - 18x + 9?