Question

Find the coordinates of the vertex of the following parabola algebraically. Write your answer as an (x,y) point.

y=-x^2-6x-16

Answer 1

`\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{-1}x^2\stackrel{\stackrel{b}{\downarrow }}{-6}x\stackrel{\stackrel{c}{\downarrow }}{-16} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ -6}{2(-1)}~~~~ ,~~~~ -16-\cfrac{ (-6)^2}{4(-1)}\right) \implies \left( - \cfrac{ -6 }{ -2 }~~,~~-16 - \cfrac{ 36 }{ -4 } \right) \\\\\\ (-3~~,~~-16+9)\implies (-3~~,~~-7)`