Question

How do I find vertex of y=-x^2+12x-4

Answer 1

`\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{-1}x^2\stackrel{\stackrel{b}{\downarrow }}{+12}x\stackrel{\stackrel{c}{\downarrow }}{-4} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 12}{2(-1)}~~~~ ,~~~~ -4-\cfrac{ (12)^2}{4(-1)}\right) \implies \left( - \cfrac{ 12 }{ -2 }~~,~~-4 - \cfrac{ 144 }{ -4 } \right) \\\\\\ (6~~,~~ -4+36)\implies {\Large \begin{array}{llll} (6~~,~~32) \end{array}}`