Question

I have failed this question a lot and I don't know what I'm doing wrong,
Find the vertex of:

Answer 1

`\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{-15}x^2\stackrel{\stackrel{b}{\downarrow }}{-60}x\stackrel{\stackrel{c}{\downarrow }}{-67} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ -60}{2(-15)}~~~~ ,~~~~ -67-\cfrac{ (-60)^2}{4(-15)}\right) \implies \left( - \cfrac{ -60 }{ -30 }~~,~~-67 - \cfrac{ 3600 }{ -60 } \right) \\\\\\ (-2~~,~~-67+60)\implies {\Large \begin{array}{llll} (-2~~,~~-7) \end{array}}`

Answer 2

(-2,-7)

Explanation:

Rewrite in vertex form and use this form to find the vertex (h,k).