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![\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ f(x)=\stackrel{\stackrel{a}{\downarrow }}{1}x^2\stackrel{\stackrel{b}{\downarrow }}{-6}x\stackrel{\stackrel{c}{\downarrow }}{-6} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left( -\cfrac{-6}{2(1)}~~,~~-6-\cfrac{(-6)^2}{4(1)}\right)\implies \left( 3~,~-6-\cfrac{36}{4} \right)\implies (3~,~-6-9) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (3,-15)~\hfill](/tpl/images/0031/9900/e299a.png)
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What is the vertex form of f(x)=x^2 -6x-6...
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