Question

Please help!
What is the vertex of the parabola?


y+1=−14(x−2)2

Answer 1

`\bf \textit{parabola vertex form with focus point distance} \\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ 4p(y- k)=(x- h)^2 \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ y+1=-14(x-2)^2\implies -\cfrac{1}{14}[y-(\stackrel{k}{-1})]=[x-\stackrel{h}{2}]^2~\hfill \stackrel{center}{(-2~,~-1)}`