AlwaysLuna0
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A parabola opening up or down has vertex (0,-2) and passes though (-4,-6). Write it's equation in vertex form

Sagot :

[tex]~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\cap}\qquad \stackrel{"a"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\stackrel{vertex}{(\stackrel{h}{0}~~,~~\stackrel{k}{-2})}\qquad y=a(x-0)^2-2\qquad \qquad \stackrel{\textit{we also know that}}{x=-4\qquad y=-6} \\\\\\ -6=a(-4-0)^2-2\implies -4=a(-4)^2\implies \cfrac{-4}{(-4)^2}=a\implies -\cfrac{1}{4}=a \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y=-\cfrac{1}{4}x^2-2~\hfill[/tex]

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