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A quadratic function y=f(x) is plotted on a graph and the vertex of the resulting parabola is (-6, 4). What is the vertex of the function defined as g(x)=f(x-2)-2

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\\\\ \stackrel{vertex}{\stackrel{h}{-6}~~,~~\stackrel{k}{4}}\qquad \implies f(x)=a[x-(-6)]^2+4\implies \underline{f(x)=a(x+6)^2+4} \\\\[-0.35em] ~\dotfill[/tex]

[tex]f(x-2)=a[(x-2)+6]^2+4\implies \underline{f(x-2)=a(x+4)^2+4} \\\\\\ f(x-2)-2\implies [a(x+4)^2+4]-2\implies \underline{a(x+4)^2+2=g(x)} \\\\\\ a[x-(\stackrel{h}{-4})]^2\stackrel{k}{+2}=g(x)\qquad \stackrel{vertex}{-4~~,~~2}[/tex]

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