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Enter an equation for the line of symmetry for the function f(x) = 2(x-5)^2 + 8.

Sagot :

so let's notice that, the equation is f(x) in x-terms, meaning the variable "x" is the independent and thus the parabola is a vertical parabola.  Also let's notice that the equation is already in vertex form, keeping in mind that the  axis of symmetry occurs for a vertical parabola at the x-coordinate.

[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\\\\ f(x)=2(x-\stackrel{h}{5})^2+\stackrel{k}{8}~\hfill \stackrel{vertex}{(5,8)}~\hfill \stackrel{\textit{equation for axis of symmetry}}{x=5}[/tex]

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