IDNLearn.com makes it easy to find answers and share knowledge with others. Find accurate and detailed answers to your questions from our experienced and dedicated community members.

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]