Question 132014
{{{f(x)=(1/4)(x+8)^2+7}}} Start with the given equation



{{{f(x)=(1/4)(x-(-8))^2+7}}} Rewrite {{{x+8}}} as {{{x-(-8)}}}


Notice how {{{f(x)=(1/4)(x-(-8))^2+7}}} is in vertex form {{{f(x)=a(x-h)^2+k}}} where h and k form the vertex. In other words, the vertex is the point (h,k)






Since h=-8 and k=7, the vertex is the point (-8,7)



In general, the line of symmetry is the equation {{{x=h}}}. So in this case the line of symmetry is {{{x=-8}}}