We are given the function
f(x)=0.25(2x−15)2+150, and we need to determine if it has an value. Notice it is not given in vertex form, since it contains
2x within the square. Let's rewrite it.
f(x)=0.25(2x−15)2+150
f(x)=0.25(2x−2⋅7.5)2+150
f(x)=0.25(2(x−7.5))2+150
f(x)=0.25⋅22(x−7.5)2+150
f(x)=(x−7.5)2+150
The final equation is written in
f(x)=a(x−h)2+k, where
a=1. Since
a>0, the opens upward and has a minimum value of
k=150.