Answer
Option C
Work Step by Step
This quadratic's equation is f(x)=$ax^{2}+c$.
Given the equation f(x)=$-0.2x^{2}+5$.
The a value is -0.2 and since it is negative, the graph opens downwards (⋂) Therefore, it has a maximum.
We calculate the maximum.
Since there is no horizontal translation the maximum y-value is at f(0)
f(0)= $-0.2(0)^{2}$ + 5
f(0) = +5
The vertex is (0, 5)
