Answer
Option D
Work Step by Step
This quadratic's equation is f(x)=$ax^{2}+c$.
Given the equation f(x)=$x^{2}+4$.
The a value is 1 and since it is positive, the graph opens upwards(U) Therefore, it has a minimum.
We calculate the minimum.
Since there is no horizontal translation the minimum y-value is at f(0)
f(0)= $(0)^{2}$ +4
f(0) = +4
The vertex is (0, 4)
