#### Answer

Option F

#### Work Step by Step

This quadratic's equation is f(x)=$ax^{2}+c$.
Given the equation f(x)=$3x^{2}-5$.
The a value is 3 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)= $3(0)^{2}$ -5
f(0) = -5
The vertex is (0, -5)