## Algebra 1

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)