## Algebra 1

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)