We were given:
[tex]\begin{gathered} f(x)=-3x^2+18x-3 \\ f(x)=y \\ \Rightarrow y=-3x^2+18x-3 \\ y=-3x^2+18x-3 \\ a=-3,b=18,c=-3 \end{gathered}[/tex]
We will calculate the minimum point as shown below:
[tex]\begin{gathered} min=c-\frac{b^2}{4a} \\ min=-3-\frac{18^2}{4(-3)} \\ min=-3-\frac{324}{-12} \\ min=-3-(-27) \\ min=-3+27 \\ min=24 \\ \text{This is the maximum value (not minimum)} \\ x=-\frac{b}{2a} \\ x=-\frac{18}{2(-3)} \\ x=\frac{-18}{-6} \\ x=3 \\ \\ \therefore Maximum\text{ point is (3, 24)} \end{gathered}[/tex]
This quadratic equation opens downward because the value of ''a'' is negative. Hence, the function only has a maximum point, it does not have a minimum point
The maximum value of the function is 24 and it occurs at x equals 3
