We were given:
![\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}](https://img.qammunity.org/2023/formulas/mathematics/college/hgpwu80kjku9lkshesm96apygmt2f4edno.png)
We will calculate the minimum point as shown below:
![\begin{gathered} min=c-(b^2)/(4a) \\ min=-3-(18^2)/(4(-3)) \\ min=-3-(324)/(-12) \\ min=-3-(-27) \\ min=-3+27 \\ min=24 \\ \text{This is the maximum value (not minimum)} \\ x=-(b)/(2a) \\ x=-(18)/(2(-3)) \\ x=(-18)/(-6) \\ x=3 \\ \\ \therefore Maximum\text{ point is (3, 24)} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/mytacl4977a3sufg8qccu5jonq8n9z46b1.png)
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