Let's find the critical points:
[tex]\begin{gathered} f^{\prime}(x)=0 \\ -9x^2+12=0 \\ x^2=\frac{12}{9} \\ x=\pm\frac{2\sqrt[]{3}}{3}\approx\pm1.155 \end{gathered}[/tex]Now:
[tex]\begin{gathered} \text{for:} \\ x<-1.155 \\ f^{\prime}(-2)=-24<0 \\ \text{for:} \\ x>-1.155 \\ f^{\prime}(-1)=3>0 \\ \text{for:} \\ x<1.155 \\ f^{\prime}(1)=3>0 \\ \text{for:} \\ x>1.155 \\ f(2)=-12<0 \end{gathered}[/tex]Therefore, the vertex of the function are:
[tex]\begin{gathered} x=-1.155,f(-1.155)=-19.238 \\ \text{and} \\ x=1.155,f(1.155)=-0.762 \end{gathered}[/tex]
