For a function of the form:
[tex]\begin{gathered} f(x)=ax^2+bx+c \\ \end{gathered}[/tex]
The vertex V(h,k) can be found using the following formula:
[tex]\begin{gathered} h=\frac{-b}{2a} \\ k=y(h) \\ \end{gathered}[/tex][tex]\begin{gathered} f(x)=2x^2-12x+24 \\ a=2 \\ b=-12 \\ c=24 \\ h=\frac{-(-12)}{2(2)} \\ h=\frac{12}{4}=3 \\ k=f(3)=2(3^2)-12(3)+24 \\ k=18-36+24 \\ k=6 \end{gathered}[/tex]
The vertex is (3,6) and the axis of symmetry is located at the vertex. So the axis of symmetry is x = 3
