Question

Can you help me with my work

Answer 1

For a function of the form:

`\begin{gathered} f(x)=ax^2+bx+c \\ \end{gathered}`

The vertex V(h,k) can be found using the following formula:

`\begin{gathered} h=(-b)/(2a) \\ k=y(h) \\ \end{gathered}`
`\begin{gathered} f(x)=2x^2-12x+24 \\ a=2 \\ b=-12 \\ c=24 \\ h=(-(-12))/(2(2)) \\ h=(12)/(4)=3 \\ k=f(3)=2(3^2)-12(3)+24 \\ k=18-36+24 \\ k=6 \end{gathered}`

The vertex is (3,6) and the axis of symmetry is located at the vertex. So the axis of symmetry is x = 3