Question

Number 9, Im not quite sure on how to do these

Answer 1

Given:

`f(x)=2(x-3)^2+1`

The equation of a parabola in vertex form is given by:

`y=a(x-h)^2+k`

The coordinates of the vertex is given by (h,k).

Hence, comparing both equations,

`\begin{gathered} f(x)=2(x-3)^2+1 \\ y=a(x-h)^2+k \\ a=2 \\ h=3 \\ k=1 \end{gathered}`

Therefore, the coordinates of the vertex is;

`(h,k)=(3,1)`