Question

Solve and explain question

Answer 1

Part (i)

For a function of the form
`y=a(x-h)^2+k`, the axis of symmetry is at
`x=h`.

So, in this case, the axis of symmetry is at
`x=2`.

Part (ii)

For a function of the form
`y=a(x-h)^2+k`, the vertex has coordinates
`(h,k)`.

So, in this case, the vertex has coordinates
`(2,-5)`.

Part (iii)

Setting
`x=0`,
`y=(1/2)(-2)^2-5=-3`.

So, the
`y`-intercept is at
`(0,-3)`.