The "standard" parabola with roots 0 and 2 is

All multiples of this parabola, i.e.

have the same roots. We can choose the factor such that the parabola passes through the desided point: if we plug 1, 5 for x, y we have

So, our claim is that the parabola

has roots 0 and 2 and vertex at (1, 5).
You can easily verify this: the roots are guaranteed by the fact that we can write the equation as

The vertex must be at x=1, because it's the midpoint of the roots. Moreover, if we evaluate the function at x=1 we have

as required.