The "standard" parabola with roots 0 and 2 is
[tex]y = (x-0)(x-2) = x^2-2x[/tex]
All multiples of this parabola, i.e.
[tex]y=a(x^2-2x)[/tex]
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
[tex]5 = a(1-2) \iff -a=5 \iff a=-5[/tex]
So, our claim is that the parabola
[tex]y=-5(x^2-2x) = -5x^2+10x[/tex]
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
[tex]y = -5x(x-2)[/tex]
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
[tex]y(1) = -5\cdot 1 \cdot (1-2) = -5 \cdot 1 \cdot (-1) = 5[/tex]
as required.
