Question

Write the equation of the quadratic function with roots 0 and 2 and a vertex at (1, 5).

Answer 1

The "standard" parabola with roots 0 and 2 is

$y = (x-0)(x-2) = x^2-2x$

All multiples of this parabola, i.e.

$y=a(x^2-2x)$

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

$5 = a(1-2) \iff -a=5 \iff a=-5$

So, our claim is that the parabola

$y=-5(x^2-2x) = -5x^2+10x$

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

$y = -5x(x-2)$

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

$y(1) = -5\cdot 1 \cdot (1-2) = -5 \cdot 1 \cdot (-1) = 5$

as required.