Question

Find the equation of the quadratic function with zeros 10 and 12 and a vertex at (11, -2).

Answer 1

The equation we seek is
$$
f(x)=(x-a)(x-b)-c
$$
Using the knowns:
$$
f(10)=0=(10-a)(10-b)-c\\
f(12)=0=(12-a)(12-b)-c\\
f(11)=2=(11-a)(11-b)-c
$$
We have 3 equations and 3 unknowns.

There are 2 options. 1 solution or no solutions.

I found no solutions.

If we didn't have the vertex requirement, then the solution would be f(x)=(x-10)(x+12). This parabola has vertex <11,-1>. Moving the parabola down so that vertex is at <11,-2> moves the zeros and hence there is no solution with vertex at this location while keeping the zeros at x=10 and at x=12.

But moving the parabola down so that the