Use the discriminant to show that $mx−y + m^2 = 0$ touches the parabola $x^2 =−4y$, for all values of m. Use the discriminant to show that $mx−y + m^2 = 0$ touches the parabola $x^2 =−4y$, for all values of m.
I attempted to solve by letting them both equal each other, but it didn't work. How do I do this question? 
Thank You in advance.

So the two functions are $y=-\frac{1}{4}x^2$ and $y=mx+m^2$.
Setting them equal, we have $-\frac{1}{4}x^2=mx+m^2$
which rearranges to $0=x^2+4mx+4m^2$
then you can use the quadratic formula (or simply complete the square) to get the result. I suspect that this is where you are supposed to use the discriminant, since in the case the discriminant would be $16m^2-16m^2=0$, showing that there is always one solution.
