Quadratic Function Help

Question

Not sure how to solve this one:

State the value $k$ such that the function $f(x) = -2 (x-3)^2 + k$ has $1$ root.

So I already know that this has to be a sideways parabola, but don't know how to continue.

Answer 1

If $k=0$, the only root is $x=3$

Answer 2

For $f(x)$ to have roots we must have $f(x)=0$. This leads to:$$-2(x-3)^2+k=0$$$$\therefore (x-3)^2=\frac{k}{2}$$$$\therefore x=3\pm\sqrt{\frac{k}{2}}$$This should help you see what $k$ must equal in order to obtain just one root.