jmc

The domain of $f$ is $\{x|a{x}^2+bx\ge 0\}$. If $a=0$, then for every positive value of $b$, the domain and range of $f$ are each equal to the interval $[0,\infty )$, so $0$ is a possible value of $a$.

If $a\ne 0$, the graph of $y=a{x}^2+bx$ is a parabola with $x$-intercepts at $x=0$ and $x=-b/a$.

If $a>0$, the domain of $f$ is $(-\infty ,-b/a]\cup [0,\infty )$, but the range of $f$ cannot contain negative numbers.

If $a<0$, the domain of $f$ is $[0,-b/a]$. The maximum value of $f$ occurs halfway between the $x$-intercepts, at $x=-b/2a$, and $$f\left(-\frac{b}{2a}\right)=\sqrt{a\left(\frac{b^2}{4a^2}\right)+b\left(-\frac{b}{2a}\right)}=\frac{b}{2\sqrt{-a}}.$$Hence, the range of $f$ is $[0,b/2\sqrt{-a}]$. For the domain and range to be equal, we must have

$$-\frac{b}{a}=\frac{b}{2\sqrt{-a}}\text{so}2\sqrt{-a}=-a.$$The only solution is $a=-4$. Thus, there are $\boxed{2}$ possible values of $a$, and they are $a=0$ and $a=-4$.