2003 AMC 12A Problem 25

If $a=0,$ then $f(x)=\sqrt{bx}$ has domain and range both $[0,\infty),$ so $a=0$ works. If $a\gt0,$ the domain is $({-}\infty,-b/a]\cup[0,\infty),$ while the range is $[0,\infty),$ so no such $b$ exists. If $a\lt0,$ the domain is $[0,-\frac{b}{a}]$ and the range is $\left[0,\dfrac{b}{2\sqrt{-a}}\right].$ Equating the right endpoints gives $-\dfrac ba=\dfrac{b}{2\sqrt{-a}},$ so $2\sqrt{-a}=-a,$ giving $a=-4.$ Thus there are $2$ values of $a,$ and the correct answer is C.