Counting solutions depending on a parameter

Variations. $g'(x)=3x^2-6x=3x(x-2)$. So $g$ is increasing on $(-\infty,0]$, decreasing on $[0,2]$ and increasing on $[2,+\infty)$. We have $g(0)=-1$ (local maximum) and $g(2)=8-12-1=-5$ (local minimum), with $g\to-\infty$ at $-\infty$ and $g\to+\infty$ at $+\infty$.

Ranges of the three monotone pieces.
- on $(-\infty,0]$, $g$ takes each value of $(-\infty,-1]$ exactly once,
- on $[0,2]$, each value of $[-5,-1]$ exactly once,
- on $[2,+\infty)$, each value of $[-5,+\infty)$ exactly once.

This uses the intermediate value theorem (every value in the range is reached) and strict monotonicity (reached only once).

Counting.
- $a>-1$: only the third piece reaches $a$, one solution.
- $a=-1$: $x=0$ and one solution in $(2,+\infty)$, two solutions.
- $-5<a<-1$: each piece contributes one, three solutions.
- $a=-5$: $x=2$ and one in $(-\infty,0)$, two solutions.
- $a<-5$: only the first piece, one solution.

Remark. The count changes exactly when $a$ crosses the local extreme values $-1$ and $-5$.