Science:Math Exam Resources/Courses/MATH102/December 2015/Question 04/Solution 1

The phase diagram represents the sign of $f'(y)$.
We need to find the regions where $f'(y)$ is positive and the regions where $f'(y)$ is negative.
The only place where $y'$ changes sign are at $y=0$, $y=1$, or $y=2$. Therefore, there are four intervals to check: $(-\infty ,0)$, $(0,1)$, $(1,2)$, $(2,\infty )$.
For y < 0, we have that y, y-1, y-2, are all negative. Thus -y(y-1)(y-2) is positive.
For 0 < y < 1, we have that y is positive, but y-1 and y-2 are negative. Thus -y(y-1)(y-2) is negative.
For 1 < y < 2, we have that y and y-1are positive, but y-2 is negative. Thus -y(y-1)(y-2) is positive.
For y > 2, we have that y, y-1, and y-2 are all positive. Thus -y(y-1)(y-2) is negative.
This is consistent with (d).
answer: $\color {blue}(d)$