Answer
See graph.
Work Step by Step
Rewrite the polynomial in factored form:
$$g(s)=s^2(s-3)$$
Finding the $s$-intercepts:
$$s=0$$
Thus, an $s$-intercept is at $(0,0)$. With multiplicity of $2$, the graph at zero of $0$ bounces off the $s$-axis.
$$s-3=0$$ $$s=3$$
Thus, an $s$intercept is at $(3,0)$. With multiplicity of $1$, the graph at zero of $1$ crosses the $s$-axis.
Notice that the leading coefficient $1$ is positive and the polynomial has a degree of $3$ which is odd, it means the end behavior at left side falls while at right side rises.
With the degree of $3$, the maximum number of turning points is $n-1=3-1=2$.
Finding one more point of the graph, set $s=2$ which is between the two zeros:
$$g(2)=(2)^3-3(2)^2=-4$$
Thus, another point is at $(2,-4)$.
Plotting the three points and drawing a curve on the points with the descriptions of the graph as indicated above, the sketch of the graph of the function is as shown.
