Answer
See graph
Work Step by Step
Notice that the leading coefficient $-\frac{1}{2}$ is negative and the polynomial has a degree of $4$ which means the end behavior of both ends of the graph fall.
With the degree of $4$, the number of turning points is $n-1=4-1=3$.
Finding the $x$-intercepts, set $f(t)=0$:
$$0=-\frac{1}{2}(t-1)^2(t+2)^2$$ $$t-1=0$$ $$t=1$$ $$t+2=0$$ $$t=-2$$
Thus, two $x$intercepts are at $(-2,0)$ and $(1,0)$.
Notice that the zeros $-2$ and $1$ have each multiplicity of $2$ which means the graph passing each of them just bounces off the $x$-axis.
Finding one more point of the graph, set $x=-\frac{1}{2}$ which is between the two zeros:
$$f\left(\frac{1}{2}\right)=-\frac{1}{2}\left(-\frac{1}{2}-1\right)^2\frac{1}{2}\left(\frac{1}{2}+2\right)^2=-\frac{81}{32}$$
Thus, another point is at $\left(-\frac{1}{2},-\frac{81}{32}\right)$.
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.
