Answer
a. $t=3$.
b. The multiplicity for zero $t=3$ is $2$ and is even.
c. The maximum possible number of turning points is $1$.
d. See graph
Work Step by Step
a.
Set $h(t)=0$:
$$0=t^2-6t+9$$ $$0=(t-3)^2$$ $$t-3=0$$ $$t=3$$ Thus, the real zero is $t=3$.
b.
$$h(t)=(t-3)^2$$ From the power of the factor $(t-3)^2$ which is $2$, the multiplicity for zero $t=3$ is $2$ and is even.
c.
Since the power of the leading coefficient of the polynomial $t^2-6t+9$ is $2$, the maximum possible number of turning points is: $$n-1=2-1=1$$
d. The graph of the function is as shown and it shows the zero is $t=3$, the multiplicity of zero $t=3$ is even since the graph bounces off the $x$-axis, and number of turning points is $1$.
