Answer
a. $x=-9,9$.
b. The multiplicity for zero $x=-9$ is $1$ and is odd.
The multiplicity for zero $x=9$ is $1$ and is odd.
c. The maximum possible number of turning points is $1$.
d. See graph
Work Step by Step
a.
Set $f(x)=0$:
$$0=81-x^2$$ $$0=-(x^2-81)$$ $$0=-(x+9)(x-9)$$ $$x+9=0$$ $$x=-9$$ $$x-9=0$$ $$x=9$$
Thus, the real zeros are $x=-9,9$.
b.
$$f(x)=-(x+9)^1(x-9)^1$$
From the power of the factor $(x+9)^1$ which is $1$, the multiplicity for zero $x=-9$ is $1$ and is odd.
From the power of the factor $(x-9)^1$ which is $1$, the multiplicity for zero $x=9$ is $1$ and is odd.
c.
Since the power of the leading coefficient of the polynomial $81-x^2$ 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 zeros are $x=-9,9$, the multiplicity of zero $x=-9$ is odd since the graph crosses the $x$-axis and the multiplicity of zero $x=9$ is odd since the graph crosses the $x$-axis , and number of turning points is $1$.
