Answer
a. $x=-6,6$.
b. The multiplicity for zero $x=-6$ is $1$ and is odd.
The multiplicity for zero $x=6$ is $1$ and is odd.
c. The maximum possible number of turning points is $1$.
d. See graph
Work Step by Step
a.
Set $f(x)=0$:
$$0=x^2-36$$ $$0=(x+6)(x-6)$$ $$x+6=0$$ $$x=-6$$ $$x-6=0$$ $$x=6$$
Thus, the real zeros are $x=-6,6$.
b.
$$f(x)=(x+6)^1(x-6)^1$$
From the power of the factor $(x+6)^1$ which is $1$, the multiplicity for zero $x=-6$ is $1$ and is odd.
From the power of the factor $(x-6)^1$ which is $1$, the multiplicity for zero $x=6$ is $1$ and is odd.
c.
Since the power of the leading coefficient of the polynomial $x^2-36$ 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 $x=-6,6$, the multiplicity of zero $x=-6$ is odd since the graph crosses the $x$-axis and the multiplicity of zero $x=6$ is odd since the graph crosses the $x$-axis , and number of turning points is $1$.
