Question

find the zeros for the polynomial function and give the multiplicity for each zero. state whether the graph crosses the x - axis or touches the x - axis and turns around at each zero. f(x)=-9(x - 7)(x + 6)^3 determine the zero(s). the zero(s) is/are . (type integers or decimals. use a comma to separate answers as needed.)

Explanation:

Step1: Set the function equal to zero

Set $f(x)=-9(x - 7)(x + 6)^{3}=0$.

Step2: Use the zero - product property

If $ab = 0$, then either $a = 0$ or $b = 0$. So we set $x-7=0$ and $(x + 6)^{3}=0$.
For $x-7=0$, we get $x=7$.
For $(x + 6)^{3}=0$, we get $x=-6$.

Step3: Determine the multiplicities

The factor $(x - 7)$ has an exponent of 1, so the zero $x = 7$ has multiplicity 1.
The factor $(x + 6)^{3}$ has an exponent of 3, so the zero $x=-6$ has multiplicity 3.

Step4: Determine the behavior at the zeros

If the multiplicity of a zero is odd, the graph crosses the x - axis at that zero. If the multiplicity is even, the graph touches the x - axis and turns around at that zero.
For $x = 7$ (multiplicity 1, odd), the graph crosses the x - axis.
For $x=-6$ (multiplicity 3, odd), the graph crosses the x - axis.

Answer:

The zeros are $7,-6$. The multiplicity of $7$ is 1 and the graph crosses the x - axis at $x = 7$. The multiplicity of $-6$ is 3 and the graph crosses the x - axis at $x=-6$.