Question

find the zeros for the following 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)=2(x - 4)(x + 6)^3
the zero(s) of f is/are 4, - 6.
(use a comma to separate answers as needed.)
the multiplicity at the smallest zero is 3.
(type a whole number.)
the multiplicity at the largest zero is 1.
(type a whole number.)
determine whether the graph crosses the x - axis or touches the x - axis and turns around, at the leftmost zero.
a. touches the x - axis and turns around
b. crosses the x - axis
determine whether the graph crosses the x - axis or touches the x - axis and turns around, at the rightmost zero.
a. crosses the x - axis
b. touches the x - axis and turns around

Explanation:

Step1: Find the zeros

Set $f(x)=0$, so $2(x - 4)(x + 6)^3=0$. Using the zero - product property, $x-4 = 0$ gives $x = 4$ and $(x + 6)^3=0$ gives $x=-6$.

Step2: Determine multiplicities

The factor $(x - 4)$ has exponent 1, so the zero $x = 4$ has multiplicity 1. The factor $(x + 6)$ has exponent 3, so the zero $x=-6$ has multiplicity 3.

Step3: Analyze graph behavior at zeros

If the multiplicity of a zero $r$ of a polynomial function $y = f(x)$ is odd, the graph of the function crosses the $x$-axis at $x = r$. If the multiplicity is even, the graph touches the $x$-axis and turns around at $x = r$. The left - most zero is $x=-6$ with multiplicity 3 (odd), so the graph crosses the $x$-axis at $x=-6$. The right - most zero is $x = 4$ with multiplicity 1 (odd), so the graph crosses the $x$-axis at $x = 4$.

Answer:

The zero(s) of $f$ is/are $4,-6$.
The multiplicity at the smallest zero ($x=-6$) is $3$.
The multiplicity at the largest zero ($x = 4$) is $1$.
At the leftmost zero ($x=-6$), the graph crosses the $x$-axis.
At the rightmost zero ($x = 4$), the graph crosses the $x$-axis.