Question

find the zeros for the given 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)=x^3 - 12x^2 + 36x
the zeros are
(use a comma to separate answers as needed.)

Explanation:

Step1: Factor the polynomial

First, factor out the greatest - common factor $x$ from $f(x)=x^{3}-12x^{2}+36x$. We get $f(x)=x(x^{2}-12x + 36)$. Then, factor the quadratic $x^{2}-12x + 36$ using the formula $(a - b)^2=a^{2}-2ab + b^{2}$, where $a = x$ and $b = 6$. So $x^{2}-12x + 36=(x - 6)^{2}$. Thus, $f(x)=x(x - 6)^{2}$.

Step2: Find the zeros

Set $f(x)=0$. Then $x(x - 6)^{2}=0$. Using the zero - product property, if $ab = 0$, then $a = 0$ or $b = 0$. So $x=0$ or $(x - 6)^{2}=0$. Solving $(x - 6)^{2}=0$ gives $x = 6$.

Step3: Determine the multiplicities

For the factor $x$, the exponent is 1, so the zero $x = 0$ has multiplicity 1. For the factor $(x - 6)^{2}$, the exponent is 2, so the zero $x = 6$ has multiplicity 2.

Step4: Analyze the graph behavior

If the multiplicity of a zero is odd, the graph of the function crosses the $x$ - axis at that zero. If the multiplicity is even, the graph of the function touches the $x$ - axis and turns around at that zero. Since the multiplicity of $x = 0$ is 1 (odd), the graph crosses the $x$ - axis at $x = 0$. Since the multiplicity of $x = 6$ is 2 (even), the graph touches the $x$ - axis and turns around at $x = 6$.

Answer:

The zeros are $0,6$. The zero $0$ has multiplicity 1 and the graph crosses the $x$-axis at $x = 0$. The zero $6$ has multiplicity 2 and the graph touches the $x$-axis and turns around at $x = 6$.