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 - 6x^2 + 9x. 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 $f(x)=x^{3}-6x^{2}+9x$. We get $f(x)=x(x^{2}-6x + 9)$. Then, factor the quadratic $x^{2}-6x + 9$ using the formula $(a - b)^2=a^{2}-2ab + b^{2}$, where $a=x$ and $b = 3$. So $x^{2}-6x + 9=(x - 3)^{2}$. Thus, $f(x)=x(x - 3)^{2}$.

Step2: Find the zeros

Set $f(x)=0$. Then $x(x - 3)^{2}=0$. Using the zero - product property, if $ab = 0$, then $a = 0$ or $b = 0$. So $x=0$ or $(x - 3)^{2}=0$. Solving $(x - 3)^{2}=0$ gives $x = 3$.

Step3: Determine the multiplicities

For the factor $x$, the exponent is 1, so the zero $x = 0$ has multiplicity 1. For the factor $(x - 3)^{2}$, the exponent is 2, so the zero $x = 3$ 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 $x = 3$ is 2 (even), the graph touches the $x$ - axis and turns around at $x = 3$.

Answer:

The zeros are $0,3$. The multiplicity of $0$ is $1$ and the graph crosses the $x$ - axis at $x = 0$. The multiplicity of $3$ is $2$ and the graph touches the $x$ - axis and turns around at $x = 3$.