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 0,6. (use a comma to separate answers as needed.)
the multiplicity at the leftmost zero is 1. (type a whole number.)
the multiplicity at the rightmost zero is 2. (type a whole number.)
determine whether the graph crosses the x - axis or touches the x - axis and turns around at the leftmost zero.
crosses the x - axis
touches the x - axis and turns around

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 $x^{2}-12x + 36=(x - 6)^{2}$ using the formula $(a - b)^2=a^{2}-2ab + b^{2}$ where $a=x$ and $b = 6$. So, $f(x)=x(x - 6)^{2}$.

Step2: Find the zeros

Set $f(x)=0$. Then $x(x - 6)^{2}=0$. By the zero - product property, if $AB = 0$, then either $A = 0$ or $B = 0$. So $x=0$ or $(x - 6)^{2}=0$. Solving $(x - 6)^{2}=0$ gives $x = 6$. The zeros are $x = 0,6$.

Step3: Determine multiplicities

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

Step4: Analyze the graph behavior

If the multiplicity of a zero $c$ of a polynomial function $y = f(x)$ is odd, the graph of the function crosses the $x$ - axis at $x = c$. If the multiplicity is even, the graph of the function touches the $x$ - axis and turns around at $x = c$. 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 multiplicity at the left - most zero ($x = 0$) is $1$.
The multiplicity at the right - most zero ($x = 6$) is $2$.
The graph crosses the $x$ - axis at the left - most zero ($x = 0$).