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 - 4x^2 + 4x

determine the zero(s), if they exist.
the zero(s) is/are .
(type integers or decimals. use a comma to separate answers as needed.)

Explanation:

Step1: Factor the polynomial

$f(x)=x(x^{2}-4x + 4)=x(x - 2)^{2}$

Step2: Set the function equal to zero

$x(x - 2)^{2}=0$

Step3: Solve for x

$x=0$ or $(x - 2)^{2}=0$. When $(x - 2)^{2}=0$, we have $x = 2$.

Step4: Determine multiplicity

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

Answer:

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