Questions: Find the zeros for the 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)=-9(x-3)(x+1)^2 Determine the behavior of the function at each zero. Select the correct choice below and, if necessary, fill in the answer boxes within your choice. A. The graph crosses the x-axis at all zeros. B. The graph touches the x-axis and turns around at all zeros. C. The graph crosses the x-axis at x=3 and touches the x-axis and turns around at x=-1.

Solution Steps

To find the zeros of the polynomial function \( f(x) = -9(x-3)(x+1)^2 \), we need to set the function equal to zero and solve for \( x \). The zeros occur where each factor of the polynomial is zero. The multiplicity of each zero is determined by the exponent of the factor. The behavior of the graph at each zero depends on the multiplicity: if the multiplicity is odd, the graph crosses the x-axis; if even, it touches the x-axis and turns around.

To find the zeros of the polynomial function \( f(x) = -9(x-3)(x+1)^2 \), we set the function equal to zero and solve for \( x \). The zeros occur where each factor of the polynomial is zero. Thus, we solve: \[ -9(x-3)(x+1)^2 = 0 \] This gives us the zeros \( x = 3 \) and \( x = -1 \).

The multiplicity of a zero is determined by the exponent of the corresponding factor in the polynomial. For \( x = 3 \), the factor is \( (x-3) \) with multiplicity 1. For \( x = -1 \), the factor is \( (x+1)^2 \) with multiplicity 2.

The behavior of the graph at each zero depends on the multiplicity:

• If the multiplicity is odd, the graph crosses the x-axis.
• If the multiplicity is even, the graph touches the x-axis and turns around.

For \( x = 3 \) with multiplicity 1 (odd), the graph crosses the x-axis.
For \( x = -1 \) with multiplicity 2 (even), the graph touches the x-axis and turns around.

Final Answer