Questions: How does the graph behave at the zeros of f(x)=x(-x+6)(-x-1)^2 ? (1 point) The graph crosses the x-axis at 0 and 6, and touches the x-axis and turns around at -1. The graph crosses the x-axis at 0 and -6, and touches the x-axis and turns around at 1. The graph touches the x-axis and turns around at 0 and 6, and touches the x-axis at 1. The graph crosses the x-axis at 6, and touches the x-axis and turns around at -1.

Solution Steps

To determine the behavior of the graph at the zeros of the function \( f(x) = x(-x+6)(-x-1)^2 \), we need to identify the zeros and analyze the multiplicity of each zero. The zeros are the values of \( x \) that make \( f(x) = 0 \). The behavior at each zero depends on whether the zero has an odd or even multiplicity:

• If the zero has an odd multiplicity, the graph crosses the x-axis at that point.
• If the zero has an even multiplicity, the graph touches the x-axis and turns around at that point.

The given function is \( f(x) = x(-x + 6)(-x - 1)^2 \).

To find the zeros, we set \( f(x) = 0 \):

\[ x(-x + 6)(-x - 1)^2 = 0 \]

This equation is satisfied when any of the factors is zero:

1. \( x = 0 \)
2. \( -x + 6 = 0 \) which simplifies to \( x = 6 \)
3. \( (-x - 1)^2 = 0 \) which simplifies to \( x = -1 \)

So, the zeros are \( x = 0 \), \( x = 6 \), and \( x = -1 \).

The factor \( x \) is linear (degree 1), so the graph crosses the \( x \)-axis at \( x = 0 \).

The factor \( -x + 6 \) is also linear (degree 1), so the graph crosses the \( x \)-axis at \( x = 6 \).

The factor \( (-x - 1)^2 \) is quadratic (degree 2), so the graph touches the \( x \)-axis and turns around at \( x = -1 \).

Final Answer