Question

Sketch f(x)=2x4−6x3−8x2. How does the graph behave at the zeros of the polynomial?(1 point) Responses The graph crosses the x-axis at 4 and −1. The graph crosses the x -axis at 4 and negative 1 . The graph crosses the x-axis at 4 and −1, and touches the x-axis and turns around at 0. The graph crosses the x -axis at 4 and negative 1 , and touches the x -axis and turns around at 0. The graph crosses the x-axis and turns around at 4 and −1. The graph crosses the x -axis and turns around at 4 and negative 1 . The graph crosses the x-axis and turns around at 4 and −1, and touches the x-axis at 0. The graph crosses the x -axis and turns around at 4 and negative 1 , and touches the x -axis at 0.

Answer 1

To analyze the behavior of the polynomial \( f(x) = 2x^4 - 6x^3 - 8x^2 \) at its zeros, we first need to find its factors and identify the zeros.

1. Finding the zeros: Factor out the common term from the polynomial: \[ f(x) = 2x^2(x^2 - 3x - 4) \] Now, we can factor the quadratic: \[ x^2 - 3x - 4 = (x - 4)(x + 1) \] Therefore, the complete factorization of \( f(x) \) is: \[ f(x) = 2x^2(x - 4)(x + 1) \] The zeros of the polynomial are:

   • \( x = 0 \) (with a multiplicity of 2),
   • \( x = 4 \) (with a multiplicity of 1),
   • \( x = -1 \) (with a multiplicity of 1).

2. Behavior at the zeros:

   • At \( x = 0\), since the multiplicity is even (2), the graph touches the x-axis and turns around.
   • At \( x = 4\), since the multiplicity is odd (1), the graph crosses the x-axis.
   • At \( x = -1\), since the multiplicity is also odd (1), the graph crosses the x-axis here as well.

3. Conclusion: The graph crosses the x-axis at \( x = 4 \) and \( x = -1 \), and it touches the x-axis and turns around at \( x = 0 \).

This matches the response: The graph crosses the x-axis at 4 and -1, and touches the x-axis and turns around at 0.