To analyze the behavior of the polynomial function \( f(x) = 2x^4 - 6x^3 - 8x^2 \) at its zeros, we first need to find the zeros of the polynomial and determine their multiplicities.
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Find the zeros:
- We can factor the polynomial: \[ f(x) = 2x^2(x^2 - 3x - 4) \] Next, we can factor the quadratic further: \[ x^2 - 3x - 4 = (x - 4)(x + 1) \] Therefore, \[ f(x) = 2x^2(x - 4)(x + 1) \] The zeros are obtained from setting each factor equal to zero:
- \(2x^2 = 0 \rightarrow x = 0\) (multiplicity 2)
- \(x - 4 = 0 \rightarrow x = 4\) (multiplicity 1)
- \(x + 1 = 0 \rightarrow x = -1\) (multiplicity 1)
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Determine the behavior at the zeros:
- At \(x = 0\): Since the factor \(2x^2\) has even multiplicity (2), the graph will touch the x-axis at \(x = 0\) and turn around.
- At \(x = 4\): The factor \((x - 4)\) has odd multiplicity (1), so the graph will cross the x-axis at \(x = 4\).
- At \(x = -1\): The factor \((x + 1)\) also has odd multiplicity (1), so the graph will cross the x-axis at \(x = -1\).
Based on this analysis, the correct statement regarding how the graph behaves at the zeros of the polynomial is:
The graph crosses the x-axis at 4 and -1, and touches the x-axis and turns around at 0.