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
, 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
, 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.

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 at 4 and −1
.

1 answer

To determine how the graph of the polynomial \( f(x) = 2x^4 - 6x^3 - 8x^2 \) behaves at its zeros, we first need to find the roots of the polynomial.

Step 1: Factoring the polynomial.

We can start by factoring out the common term: \[ f(x) = 2x^2(x^2 - 3x - 4). \]

Next, we can factor the quadratic further: \[ x^2 - 3x - 4 = (x - 4)(x + 1). \]

Thus, we have: \[ f(x) = 2x^2(x - 4)(x + 1). \]

Step 2: Identify the zeros.

The zeros of the polynomial are:

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

Step 3: Determine the behavior at the zeros:

  1. At \( x = 0 \) (multiplicity 2):

    • Since the multiplicity is even, the graph will touch the x-axis at this point and turn around.
  2. At \( x = 4 \) (multiplicity 1):

    • Since the multiplicity is odd, the graph will cross the x-axis at this point.
  3. At \( x = -1 \) (multiplicity 1):

    • Similarly, since the multiplicity is odd, the graph will also cross the x-axis at this point.

Putting this all together, we can conclude:

  • The graph touches the x-axis and turns around at \( x = 0 \).
  • The graph crosses the x-axis at \( x = 4 \) and \( x = -1 \).

Based on the analysis, the correct response is: The graph crosses the x-axis at 4 and -1, and touches the x-axis and turns around at 0.

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