Question

Sketch  f(x)=2x4−6x3−8x2 . How does the graph behave at the zeros of the polynomial?(1 point)ResponsesThe 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, 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 −

Answer 1

To analyze the behavior of the graph of the polynomial function \( f(x) = 2x^4 - 6x^3 - 8x^2 \), we first need to find its zeros. The zeros of the polynomial can be found by factoring it or using the Rational Root Theorem.

1. **Finding the Zeros:**
Start by factoring out the common term:
\[
f(x) = 2x^2(x^2 - 3x - 4)
\]

Now, we can factor \( x^2 - 3x - 4 \):
\[
x^2 - 3x - 4 = (x - 4)(x + 1)
\]
So, we rewrite \( f(x) \):
\[
f(x) = 2x^2(x - 4)(x + 1)
\]

This gives us the zeros:
\[
x = 0 \quad (multiplicity \ 2), \quad x = 4 \quad (multiplicity \ 1), \quad x = -1 \quad (multiplicity \ 1)
\]

2. **Behavior at the Zeros:**
- At **\( x = 0 \)** (multiplicity 2): The graph touches the x-axis and turns around because the polynomial has an even multiplicity at this zero.
- At **\( x = 4 \)** (multiplicity 1): The graph crosses the x-axis since the polynomial has an odd multiplicity at this zero.
- At **\( x = -1 \)** (multiplicity 1): Similar to \( x = 4 \), the graph crosses the x-axis because this zero also has an odd multiplicity.

**Summary of behavior at the zeros:**
- Touching and turning around at \( x = 0 \)
- Crossing at \( x = 4 \)
- Crossing at \( x = -1 \)

From the responses given in the question, the appropriate choice is:
- **The graph crosses the x-axis at 4 and -1, and touches the x-axis and turns around at 0.**