Question

Sketch f(x)=2x
4
−6x
3
−8x
2
. How does the graph behave at the zeros of the polynomial?
A. The graph crosses the x-axis at 4 and -1, and touches the x-axis and turns around at 0.
B. The graph crosses the x-axis at 4 and -1.
C. The graph crosses the x-axis and turns around at 4 and -1.
D. The graph crosses the x-axis and turns around at 4 and -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 the zeros (roots) of this polynomial.

### Step 1: Factor the polynomial

First, we can factor out the common term:

\[
f(x) = 2x^2(x^2 - 3x - 4)
\]

Next, we can factor the quadratic \( x^2 - 3x - 4 \):

\[
x^2 - 3x - 4 = (x - 4)(x + 1)
\]

So, the fully factored form of the polynomial is:

\[
f(x) = 2x^2(x - 4)(x + 1)
\]

### Step 2: Find the zeros

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

1. From \( 2x^2 = 0 \) → \( x = 0 \) (with a multiplicity of 2)
2. From \( (x - 4) = 0 \) → \( x = 4 \) (with a multiplicity of 1)
3. From \( (x + 1) = 0 \) → \( x = -1 \) (with a multiplicity of 1)

So, the zeros are at \( x = 0 \), \( x = 4 \), and \( x = -1 \).

### Step 3: Determine the behavior at each zero

- **At \( x = 0 \)**:
- The zero has a multiplicity of \( 2 \) (even). This means the graph will touch the x-axis at \( x = 0 \) and turn around.

- **At \( x = 4 \)**:
- The zero has a multiplicity of \( 1 \) (odd). This means the graph will cross the x-axis at \( x = 4 \).

- **At \( x = -1 \)**:
- The zero has a multiplicity of \( 1 \) (odd). This means the graph will also cross the x-axis at \( x = -1 \).

### Conclusion

Based on our analysis, the correct option describing the behavior of the graph at the zeros is:

**A. The graph crosses the x-axis at 4 and -1, and touches the x-axis and turns around at 0.**