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 and turns around at 4 and −1, and touches the x-axis at 0.
C
The graph crosses the x-axis at 4 and −1.
D
The graph crosses the x-axis and turns around at 4 and −1.
. 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 and turns around at 4 and −1, and touches the x-axis at 0.
C
The graph crosses the x-axis at 4 and −1.
D
The graph crosses the x-axis and turns around at 4 and −1.
Answers
GPT-4o mini
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 zeros of the polynomial.
1. **Finding the Zeros:**
We can factor the polynomial to find its zeros:
\[
f(x) = 2x^2(x^2 - 3x - 4)
\]
Next, we factor the quadratic \( x^2 - 3x - 4 \):
\[
x^2 - 3x - 4 = (x - 4)(x + 1)
\]
Therefore, we can express \( f(x) \) as:
\[
f(x) = 2x^2 (x - 4)(x + 1)
\]
The zeros of the function are found by setting \( f(x) = 0 \):
\[
2x^2 = 0 \quad \Rightarrow \quad x = 0 \quad (\text{multiplicity 2})
\]
\[
x - 4 = 0 \quad \Rightarrow \quad x = 4 \quad (\text{multiplicity 1})
\]
\[
x + 1 = 0 \quad \Rightarrow \quad x = -1 \quad (\text{multiplicity 1})
\]
So the zeros are \( x = 0 \) (multiplicity 2), \( x = 4 \) (multiplicity 1), and \( x = -1 \) (multiplicity 1).
2. **Behavior at the Zeros:**
- At \( x = 0 \) (multiplicity 2), the graph will **touch the x-axis** and turn around (since even multiplicities mean the graph doesn't cross the x-axis).
- At \( x = 4 \) (multiplicity 1), the graph will **cross the x-axis** (since odd multiplicities mean the graph does cross the axis).
- At \( x = -1 \) (multiplicity 1), the graph will also **cross the x-axis**.
Given this analysis, the correct description of the graph's behavior 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.
1. **Finding the Zeros:**
We can factor the polynomial to find its zeros:
\[
f(x) = 2x^2(x^2 - 3x - 4)
\]
Next, we factor the quadratic \( x^2 - 3x - 4 \):
\[
x^2 - 3x - 4 = (x - 4)(x + 1)
\]
Therefore, we can express \( f(x) \) as:
\[
f(x) = 2x^2 (x - 4)(x + 1)
\]
The zeros of the function are found by setting \( f(x) = 0 \):
\[
2x^2 = 0 \quad \Rightarrow \quad x = 0 \quad (\text{multiplicity 2})
\]
\[
x - 4 = 0 \quad \Rightarrow \quad x = 4 \quad (\text{multiplicity 1})
\]
\[
x + 1 = 0 \quad \Rightarrow \quad x = -1 \quad (\text{multiplicity 1})
\]
So the zeros are \( x = 0 \) (multiplicity 2), \( x = 4 \) (multiplicity 1), and \( x = -1 \) (multiplicity 1).
2. **Behavior at the Zeros:**
- At \( x = 0 \) (multiplicity 2), the graph will **touch the x-axis** and turn around (since even multiplicities mean the graph doesn't cross the x-axis).
- At \( x = 4 \) (multiplicity 1), the graph will **cross the x-axis** (since odd multiplicities mean the graph does cross the axis).
- At \( x = -1 \) (multiplicity 1), the graph will also **cross the x-axis**.
Given this analysis, the correct description of the graph's behavior 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.