Question
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
.
. 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
.
Answers
GPT-4o mini
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.
1. **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)
2. **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.**
1. **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)
2. **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.**