Asked by burnout
Algebra 2A Semester Exam
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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
.
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
.
The graph crosses the x -axis and turns around at 4 and negative 1 .
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.
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Answers
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Answered by
GPT-4o mini
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 of the polynomial.
1. **Finding Zeros**:
We can factor the polynomial \( f(x) \).
\[
f(x) = 2x^2 (x^2 - 3x - 4)
\]
Now, we can factor \( x^2 - 3x - 4 \) further:
\[
x^2 - 3x - 4 = (x - 4)(x + 1)
\]
Thus,
\[
f(x) = 2x^2 (x - 4)(x + 1)
\]
2. **Identifying the Zeros**:
Setting \( f(x) = 0 \), we find the zeros:
- From \( 2x^2 = 0 \): \( x = 0 \) (with multiplicity 2)
- From \( x - 4 = 0 \): \( x = 4 \) (with multiplicity 1)
- From \( x + 1 = 0 \): \( x = -1 \) (with multiplicity 1)
3. **Behavior of the Graph at Zeros**:
- At \( x = 0 \): Since it has an even multiplicity (2), the graph will **touch** the x-axis at this point and turn around.
- At \( x = 4 \) and \( x = -1 \): Both have odd multiplicities (1), so the graph will **cross** the x-axis at these points.
Thus, to summarize the behavior of the polynomial:
- The graph **touches the x-axis and turns around at \( x = 0 \)**.
- The graph **crosses the x-axis at \( x = 4 \) and \( x = -1 \)**.
Given these analyses, the correct response to 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. **Finding Zeros**:
We can factor the polynomial \( f(x) \).
\[
f(x) = 2x^2 (x^2 - 3x - 4)
\]
Now, we can factor \( x^2 - 3x - 4 \) further:
\[
x^2 - 3x - 4 = (x - 4)(x + 1)
\]
Thus,
\[
f(x) = 2x^2 (x - 4)(x + 1)
\]
2. **Identifying the Zeros**:
Setting \( f(x) = 0 \), we find the zeros:
- From \( 2x^2 = 0 \): \( x = 0 \) (with multiplicity 2)
- From \( x - 4 = 0 \): \( x = 4 \) (with multiplicity 1)
- From \( x + 1 = 0 \): \( x = -1 \) (with multiplicity 1)
3. **Behavior of the Graph at Zeros**:
- At \( x = 0 \): Since it has an even multiplicity (2), the graph will **touch** the x-axis at this point and turn around.
- At \( x = 4 \) and \( x = -1 \): Both have odd multiplicities (1), so the graph will **cross** the x-axis at these points.
Thus, to summarize the behavior of the polynomial:
- The graph **touches the x-axis and turns around at \( x = 0 \)**.
- The graph **crosses the x-axis at \( x = 4 \) and \( x = -1 \)**.
Given these analyses, the correct response to 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.**
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