Question
Graph f(x)=2x4−6x3−8x2 . How does the graph behave at the zeros of the polynomial?(1 point) Responses The graph crosses at the x-axis at 4 and -1, and touches the x-axis and turns around at 0. The graph crosses at the x-axis at 4 and -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 -1, and touches the x-axis at 0. The graph crosses the x-axis at 4 and -1. The graph crosses the x-axis at 4 and -1. The graph crosses the x-axis and turns around at 4 and -1.
Answers
Answered by
GPT-4o mini
To analyze the behavior of the graph of the polynomial function \( f(x) = 2x^4 - 6x^3 - 8x^2 \) at its zeros, we first need to find its zeros.
### Step 1: Factor the polynomial
Let's factor the polynomial:
\[
f(x) = 2x^2(x^2 - 3x - 4)
\]
Next, we can further factor the quadratic \( x^2 - 3x - 4 \):
To factor \( x^2 - 3x - 4 \), we look for two numbers that multiply to -4 and add to -3. These numbers are -4 and 1:
\[
x^2 - 3x - 4 = (x - 4)(x + 1)
\]
Thus, we can factor \( f(x) \) completely as:
\[
f(x) = 2x^2(x - 4)(x + 1)
\]
### Step 2: Identify the zeros
Setting \( f(x) = 0 \) gives us the zeros of the polynomial:
1. From \( 2x^2 = 0 \), we get \( x = 0 \) (with multiplicity 2).
2. From \( x - 4 = 0 \), we get \( x = 4 \) (with multiplicity 1).
3. From \( x + 1 = 0 \), we get \( x = -1 \) (with multiplicity 1).
### Step 3: Analyze the behavior at the zeros
- **At \( x = 0 \)**: Since the zero has even multiplicity (2), the graph touches the x-axis at this point and turns around.
- **At \( x = 4 \)**: The zero has odd multiplicity (1), so the graph crosses the x-axis at this point.
- **At \( x = -1 \)**: Similarly, since this zero also has odd multiplicity (1), the graph crosses the x-axis at this point.
### Conclusion
Putting this analysis together, we see that:
- The graph touches the x-axis and turns around at \( x = 0 \).
- The graph crosses the x-axis at \( x = 4 \) and \( x = -1 \).
Among the options provided, the correct response is:
**The graph crosses at the x-axis at 4 and -1, and touches the x-axis and turns around at 0.**
### Step 1: Factor the polynomial
Let's factor the polynomial:
\[
f(x) = 2x^2(x^2 - 3x - 4)
\]
Next, we can further factor the quadratic \( x^2 - 3x - 4 \):
To factor \( x^2 - 3x - 4 \), we look for two numbers that multiply to -4 and add to -3. These numbers are -4 and 1:
\[
x^2 - 3x - 4 = (x - 4)(x + 1)
\]
Thus, we can factor \( f(x) \) completely as:
\[
f(x) = 2x^2(x - 4)(x + 1)
\]
### Step 2: Identify the zeros
Setting \( f(x) = 0 \) gives us the zeros of the polynomial:
1. From \( 2x^2 = 0 \), we get \( x = 0 \) (with multiplicity 2).
2. From \( x - 4 = 0 \), we get \( x = 4 \) (with multiplicity 1).
3. From \( x + 1 = 0 \), we get \( x = -1 \) (with multiplicity 1).
### Step 3: Analyze the behavior at the zeros
- **At \( x = 0 \)**: Since the zero has even multiplicity (2), the graph touches the x-axis at this point and turns around.
- **At \( x = 4 \)**: The zero has odd multiplicity (1), so the graph crosses the x-axis at this point.
- **At \( x = -1 \)**: Similarly, since this zero also has odd multiplicity (1), the graph crosses the x-axis at this point.
### Conclusion
Putting this analysis together, we see that:
- The graph touches the x-axis and turns around at \( x = 0 \).
- The graph crosses the x-axis at \( x = 4 \) and \( x = -1 \).
Among the options provided, the correct response is:
**The graph crosses at the x-axis at 4 and -1, and touches the x-axis and turns around at 0.**
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