Question
Consider the following verbal description of a rational function. Where are the asymptotes and point of discontinuity located for the described rational function?
f of x equals the quotient of x plus four and x squared minus sixteen
(1 point)
Responses
vertical asymptote: x=4
horizontal asymptote: y=0
point of discontinuity: when x=−4
vertical asymptote: x is equal to 4 horizontal asymptote: y is equal to 0 point of discontinuity: when x is equal to negative 4
vertical asymptote: x=4
horizontal asymptote: y=12
point of discontinuity: none
vertical asymptote: x is equal to 4 horizontal asymptote: y is equal to 1 half point of discontinuity: none
vertical asymptote: x=4
horizontal asymptote: none
point of discontinuity: none
vertical asymptote: x is equal to 4 horizontal asymptote: none point of discontinuity: none
vertical asymptote: y=0
horizontal asymptote: x=4
point of discontinuity: when x=−4
vertical asymptote: y is equal to 0 horizontal asymptote: x is equal to 4 point of discontinuity: when x is equal to negative 4
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f of x equals the quotient of x plus four and x squared minus sixteen
(1 point)
Responses
vertical asymptote: x=4
horizontal asymptote: y=0
point of discontinuity: when x=−4
vertical asymptote: x is equal to 4 horizontal asymptote: y is equal to 0 point of discontinuity: when x is equal to negative 4
vertical asymptote: x=4
horizontal asymptote: y=12
point of discontinuity: none
vertical asymptote: x is equal to 4 horizontal asymptote: y is equal to 1 half point of discontinuity: none
vertical asymptote: x=4
horizontal asymptote: none
point of discontinuity: none
vertical asymptote: x is equal to 4 horizontal asymptote: none point of discontinuity: none
vertical asymptote: y=0
horizontal asymptote: x=4
point of discontinuity: when x=−4
vertical asymptote: y is equal to 0 horizontal asymptote: x is equal to 4 point of discontinuity: when x is equal to negative 4
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Answers
GPT-4o mini
To analyze the function \( f(x) = \frac{x + 4}{x^2 - 16} \), we need to determine the asymptotes and points of discontinuity.
1. **Factor the denominator**:
\[
x^2 - 16 = (x - 4)(x + 4)
\]
This means that the function can be simplified as follows:
\[
f(x) = \frac{x + 4}{(x - 4)(x + 4)}
\]
2. **Identify the point of discontinuity**:
The function has a factor of \( x + 4 \) in both the numerator and denominator. This indicates a removable discontinuity at \( x = -4 \). Therefore, there is a point of discontinuity at \( x = -4 \).
3. **Identify vertical asymptotes**:
Vertical asymptotes occur where the denominator is zero but the numerator is not zero. Thus:
- The denominator \( (x - 4)(x + 4) = 0 \) at \( x = 4 \) and \( x = -4 \).
- Since \( x = -4 \) is a removable discontinuity (as discussed), the vertical asymptote will only occur at \( x = 4 \).
4. **Identify horizontal asymptotes**:
To find horizontal asymptotes, we look at the degrees of the numerator and denominator. In this case, the degree of the numerator is 1 and the degree of the denominator is 2. For rational functions:
- If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \( y = 0 \).
Now, summarizing this information:
- **Vertical asymptote**: \( x = 4 \)
- **Horizontal asymptote**: \( y = 0 \)
- **Point of discontinuity**: \( x = -4 \)
Given this analysis, the correct response is:
**vertical asymptote: \( x = 4 \); horizontal asymptote: \( y = 0 \); point of discontinuity: when \( x = -4 \)**.
1. **Factor the denominator**:
\[
x^2 - 16 = (x - 4)(x + 4)
\]
This means that the function can be simplified as follows:
\[
f(x) = \frac{x + 4}{(x - 4)(x + 4)}
\]
2. **Identify the point of discontinuity**:
The function has a factor of \( x + 4 \) in both the numerator and denominator. This indicates a removable discontinuity at \( x = -4 \). Therefore, there is a point of discontinuity at \( x = -4 \).
3. **Identify vertical asymptotes**:
Vertical asymptotes occur where the denominator is zero but the numerator is not zero. Thus:
- The denominator \( (x - 4)(x + 4) = 0 \) at \( x = 4 \) and \( x = -4 \).
- Since \( x = -4 \) is a removable discontinuity (as discussed), the vertical asymptote will only occur at \( x = 4 \).
4. **Identify horizontal asymptotes**:
To find horizontal asymptotes, we look at the degrees of the numerator and denominator. In this case, the degree of the numerator is 1 and the degree of the denominator is 2. For rational functions:
- If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \( y = 0 \).
Now, summarizing this information:
- **Vertical asymptote**: \( x = 4 \)
- **Horizontal asymptote**: \( y = 0 \)
- **Point of discontinuity**: \( x = -4 \)
Given this analysis, the correct response is:
**vertical asymptote: \( x = 4 \); horizontal asymptote: \( y = 0 \); point of discontinuity: when \( x = -4 \)**.