Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Connect with professionals on our platform to receive accurate answers to your questions quickly and efficiently. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
To determine which function among the given options has a vertical asymptote at \(x = -1\), a horizontal asymptote at \(f(x) = 5\), and a root at \(x = -3\), we need to analyze each function one by one.
### Vertical Asymptote
A vertical asymptote at \(x = -1\) implies that the function's denominator becomes zero when \(x = -1\).
### Horizontal Asymptote
A horizontal asymptote at \(f(x) = 5\) implies that the function approaches the value 5 as \(x\) approaches \( \infty \) or \(-\infty\).
### Root
A root at \(x = -3\) means \(f(-3) = 0\).
Let's consider each function in turn:
#### Option A: \( f(x) = \frac{-10}{x+1} - 5 \)
1. Vertical Asymptote:
- The denominator \(x + 1 = 0\) when \(x = -1\). So, there is a vertical asymptote at \(x = -1\).
2. Horizontal Asymptote:
- As \(x \to \infty\) or \(x \to -\infty\), \(\frac{-10}{x+1} \to 0\), so \(f(x) \to -5\). Hence, the horizontal asymptote is \(y = -5\), not \(y = 5\).
3. Root:
- Setting \(f(-3) = 0\):
[tex]\[ \frac{-10}{-3 + 1} - 5 = 0 \implies \frac{-10}{-2} - 5 = 0 \implies 5 - 5 = 0 \][/tex]
- This is satisfied.
Even though the function has a root at \(x = -3\) and a vertical asymptote at \(x = -1\), the horizontal asymptote does not match. So, this is not the correct function.
#### Option B: \( f(x) = \frac{10}{x+1} + 5 \)
1. Vertical Asymptote:
- The denominator \(x + 1 = 0\) when \(x = -1\). So, there is a vertical asymptote at \(x = -1\).
2. Horizontal Asymptote:
- As \(x \to \infty\) or \(x \to -\infty\), \(\frac{10}{x+1} \to 0\), so \(f(x) \to 5\). Hence, the horizontal asymptote is \(y = 5\).
3. Root:
- Setting \(f(-3) = 0\):
[tex]\[ \frac{10}{-3 + 1} + 5 = 0 \implies \frac{10}{-2} + 5 = 0 \implies -5 + 5 = 0 \][/tex]
- This is satisfied.
Thus, this function has a vertical asymptote at \(x = -1\), a horizontal asymptote at \(y = 5\), and a root at \(x = -3\). Hence, this is the correct function.
#### Option C: \( f(x) = \frac{-10}{x-1} + 5 \)
1. Vertical Asymptote:
- The denominator \(x - 1 = 0\) when \(x = 1\). So, there is a vertical asymptote at \(x = 1\), not \(x = -1\).
Since the vertical asymptote does not match, we do not need to check further. This option is not correct.
#### Option D: \( f(x) = \frac{10}{x-1} + 5 \)
1. Vertical Asymptote:
- The denominator \(x - 1 = 0\) when \(x = 1\). So, there is a vertical asymptote at \(x = 1\), not \(x = -1\).
Since the vertical asymptote does not match, we do not need to check further. This option is not correct.
After analyzing all the options, we determine that Option B is the correct function:
[tex]\[ \boxed{f(x) = \frac{10}{x+1} + 5} \][/tex]
### Vertical Asymptote
A vertical asymptote at \(x = -1\) implies that the function's denominator becomes zero when \(x = -1\).
### Horizontal Asymptote
A horizontal asymptote at \(f(x) = 5\) implies that the function approaches the value 5 as \(x\) approaches \( \infty \) or \(-\infty\).
### Root
A root at \(x = -3\) means \(f(-3) = 0\).
Let's consider each function in turn:
#### Option A: \( f(x) = \frac{-10}{x+1} - 5 \)
1. Vertical Asymptote:
- The denominator \(x + 1 = 0\) when \(x = -1\). So, there is a vertical asymptote at \(x = -1\).
2. Horizontal Asymptote:
- As \(x \to \infty\) or \(x \to -\infty\), \(\frac{-10}{x+1} \to 0\), so \(f(x) \to -5\). Hence, the horizontal asymptote is \(y = -5\), not \(y = 5\).
3. Root:
- Setting \(f(-3) = 0\):
[tex]\[ \frac{-10}{-3 + 1} - 5 = 0 \implies \frac{-10}{-2} - 5 = 0 \implies 5 - 5 = 0 \][/tex]
- This is satisfied.
Even though the function has a root at \(x = -3\) and a vertical asymptote at \(x = -1\), the horizontal asymptote does not match. So, this is not the correct function.
#### Option B: \( f(x) = \frac{10}{x+1} + 5 \)
1. Vertical Asymptote:
- The denominator \(x + 1 = 0\) when \(x = -1\). So, there is a vertical asymptote at \(x = -1\).
2. Horizontal Asymptote:
- As \(x \to \infty\) or \(x \to -\infty\), \(\frac{10}{x+1} \to 0\), so \(f(x) \to 5\). Hence, the horizontal asymptote is \(y = 5\).
3. Root:
- Setting \(f(-3) = 0\):
[tex]\[ \frac{10}{-3 + 1} + 5 = 0 \implies \frac{10}{-2} + 5 = 0 \implies -5 + 5 = 0 \][/tex]
- This is satisfied.
Thus, this function has a vertical asymptote at \(x = -1\), a horizontal asymptote at \(y = 5\), and a root at \(x = -3\). Hence, this is the correct function.
#### Option C: \( f(x) = \frac{-10}{x-1} + 5 \)
1. Vertical Asymptote:
- The denominator \(x - 1 = 0\) when \(x = 1\). So, there is a vertical asymptote at \(x = 1\), not \(x = -1\).
Since the vertical asymptote does not match, we do not need to check further. This option is not correct.
#### Option D: \( f(x) = \frac{10}{x-1} + 5 \)
1. Vertical Asymptote:
- The denominator \(x - 1 = 0\) when \(x = 1\). So, there is a vertical asymptote at \(x = 1\), not \(x = -1\).
Since the vertical asymptote does not match, we do not need to check further. This option is not correct.
After analyzing all the options, we determine that Option B is the correct function:
[tex]\[ \boxed{f(x) = \frac{10}{x+1} + 5} \][/tex]
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
