Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our 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]
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
