At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
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 trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
