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]
