Which of the following functions has a vertical asymptote at [tex]$x=-1$[/tex], a horizontal asymptote at [tex]f(x)=5[/tex], and a root at [tex]x=-3[/tex]?

A. [tex]f(x)=\frac{-10}{x+1}-5[/tex]

B. [tex]f(x)=\frac{10}{x+1}+5[/tex]

C. [tex]f(x)=\frac{-10}{x-1}+5[/tex]

D. [tex]f(x)=\frac{10}{x-1}+5[/tex]



Answer :

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]

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