Question

Which function has an asymptote at x = 5 and an x-intercept of (6,0)?
A.
f(x) = log(x − 5)
B.
f(x) = log(x + 5)
C.
f(x) = log x − 5
D.
f(x) = log x + 5

Answer 1

To find the function that has an asymptote at \( x = 5 \) and an x-intercept at \( (6, 0) \), let's analyze the options given.

1. **Asymptote at \( x = 5 \)**: This means that the function will approach but never reach the line \( x = 5 \). This is typical for logarithmic functions that have been transformed. Specifically, the function must be of the form \( f(x) = \log(x - 5) \) because a vertical asymptote occurs where the argument inside the log becomes zero, which happens at \( x = 5 \).

2. **X-intercept at \( (6, 0) \)**: An x-intercept of \( (6, 0) \) means that when \( x = 6 \), \( f(x) = 0 \). Therefore, we can set up the equation:
\[
f(6) = \log(6 - 5) = \log(1) = 0
\]
This is indeed satisfied by \( f(x) = \log(x - 5) \).

Now let's examine each option to see which one matches:

- **A**. \( f(x) = \log(x - 5) \)
This has an asymptote at \( x = 5 \) and an x-intercept at \( (6, 0) \). **This is a correct option.**

- **B**. \( f(x) = \log(x + 5) \)
This does not have an asymptote at \( x = 5 \) (the asymptote is vertical at \( x = -5 \) instead).

- **C**. \( f(x) = \log x - 5 \)
This is a transformed logarithm and does not have an asymptote at \( x = 5 \).

- **D**. \( f(x) = \log x + 5 \)
Similar to option C, this does not have an asymptote at \( x = 5 \).

Thus, the only function that satisfies both conditions is:

**A. \( f(x) = \log(x - 5) \)**