Which function has an asymptote at x = 39?

Question

(1 point)
Responses

f left-parenthesis x right-parenthesis equals 7 superscript x minus 39 baseline
Image with alt text: f left-parenthesis x right-parenthesis equals 7 superscript x minus 39 baseline

f left-parenthesis x right-parenthesis equals log subscript 7 baseline left-parenthesis x minus 39 right-parenthesis
Image with alt text: f left-parenthesis x right-parenthesis equals log subscript 7 baseline left-parenthesis x minus 39 right-parenthesis

f left-parenthesis x right-parenthesis equals 7 superscript x plus 39 baseline
Image with alt text: f left-parenthesis x right-parenthesis equals 7 superscript x plus 39 baseline

f left-parenthesis x right-parenthesis equals log subscript 7 baseline left-parenthesis x plus 39 right-parenthesis

Answer 1

To determine which function has an asymptote at \( x = 39 \), we need to analyze the given functions:

1. \( f(x) = 7^x - 39 \): This function does not have a vertical asymptote at \( x = 39 \). It is an exponential function shifted vertically.

2. \( f(x) = \log_7(x - 39) \): Logarithmic functions have vertical asymptotes where their arguments equal zero. Here, the asymptote occurs when \( x - 39 = 0 \), thus at \( x = 39 \).

3. \( f(x) = 7^x + 39 \): Similar to the first function, this exponential function does not have an asymptote at \( x = 39 \).

4. \( f(x) = \log_7(x + 39) \): This function has a vertical asymptote where \( x + 39 = 0 \), at \( x = -39 \).

From the analysis, the function that has a vertical asymptote at \( x = 39 \) is:

\( f(x) = \log_7(x - 39) \).