Question

graph the function $f(x) = -log_3 (x - 4)$ on the axes below. you must plot the asymptote and any two points with integer coordinates.
asymptote:
vertical
horizontal

Explanation:

Step1: Determine the asymptote type

For a logarithmic function of the form \( y = \log_b(x - h) + k \), the vertical asymptote occurs where the argument of the logarithm is zero, i.e., \( x - h = 0 \). In the function \( f(x)=-\log_3(x - 4) \), the argument is \( x - 4 \). Setting \( x - 4 = 0 \) gives \( x = 4 \). Logarithmic functions have vertical asymptotes (they do not have horizontal asymptotes in the typical sense for basic log functions), so the asymptote is vertical.

Step2: Find two points with integer coordinates

• Let's find when \( x - 4 = 1 \) (since \( \log_b(1)=0 \) for any \( b>0, b

eq1 \)). If \( x - 4 = 1 \), then \( x = 5 \). Substituting \( x = 5 \) into \( f(x) \): \( f(5)=-\log_3(5 - 4)=-\log_3(1)=0 \). So one point is \( (5, 0) \).

• Let's find when \( x - 4 = 3 \) (since \( \log_3(3)=1 \)). If \( x - 4 = 3 \), then \( x = 7 \). Substituting \( x = 7 \) into \( f(x) \): \( f(7)=-\log_3(7 - 4)=-\log_3(3)= - 1 \). So another point is \( (7, - 1) \). We could also use \( x - 4=\frac{1}{3} \) etc., but these two have integer coordinates.

Answer:

Asymptote: Vertical (and points like \( (5, 0) \) and \( (7, - 1) \) can be plotted)