Question

use transformations of ( f(x) = \frac{1}{x} ) to graph ( g(x) = \frac{1}{x - 5} ).
select the correct graph.
a. graph a
b. graph b
c. graph c
d. graph d

Explanation:

Step1: Analyze the transformation

The function \( f(x)=\frac{1}{x} \) has a vertical asymptote at \( x = 0 \) and a horizontal asymptote at \( y=0 \). For the function \( g(x)=\frac{1}{x - 5}\), this is a horizontal shift of the graph of \( f(x) \). The rule for horizontal shifts is: if we have \( f(x - h) \), the graph of \( f(x) \) is shifted \( h \) units to the right (if \( h>0 \)) or \( |h| \) units to the left (if \( h < 0 \)). Here, \( h = 5 \), so the graph of \( f(x)=\frac{1}{x} \) is shifted 5 units to the right. This means the vertical asymptote of \( g(x) \) will be at \( x=5 \) (since the vertical asymptote of \( f(x) \) at \( x = 0 \) is shifted 5 units to the right) and the horizontal asymptote remains \( y = 0 \).

Step2: Identify the correct graph

We need to look for the graph where the vertical asymptote is at \( x = 5 \) and the horizontal asymptote is at \( y=0 \), and the hyperbola is a shift of \( \frac{1}{x} \) 5 units to the right. Let's analyze each option:

• Option A: Check the vertical asymptote. If it's not at \( x = 5 \), eliminate.
• Option B: Check the vertical asymptote. If it's not at \( x = 5 \), eliminate.
• Option C: Check the vertical asymptote. If it's not at \( x = 5 \), eliminate.
• Option D: The vertical asymptote should be at \( x = 5 \), and the graph should be the hyperbola \( \frac{1}{x} \) shifted 5 units to the right. So we look for the graph where the two branches of the hyperbola are shifted right, with vertical asymptote at \( x = 5 \) and horizontal at \( y = 0 \). From the visual (assuming the graphs are as per the standard transformation), the correct graph should have the vertical asymptote at \( x=5 \). So we identify the graph with vertical asymptote \( x = 5 \).

Answer:

D (assuming D is the graph with vertical asymptote at \( x = 5 \) and the hyperbola shifted 5 units right from \( \frac{1}{x} \))