Question

question
graph the function $f(x) = \log_{2} x - 1$ on the axes below. you must plot the asymptote and any two points with integer coordinates.
asymptote:

Explanation:

Step1: Recall the vertical asymptote of logarithmic functions

The general form of a logarithmic function is \( y = \log_b(x - h) + k \), and its vertical asymptote is \( x = h \). For the function \( f(x)=\log_2 x - 1 \), we can rewrite it as \( f(x)=\log_2(x - 0)-1 \), so \( h = 0 \). Thus, the vertical asymptote is \( x = 0 \) (the y - axis).

Step2: Find two points with integer coordinates

• When \( x = 1 \):

Substitute \( x = 1 \) into the function \( f(x)=\log_2 x-1 \). We know that \( \log_2 1=0 \) (since \( 2^0 = 1 \)), so \( f(1)=0 - 1=-1 \). So the point is \( (1,-1) \).

• When \( x = 2 \):

Substitute \( x = 2 \) into the function \( f(x)=\log_2 x - 1 \). We know that \( \log_2 2 = 1 \) (since \( 2^1=2 \)), so \( f(2)=1 - 1 = 0 \). So the point is \( (2,0) \).

• We can also check \( x = 4 \):

Substitute \( x = 4 \) into the function \( f(x)=\log_2 x-1 \). We know that \( \log_2 4=2 \) (since \( 2^2 = 4 \)), so \( f(4)=2 - 1=1 \). So the point is \( (4,1) \). (We can choose any two of these points, for example, \( (1,-1) \) and \( (2,0) \) or \( (1,-1) \) and \( (4,1) \) etc.)

Answer:

The vertical asymptote is \( x = 0 \). Two points with integer coordinates can be \( (1,-1) \) and \( (2,0) \) (or other valid points like \( (1,-1) \) and \( (4,1) \) etc.). The asymptote is \( x = 0 \).