To determine the correct statement about the graph of the function [tex]\( f(x) = \log_2 x \)[/tex], let's analyze the graph step-by-step.
1. Vertical Asymptote:
- The logarithmic function [tex]\( \log_2 x \)[/tex] has a vertical asymptote where the argument of the logarithm (in this case, [tex]\( x \)[/tex]) approaches zero from the right.
- Therefore, the vertical asymptote for [tex]\( f(x) = \log_2 x \)[/tex] is at [tex]\( x = 0 \)[/tex].
2. Behavior of the Graph:
- As [tex]\( x \)[/tex] approaches 0 from the right ([tex]\( x \to 0^+ \)[/tex]), the value of [tex]\( \log_2 x \)[/tex] tends towards negative infinity. This confirms that the graph has a vertical asymptote at [tex]\( x = 0 \)[/tex].
- The function [tex]\( f(x) = \log_2 x \)[/tex] is defined only for positive [tex]\( x \)[/tex] values.
3. Sign of the Function on the Interval [tex]\( (0, 1) \)[/tex]:
- We next investigate the value of [tex]\( f(x) \)[/tex] on the interval [tex]\( (0, 1) \)[/tex]. We know that for [tex]\( x \)[/tex] in [tex]\( (0, 1) \)[/tex], [tex]\( 0 < x < 1 \)[/tex].
- For any [tex]\( x \)[/tex] between 0 and 1, [tex]\( \log_2 x \)[/tex] is negative because we are taking the logarithm base 2 of a number less than 1.
We can now address each of the statements given in the question:
A. The graph has an asymptote of [tex]\( y = 0 \)[/tex] and is increasing as [tex]\( x \)[/tex] approaches positive infinity. - This is incorrect. The asymptote is at [tex]\( x = 0 \)[/tex], not [tex]\( y = 0 \)[/tex].
B. The graph has an asymptote of [tex]\( x = 0 \)[/tex] and is negative over the interval [tex]\( (0, 1) \)[/tex]. - This statement is correct.
C. The graph has an asymptote of [tex]\( x = 0 \)[/tex] and is positive over the interval [tex]\( (0, 1) \)[/tex]. - This is incorrect. The function [tex]\( \log_2 x \)[/tex] is negative over the interval [tex]\( (0, 1) \)[/tex].
D. The graph has an asymptote of [tex]\( y = 0 \)[/tex] and is decreasing as [tex]\( x \)[/tex] approaches positive infinity. - This is incorrect. The asymptote is at [tex]\( x = 0 \)[/tex], not [tex]\( y = 0 \)[/tex], and the function is increasing as [tex]\( x \)[/tex] approaches positive infinity.
Based on this analysis, the correct answer is B. The graph has an asymptote of [tex]\( x=0 \)[/tex] and is negative over the interval [tex]\( (0,1) \)[/tex].
