Let's analyze the function [tex]\( f(x) = 2^x \)[/tex] and its transformed version [tex]\( g(x) = f(x-4) \)[/tex].
1. Horizontal Asymptote Analysis:
- The original function [tex]\( f(x) = 2^x \)[/tex] is an exponential function. Exponential functions of the form [tex]\( 2^x \)[/tex] have a horizontal asymptote at [tex]\( y = 0 \)[/tex]. This is because as [tex]\( x \)[/tex] approaches negative infinity, [tex]\( 2^x \)[/tex] approaches 0.
- When we modify the function to [tex]\( g(x) = f(x-4) \)[/tex], we are introducing a horizontal shift of the original function [tex]\( f(x) \)[/tex] to the right by 4 units. A horizontal shift does not affect the horizontal asymptote, so the horizontal asymptote remains [tex]\( y = 0 \)[/tex].
2. Y-Intercept Analysis:
- The y-intercept of a function is the point where the graph intersects the y-axis. For [tex]\( f(x) = 2^x \)[/tex], the y-intercept is found by plugging in [tex]\( x = 0 \)[/tex], which gives us [tex]\( f(0) = 2^0 = 1 \)[/tex]. Therefore, the y-intercept of [tex]\( f(x) \)[/tex] is at [tex]\( (0,1) \)[/tex].
- For the function [tex]\( g(x) = f(x-4) \)[/tex], to find the y-intercept, we again plug in [tex]\( x = 0 \)[/tex]. We get:
[tex]\[
g(0) = f(0-4) = f(-4) = 2^{-4} = \frac{1}{2^4} = \frac{1}{16}
\][/tex]
So, the y-intercept of [tex]\( g(x) \)[/tex] is at [tex]\( (0, \frac{1}{16}) \)[/tex], which is approximately [tex]\( (0, 0.0625) \)[/tex].
Given the options in the question:
- Option A states that the horizontal asymptote is [tex]\( y - 4 \)[/tex], which is incorrect.
- Option B states that the horizontal asymptote is [tex]\( y = 0 \)[/tex], which is correct.
- Option C states the y-intercept at [tex]\( (0, 3) \)[/tex], which is incorrect.
- Option D states the y-intercept at [tex]\( (0, 1) \)[/tex], which is incorrect.
Thus, the correct answer is:
[tex]\[ \boxed{B} \][/tex]
