To solve the problem, consider the function [tex]\( f(x) = 2^x \)[/tex] and the modified function [tex]\( g(x) = 2 f(x) \)[/tex], which simplifies to [tex]\( g(x) = 2 \cdot 2^x \)[/tex].
First, let's analyze the graph and characteristics of [tex]\( f(x) = 2^x \)[/tex]:
1. The y-intercept occurs where [tex]\( x = 0 \)[/tex]:
[tex]\[
f(0) = 2^0 = 1
\][/tex]
Therefore, the y-intercept of [tex]\( f(x) \)[/tex] is at [tex]\( (0, 1) \)[/tex].
2. Since [tex]\( f(x) = 2^x \)[/tex] is an exponential function, it has a horizontal asymptote at [tex]\( y = 0 \)[/tex] as [tex]\( x \to -\infty \)[/tex].
Now, let's consider the modifications made for [tex]\( g(x) \)[/tex]:
[tex]\[
g(x) = 2 f(x) = 2 \cdot 2^x = 2^{x+1}
\][/tex]
1. To find the y-intercept of [tex]\( g(x) \)[/tex]:
[tex]\[
g(0) = 2 \cdot 2^0 = 2 \cdot 1 = 2
\][/tex]
This means the y-intercept of [tex]\( g(x) \)[/tex] is at [tex]\( (0, 2) \)[/tex].
2. [tex]\( g(x) \)[/tex] is still an exponential function, and it will not change the horizontal asymptote from [tex]\( y = 0 \)[/tex] since the horizontal asymptote of the exponential function [tex]\( 2^x \)[/tex] is [tex]\( y = 0 \)[/tex] and this property remains unchanged under scaling by a constant factor.
Given these points:
- The y-intercept of [tex]\( g(x) \)[/tex] is [tex]\((0, 2)\)[/tex].
Therefore, the correct answer is:
[tex]\[ \text{B. } y\text{-intercept at } (0, 2) \][/tex]
