To determine the asymptotes of the function [tex]\( f(x) = \frac{15}{1 + 4 e^{-0.2 x}} \)[/tex], we analyze the behavior of the function as [tex]\( x \)[/tex] approaches both [tex]\( +\infty \)[/tex] and [tex]\( -\infty \)[/tex].
1. As [tex]\( x \)[/tex] approaches [tex]\( +\infty \)[/tex]:
- When [tex]\( x \to +\infty \)[/tex], the exponent [tex]\( -0.2x \)[/tex] will become very large negative. Therefore, the term [tex]\( e^{-0.2x} \)[/tex] will approach 0 because the exponential function [tex]\( e^{-k} \)[/tex] goes to 0 as [tex]\( k \)[/tex] becomes large.
- Substituting [tex]\( e^{-0.2x} \approx 0 \)[/tex] into the function, we get:
[tex]\[
f(x) \approx \frac{15}{1 + 4 \cdot 0} = \frac{15}{1} = 15
\][/tex]
- Thus, [tex]\( y = 15 \)[/tex] is a horizontal asymptote.
2. As [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex]:
- When [tex]\( x \to -\infty \)[/tex], the exponent [tex]\( -0.2x \)[/tex] will become very large positive. Therefore, the term [tex]\( e^{-0.2x} \)[/tex] will grow significantly and approach infinity.
- Substituting [tex]\( e^{-0.2x} \to \infty \)[/tex] into the function, we get:
[tex]\[
f(x) \approx \frac{15}{1 + 4 \cdot \infty} = \frac{15}{\infty} \approx 0
\][/tex]
- Thus, [tex]\( y = 0 \)[/tex] is a horizontal asymptote.
3. Verification of Other Given Options:
- [tex]\( y = 4 \)[/tex] and [tex]\( y = -0.2 \)[/tex] are not asymptotes of the function. These values do not represent any limiting behavior of the function [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches either [tex]\( +\infty \)[/tex] or [tex]\( -\infty \)[/tex].
In conclusion, the asymptotes of the graph of the function [tex]\( f(x) = \frac{15}{1+4 e^{-0.2 x}} \)[/tex] are:
- [tex]\( y = 15 \)[/tex]
- [tex]\( y = 0 \)[/tex]
The correct answers are [tex]\( y = 15 \)[/tex] and [tex]\( y = 0 \)[/tex].
