Answer: (A) y = 15 (B) y = 0
Step-by-step explanation:
[tex]f(x) = \dfrac{15}{1+4e^{-0.2x}}[/tex]
The vertical asymptote is the restriction on the x-value. Since the denominator cannot be equal to zero, then
1 + 4e⁻⁰⁻²ˣ ≠ 0
→ 4e⁻⁰⁻²ˣ ≠ -1
→ e⁻⁰⁻²ˣ ≠ [tex]-\dfrac{1}{4}[/tex]
→ ln e⁻⁰⁻²ˣ ≠ [tex]ln\bigg(-\dfrac{1}{4}\bigg)[/tex]
Note: ln cannot be negative. So, there is no vertical asymptote.
The horizontal asymptote can be found by finding the limit as x approaches positive and negative infinity.
[tex]\lim_{x \to \infty} \dfrac{15}{1+4e^{-0.2x}}[/tex]
[tex]= \dfrac{15}{1+4(0)}[/tex]
[tex]= \dfrac{15}{1}[/tex]
[tex]= 15[/tex]
So, the horizontal asymptote when x → +∞ is y = 15
[tex]\lim_{x \to -\infty} \dfrac{15}{1+4e^{-0.2x}}[/tex]
[tex]= \dfrac{15}{1+4(\infty)}[/tex]
[tex]= \dfrac{15}{\infty}[/tex]
[tex]= 0[/tex]
So, the horizontal asymptote when x → -∞ is y = 0
