To find vertical asymptotes, equate the denominator to zero, and solve for x.
To find horizontal asymptotes, compute:
What do you find?
Horizontal:
6 /(1 + (2/∞))= 6
vertical:
1+2e^{-5x}=0
2*(-5x)=-ln1
x= ln1/10 = 0.
These are the results i got when i tried earlier, however with f(x) = 0 i get x as 2, so the vertical cant be 0. Was wondering if there might be two horizontal
You have found one horizontal asymptote, try the limit as x goes to negative infinity as well.
You are incorrectly solving for x to find any possible vertical asymptotes:
Without taking logs, we should know that:
for all real , hence there are no vertical asymptotes.
Since the exponent is negative, the denominator goes to 1 as goes go to infinity and goes to infinity as x goes to negative infinity.
6/1 = 6 so y = 6 is the horizontal asymptote on the right.
6/∞ = 0 so y = 0 is the horizontal asymptote to the left.