Sorry to be a bother at this time around. I have a question though about a wierd looking function : e^(6-2x) + 5. (The 5 is outside of the parenthesis, so it is a coefficient, not an exponent). I am interested in seeing how to use limits to construct its graph. In particular, I want to find out what it looks like to the left of 3. (Its vertical asymptote in short)(adsbygoogle = window.adsbygoogle || []).push({});

y - intercepts

y = e^(6) + 5

(0, e^(6) + 5)

x - intercepts

0 = e^(6) / e^(2x) + 5

-5e^(2x) = e^(6)

Since natural logging it would give an undefined answer there is no x-intercept. At x --> infinity (e^(6)/ e^(2x) + 5) tends to 5. Then the horizontal asymptote is 5.

Applying the derivative test, e^(6-2x) + 5 becomes (-2)e^(6-2x) + 5. Its only critical point or when f'(x) = 0 is at x = 3. Evaluating at x=2 and x=4 it appears to have negative values indicating that the function is decreasing. When x = 3 is put back into the original f(x) it gives 5, indicating that the asymptote starts at 3.

Since the horizontal asymptote starts at x = 3. At x --> 3- (e^(6-2x) + 5) = + infinity. The function from the left of 3 is a monotic function because like stated before, the MVT says that at f'(c) = 0 and the only critical point is at x = 3 so it is safe to assume that it has either an increasing or decreasing pattern.

I know that at x = 3 it has a HA = 5, but how can you find the exact point where the function begins to to "blow up" vertically?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Applying Limits to Find the Vertical Asymptote

**Physics Forums | Science Articles, Homework Help, Discussion**