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) 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?