Applying Limits to Find the Vertical Asymptote

[The_ArtofScience]

Sorry to be a bother at this time around. I have a question though about a weird 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?

 

[HallsofIvy]

Sorry to be a bother at this time around. I have a question though about a weird 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.

NO. The derivative of "5" is 0.

Its only critical point or when f'(x) = 0 is at x = 3.

Again, no. Even using your "derivative", if x= 3, 6- 2x= 0 but -3e⁰+ 5= 3, not 0. f'= -2e^{6- 2x} is never 0.

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?

you can't- it doesn't. There is NO vertical asymptote.

 

[The_ArtofScience]

Hi Halls,

Thanks for catching the derivative error I made. I did put the function in an online graphing calculator and did see a vertical asymptote.

http://www.coolmath.com/graphit/

 

[Mark44]

That's not a vertical asymptote. A vertical asymptote is a line, x = a, where a is some finite number. On either side of this line the graph of a function becomes unbounded in some way (i.e., approaches infinity or negative infinity).

Your exponential function grows very large as x becomes negative. What you probably saw was the graph getting very steep for some negative value of x. In a similar way, the graph of, say, f(x) = x^3 gets pretty steep for large values of x, but it too does not have a vertical asymptote.