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Find the horizontal and vertical asymptotes of each curve. If you have a graphing device, check your work by graphing the curve and estimating the asymptotes.

$ y = \dfrac{2e^x}{e^x - 5} $

$x=\ln 5$

03:51

Daniel J.

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 6

Limits at Infinity: Horizontal Asymptotes

Limits

Derivatives

Oregon State University

Baylor University

University of Nottingham

Lectures

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to this function. And we want to see if we have any vertical assam. Tokes or horizontal assessment. Oops, vertical assam toques are going to happen, vertical ascent up is going to be a vertical line and they're going to occur when your function uh as you get closer and closer to that line, uh your function is either going to be uh you know, shooting higher and higher and higher and higher towards positive infinity or possibly lower and lower towards negative infinity. And one of the best places to look for vertical assam tops is when the denominator might be zero. Because if the denominator is getting close to zero, that means you're dividing by a tiny tiny number uh meaning you're getting a very large number. Uh And of course if the denominator is getting close to zero from the negative side, uh you're still dividing by a tiny tiny number but a tiny negative number, which means you're getting very large uh negative meaning you're heading towards negative infinity. So, to look for vertical isotopes, uh let's see when the denominator will be zero. So we want to solve uh for E T D X attract five equals zero. What value of X is going to make the denominator equal to zero? Well, if we add five to both sides, well, we get E to the X equals five. And then if you want to solve for X, simply take the natural log of both sides of this equation. Uh natural log and uh e function or are in verses. So the log of E T T X is just X. So X will equal The log of five. So when x equals the log of five, we're going to have a vertical ass from tow pin. We'll check that out. We'll check that out graphically in the moment, horizontal assam temps to find horizontal assam tops. Okay, let me just remind you vertical as from Tokyo. Want to see when the denominator might be zero to find horizontal assam tops. Look for when X moves towards positive infinity and when X moves towards negative infinity in other words, uh if you are looking for horizontal, ask them to look to see how the graph of your function is going to behave as ex moves uh far in the direction of the positive X axis and then again his ex moves uh far to the left, in the direction of the negative X axis. Now as X goes towards infinity to E to the X Over E. To the extract five is basically going to be dominated by the two et de X over E T D. Ex uh ex is very large heading towards infinity. So et de X is very large. Subtracting five is not really going to change that much. So this function is basically going to be dominated by two et de except top divided by each. The excellent bottom, but any excess will cancel. So basically as X goes towards infinity. Uh this function will be headed towards two. So Y equals two appears to be a horizontal aceto. We'll check this graphically. Now, how about effects is moving in the direction of the negative X. Axis towards negative infinity? What's going to happen to this expression? Well to E. T. D. X. Over E. To the X minus five. If X is headed towards negative infinity uh Think of E being raised to a very large negative number as X becomes very large and negative E. T. D. X will be close to zero. Think of like each and the negative 100 E. To the negative 1000 very close to zero. Uh So this term right here Is going to approach zero and the same of course with the T. D. X. Over here. So the numerator is going to be approaching two times zero which is zero. The denominator is going to be approaching zero minus five which is negative five. And so if the numerator uh is approaching zero and the denominator is approaching negative five, this whole function is approaching zero Over negative five which is going to be zero. So as ex heads towards negative infinity. Uh Two times et de X over 80 x minus five is going to head towards zero. And so we expect another horizontal ass from tow at the line Y equals zero. So what I'm going to do next is used is most graphing calculator. We're gonna graph our function and we're going to graph uh This vertical line and these two horizontal lines and show that they indeed are vertical and horizontal. Ask Youtube's. Okay, so in purple you have ah the two pieces of the graph of our function. And uh this black vertical line is going to be our X equals log of five. You can see that that is going to be a vertical ascent to as X approaches uh log of five from the right side of the graph gets closer and closer to that vertical assam. Topaz ahead toward positive infinity. And as X approaches log of five from the left side of the graph is going to get closer and closer to that vertical asking Topaz A heads down towards negative infinity. So x equals log of five is a vertical ascent. Open the horizontal line. The red line Y equals zero is a horizontal assume. Tote as ex moves along its axis towards negative infinity. Uh The graph will get closer and closer to the horizontal asientos line Y equals zero. Now, as Ex moves along guitars on tal access towards positive infinity. Uh The graph will get closer and closer to purple graft will get closer and closer to the other horizontal Assen took the blue line Y equals two. So, as X is approaching positive infinity, The graph of the function will get closer and closer to the horizontal. Ask them to y equals two

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