# Other Asymptotes

A few days ago, a reader asked if you could find the location of horizontal asymptotes from the derivative of a function. The answer, alas, is no. You can determine that a function has a horizontal asymptote from its derivative, but not where it is. This is because a function and its many possible vertical translations have the same derivative but different horizontal asymptotes. To locate it precisely, we need more information, an initial condition (a point on the curve).

I have written a post about the relationship between vertical asymptotes and the derivative of the function here. Today’s post will discuss horizontal asymptotes and their derivatives.

Definition: a straight line associated with a curve such that as a point moves along an infinite branch of the curve the distance from the point to the line approaches zero and the slope of the curve at the point approaches the slope of the line. (Merriam-Webster dictionary). Thus, asymptotes can be vertical, horizontal, or slanted.

Horizontal Asymptotes

Horizontal asymptotes are a form of end behavior: they appear as $\displaystyle x\to \infty$ or $\displaystyle x\to -\infty$. Specifically, if a function has a horizontal asymptote(s), then $\displaystyle \underset{{x\to \pm \infty }}{\mathop{{\lim }}}\,{f}'\left( x \right)=0$,  At an asymptote, the curve must approach a horizontal line, therefore its slope must be approaching zero as $\displaystyle x\to \infty$ or $\displaystyle x\to -\infty$.  The converse is not true: If the derivative approaches zero, the function may not have a horizontal asymptote. A counterexample is $\displaystyle f\left( x \right)=\sqrt{x}$.

In the first four examples that follow, the derivative approaches 0 as $\displaystyle x\to \infty$ and/or $\displaystyle x\to -\infty$

Example 1: $\displaystyle {f}'\left( x \right)=\frac{1}{{1+{{x}^{2}}}}$  (Figure 1 in blue).This is the derivative of $\displaystyle f\left( x \right)={{\tan }^{{-1}}}(x),\ -\tfrac{\pi }{2}. (Figure 1 in red). We see that the derivative approaches zero in both directions, this tells us that there may be horizontal asymptote(s). We must look at the function to find where they are: $\displaystyle y=\tfrac{\pi }{2}$ and $\displaystyle y=-\tfrac{\pi }{2}$ Figure 1

Example 2: 2008 AB 19 $\displaystyle p\left( x \right)=\frac{{5+{{2}^{x}}}}{{1={{2}^{x}}}}$. (Figure 2 in Red)

So many students missed this (pre-calculus ?) question on the exam that the question was not counted! As $\displaystyle x\to \infty$ the 2x terms approach infinity at the same rate so the limit is the ratio of their coefficients: the asymptote is y = –1. And as $\displaystyle x\to -\infty$ the 2x terms approach zero so the fractions approaches an asymptote at y = 5. (There is a vertical asymptote at x = 0, not shown.)

The derivative is $\displaystyle {p}'\left( x \right)=\frac{{6\ln (2)\cdot {{2}^{x}}}}{{{{{\left( {1-{{2}^{x}}} \right)}}^{2}}}}$ (Figure 2 blue). The derivative approaching zero in both directions, but the location of the asymptotes must be determined from the equation of the function.

Example 3: $\displaystyle g\left( x \right)=\frac{{-3}}{{{{{\left( {x+1} \right)}}^{2}}}}$ (Figure 3 in blue). This is the derivative of $\displaystyle g(x)=\frac{3}{{x+1}}+2$ (Figure 2 in red). Notice that replacing the “+2” with a different constant will translate the curve and its asymptote up or down. Here the function has the same asymptote in both directions. (There is a vertical asymptote at x = –1, not shown)

Example 4: Consider the function $\displaystyle s\left( x \right)={{1.1}^{=x}}\sin \left( x \right)$See figure 4.

As $\displaystyle x\to \infty$ this function’s values get closer to zero and the graph appears to have the x-axis as a horizontal asymptote. The interesting thing is that the function’s value at every integral  multiple of π is zero and the function changes sign there. In other words it crosses and re-crosses its asymptote and moves away from and back towards the asymptote. Each time it crosses the asymptote it moves less far from the asymptote, but continues to move towards and away from it. It’s derivative (not shown) exhibits the same phenomenon.

This feature is difficult to see on a graph since the values become exceedingly small in absolute value very quickly.

Slant Asymptotes

Other lines lines may also be asymptotes, called slant asymptotes. If as $\displaystyle x\to \infty$ or $\displaystyle x\to -\infty$, the function approaches a non-horizontal line, which appears as a slant asymptote. In this case the derivative approaches a constant equal to the slope of the line.

Example 5: $\displaystyle {h}'\left( x \right)=1-\frac{1}{{{{{\left( {x-2} \right)}}^{2}}}}$ (Figure 5 in blue) is the derivative of $\displaystyle h\left( x \right)=\frac{{{{x}^{2}}+x-5}}{{(x-2)}}=x+3+\frac{1}{{\left( {x-2} \right)}}$.

The derivative shows a horizontal asymptote at y = 1, so the function will have a slant asymptote with a slope of 1. This is obvious (I hope) from the second form of the function where the final term approaches zero as $\displaystyle x\to \infty$ and $\displaystyle x\to -\infty$. For positive numbers, the last term is positive and the function is therefore above line; for negative numbers the last term is negative and the function is below the line. Vertical translations will move the function and its slant asymptote, but not change the asymptote’s slope. (There is a vertical asymptote at x = 2, not shown)

Other “Asymptotes”

Example 6: Playing with the idea from examples 5, if the function were $\displaystyle k\left( x \right)={{x}^{2}}+x+\frac{1}{{\left( {x-2} \right)}}$. (Red in figure 6). The curve will approach the parabola $\displaystyle y={{x}^{2}}+x$ (Figure 6 black dotted) as an “asymptote.” Its derivative $\displaystyle {k}'\left( x \right)=2x+1-\frac{1}{{{{{\left( {x-2} \right)}}^{2}}}}$ (Figure 6 blue) approaches the line (its own slant asymptote) $\displaystyle y=2x+1$ (not shown). There is a vertical asymptote at x = 2 (not shown).

Another interesting feature of this kind of function is this: $\displaystyle k\left( x \right)={{x}^{2}}+x+a+\frac{1}{{x-2}}$ is the same function translated up or down by an amount a. The resulting functions have the same derivative and the same parabola for their asymptote! Try investigating this situation using Desmos or GeoGebra or some other graphing utility with a slider.

This type of “asymptote” might make a good topic for an exploration, project, or investigation by a student.

Revised March 23, 2021.

# Asymptotes and the Derivative

What does an asymptote of the derivative tell you about the function? How do asymptotes of a function appear in the graph of the derivative?

One of my most read posts is Reading the Derivative’s Graph, first published seven years ago. The long title is “Here’s the graph of the derivative; tell me about the function.” It tells how to quickly find information about the graph of a function from the derivative’s graph. I received a question from a reader recently that asked about asymptotes and the derivative, a topic that I did not cover in that post. So, I tried to find the relationship. The short answer is that an asymptote of the derivatives does not tell you much about the graph of the function. (This is probably why the idea has never been tested on the AP Calculus Exams.)

Nevertheless, there is some calculus to be learned here. You may be able to find some questions for your students to investigate on this topic. Here’s what i determined.

Before we start, here are two terms that are useful, but not commonly used.

• An even vertical asymptote is one for which the function increases or decreases without limit on both sides of the asymptote. In other words, as x approaches a the function approaches infinity or negative infinity from both sides. The function $f\left( x \right)=\frac{1}{{{{{\left( {x-2} \right)}}^{2}}}}$ has an even vertical asymptote at x = 2. (Figure 1)
• An odd vertical asymptote is one for which the function increases without bound on one side and decreases without bound on the other. The function $g\left( x \right)=\frac{1}{{x-2}}$ has an odd vertical asymptote at x = 2. (Figure 2) Likewise, the tangent, cotangent, secant, and cosecant functions have odd vertical asymptotes.

If a function has an odd vertical asymptote, then its derivative will have an even vertical asymptote. (Ask your students to explain why.)

If a function has an even vertical asymptote, then its derivative will have an odd vertical asymptote. (Ask your students to explain why.)

The converses of these two statements are false. That is, a vertical asymptote of the derivative does not necessarily indicate an asymptote of the function. The catch is continuity.

### Continuous Functions

If the derivative exists at xa, then the function is continuous there. But, since we are considering asymptotes of the derivative, we cannot know from the derivative alone if the function is continuous where the derivative has an asymptote.

A simple cusp is a situation in which at an extreme point the graph is tangent to a vertical line. See Figure 3. (Or, you could say, the tangent lines from each side are coincident. Here we will limit the discussion to a vertical tangent line.) A continuous function that has a cusp will show an odd vertical asymptote on its derivative’s graph.

An example is $\displaystyle h\left( x \right)=\sqrt{{{{{\left( {x-2} \right)}}^{2}}}}+1$  that has a cusp at the point (2,1). (Figure 3).

A continuous function that has a vertical tangent line not a cusp, has an even vertical asymptote on its derivative’s graph. For example, $k\left( x \right)={{\left( {x-2} \right)}^{{1/3}}}+1$ at (2,0) (Figure 4).

If you are given the graph of the derivative and it shows a vertical asymptote at x = a, and you know the function is continuous there, then

• an odd vertical asymptote of the derivative indicates cusp on the graph of the function. This will also be an extreme value. (Ask your students to explain why.)
• an even vertical asymptote of the derivative indicates vertical tangent line on the graph of the function, but not an extreme value. (Ask your students to explain why.)

Other than these, there is no easy way to tell what situation produced an asymptote on the derivative.

### Functions that are not continuous

If the function is not continuous at xa, then things get a lot more complicated.

• If the function is not continuous at x = a, then an even vertical asymptote of the derivative may indicate an odd vertical asymptote on the graph of the function. There is no way to be sure.
• If the function is not continuous at x = a, then an odd vertical asymptote of the derivative may indicate an even vertical asymptote on the graph of the function. There is no way to be sure.

Consider this function: $\displaystyle z\left( x \right)=\left\{ {\begin{array}{*{20}{c}} {\sqrt{{{{{\left( {x-2} \right)}}^{2}}}}-1} & {x<2} \\ {\sqrt{{{{{\left( {x-2} \right)}}^{2}}}}+1} & {x\ge 2} \end{array}} \right.$

See Figure 5. The function is defined for all Real numbers and has a jump discontinuity at x = 2. The derivative has an odd vertical asymptote there: $\displaystyle \underset{{x\to 2-}}{\mathop{{\lim }}}\,{z}'\left( x \right)=-\infty$ and $\displaystyle \underset{{x\to 2+}}{\mathop{{\lim }}}\,{z}'\left( x \right)=-\infty$. Compare this with h(x) above.

In fact, $\displaystyle {h}'\left( x \right)={z}'\left( x \right)=\tfrac{2}{3}{{\left( {x-2} \right)}^{{-1/3}}},x\ne 2$. Go figure.

For considerations of horizontal asymptotes and the derivative see Other Asymptotes

# An Exploration in Differential Equations

This is an exploration based on the AP Calculus question 2018 AB 6. I originally posed it for teachers last summer. This will make, I hope, a good review of many of the concepts and techniques students have learned during the year. The exploration, which will take an hour or more, includes these topics:

• Finding the general solution of the differential equation by separating the variables
• Checking the solution by substitution
• Using a graphing utility to explore the solutions for all values of the constant of integration, C
• Finding the solutions’ horizontal and vertical asymptotes
• Finding several particular solutions
• Finding the domains of the particular solutions
• Finding the extreme value of all solutions in terms of C
• Finding the second derivative (implicit differentiation)
• Considering concavity
• Investigating a special case or two

I also hope that in working through this exploration students will learn not so much about this particular function, but how to use the tools of algebra, calculus, and technology to fully investigate any function and to find all its foibles. The exploration is here in a PDF file. Here are the solutions.

The College Board is pleased to offer a new live online event for new and experienced AP Calculus teachers on March 5th at 7:00 PM Eastern.

I will be the presenter.

The topic will be AP Calculus: How to Review for the Exam:  In this two-hour online workshop, we will investigate techniques and hints for helping students to prepare for the AP Calculus exams. Additionally, we’ll discuss the 10 type questions that appear on the AP Calculus exams, and what students need know and to be able to do for each. Finally, we’ll examine resources for exam review.

Registration for this event is $30/members and$35/non-members. You can register for the event by following this link: http://eventreg.collegeboard.org/d/xbqbjz

# Limits – They Make the Calculus Work.

In an ideal world, I would like to have all students study limits in their pre-calculus course and know all about them when they get to calculus. Certainly, this would be better than teaching how to calculate derivatives in pre-calculus (after all derivatives are calculus, not pre-calculus).

Limits are the foundation of the calculus. Continuity, an important property of functions, depends on limits. All derivatives and all definite integrals are limits. For AP Calculus students need a good intuitive understanding of limits, what they mean, and how to find them The formal (delta-epsilon) definition is not tested and need not be taught, however, do not feel that you have to avoid it. If your students can handle it, let them try.

Here are a few of my previous posts on limits.

Why Limits?

Finding Limits  How to … and the use of “infinity” vs “DNE”

Dominance  Finding limits the easy way.

Deltas and Epsilons Not tested on the AP Exams; here’s why.

Asymptotes The graphical manifestation of limits at or equal to infinity.

Next Week: Continuity

Update of my post of August 15, 2017.

# Summer Fun

Every Spring I have a lot of fun proofreading Audrey Weeks’ new Calculus in Motion illustrations for the most recent AP Calculus Exam questions. These illustrations run on Geometers’ Sketchpad. In addition to the exam questions Calculus in Motion (and its companion Algebra in Motion) include separate animations illustrating most of the concepts in calculus and algebra. This is a great resource for your classes.

This year, I really got into 2018 AB 6, the differential equation question. I wrote an exploration (or as the kids would say “worksheet”) on a function very similar to the differential equation in that question. The exploration, which is rather long, includes these topics:

• Finding the general solution of the differential equation by separating the variables
• Checking the solution by substitution
• Using a graphing utility to explore the solutions for all values of the constant of integration, C
• Finding the solutions’ horizontal and vertical asymptotes
• Finding several particular solutions
• Finding the domains of the particular solutions
• Finding the extreme value of all solutions in terms of C
• Finding the second derivative (implicit differentiation)
• Considering concavity
• Investigating a special case or two

I also hope that in working through this exploration students will learn not so much about this particular function, but how to use the tools of algebra, calculus, and technology to fully investigate any function and to find all its foibles.

Students will need to have studied calculus through differential equations before they start the exploration. I will repost it next January for them.

The exploration is here for you to try. Try it before you look at the solutions. It will give you something to do over the summer – well not all summer, only an hour or so.

There will be only occasional, very occasional, posts over the Summer. More regular posting will begin again in August. Enjoy the Explorations, and, more important, enjoy the Summer!

# How to Tell your Asymptote from a Hole in the Graph.

The fifth in the Graphing Calculator / Technology series

(The MPAC discussion will continue next week)

Seeing discontinuities on a graphing calculator is possible; but you need to know how a calculator graphs to do it. Here’s the story:

The number you choose for XMIN becomes the x-coordinate of the (center of) the pixels in the left most column of pixels. The number you choose for XMAX is the x-coordinate of the right most column of pixels. The distance between XMIN and XMAX is divided evenly between the remaining pixels so that all the pixels are evenly spaced across the screen (the same distance apart). The rows of pixels are done the same way evenly spacing them between YMIN and YMAX.

This spacing is usually not at “nice” values as can be seen by just moving the cursor across the screen and noticing the x-values or y-values at the bottom of the screen. The cursor is located one pixel to the right of the y-axis and one pixel above the x-axis in the “standard” window of a TI-8x. Note the coordinates of that pixel at the bottom of the screen. These are the distances between the pixels.

To draw a graph, the calculator takes the x-coordinate of each pixel, calculates the corresponding y-value and turns on the pixel in that column with closest y-pixel-coordinate. If set in a connect mode, the calculator turns on several pixels in adjacent columns so that the y-values seem to connect; this is why the graph often looks jagged in steep sections of the graph. If you are in DOT mode, this does not happen and only one pixel in each column is on.

If you move the cursor over one of the points on a graph, you will see the pixel coordinates, NOT the actual y-coordinates. Use TRACE to see the actual y-coordinate. This is why when finding intersections, you should not just move the cursor over the point, but rather use “intersect” to see the actual y-value of the function.

If the function is undefined for some x-pixel value, then no pixel will turn on in that column. If the function is undefined for some value between the pixel values, then nothing happens because the calculator has not evaluated the function there, so the graph seems to be continuous.

Vertical “asymptotes” are the result of the calculator not evaluating the function at the undefined value; rather it connects the value on one side of the asymptote off the bottom of the screen with the next value on the other side of the asymptote off the top of the screen. If the asymptote appears exactly at a pixel value, then no “asymptote” will appear and that column of pixels will have no pixel turned on. (Some newer calculators and newer operating systems on older calculators have made adjustments so that the “asymptotes” do not show up. In some systems this feature can be turned on or off.) The function $\displaystyle y=\frac{3\left( x-2 \right)}{\left( x-2 \right)\left( x+2 \right)}$ in the standard window. The vertical line is not really the asymptote and the “hole” at (2, 0.75) is not seen.

A removable discontinuity, a hole in the graph (really a skipped pixel), can be seen, if it occurs at a pixel value. Since in most examples the hole is at an integer or other “nice” number, you will not see them in the “standard” window. Use a “decimal” window, which has been chosen in advance so the x-values of the pixels are integers and nice decimals. (To see this, in a decimal window move the cursor around and notice the pixel coordinates).

The other thing you can do is adjust the XMIN and XMAX values so that the distance between them will land on integer values. (Nice project for your class – the number of pixels can be found in the guidebook, or you can count them. In the old days, before decimal windows, this was necessary – it was called finding a “friendly window.”) The function $\displaystyle y=\frac{3\left( x-2 \right)}{\left( x-2 \right)\left( x+2 \right)}$ in the “decimal” window. The “asymptote” has disappeared and the “hole” at (2, 0.75) is now visible.

Zooming in or out may change these values so the hole or asymptote disappears.

For a related idea see the post My Favorite Function

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# Continuity

Limits logically come before continuity since the definition of continuity requires using limits. But practically and historically, continuity comes first. The concept of a limit is used to explain the various kinds of discontinuities and asymptotes. Start by studying discontinuities.

Types of discontinuities to consider: removable (a gap or hole in the graph), jump, infinite (vertical asymptotes), oscillating, and end behavior (horizontal asymptotes).

Numerically: Make a table for the value of $\displaystyle \frac{1}{x-3}$ near x = 3 and as $\displaystyle x\to \pm \infty$. Relate the values and their signs to the graph. (Divide by a small number get a big number; divide by a big number, get a small number.)

Use the vocabulary of limits to explain the features of graphs. Example: The function $\displaystyle \frac{{{x}^{2}}-4}{x-2}$ has no value at x = 2 (f(2) does not exist), but as you get closer to x = 2 the function value gets closer to 4 ( $\displaystyle \underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)=4$).

Relate the limit, value and graph of the function. In the example above, the graph looks like the line $y=x+2$ with a gap or hole at the point (2, 4). Another example: $\displaystyle \underset{x\to \infty }{\mathop{\lim }}\,\frac{3{{x}^{2}}+x+8}{{{x}^{2}}+1}=3$ since, the graph gets closer to y = 3 as you go farther to the right. The line y = 3 is a horizontal asymptote.

Do this numerically as well: $\displaystyle \frac{3{{x}^{2}}+x+8}{{{x}^{2}}+1}=3+\frac{x+5}{{{x}^{2}}+1}$ and since the fraction gets smaller as |x| gets larger, the function approaches 3 from above when x > 0 and from below when x < 0 (why?)

Extra for your experts: Discuss the reason for the jump discontinuity of $\displaystyle f\left( x \right)=\frac{\cos \left( x \right)\sqrt{{{x}^{2}}-2x+1}}{x-1}$