This deck covers domains, holes, and vertical, horizontal, and slant asymptotes, then sign charts and full sketches of rational functions, and finishes with polynomial and rational inequalities. It targets the classic errors: calling a cancelled factor a vertical asymptote, reading a horizontal asymptote off the constant terms, multiplying a rational inequality by a denominator whose sign is unknown, and including a zero of the denominator in a non-strict solution set.
Subject: College Algebra · 134 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
College Algebra - Deck 15
Domains, holes, vertical / horizontal / slant asymptotes, sign charts, and rational inequalities.
Objectives
By the end of this deck you can:
Warm-up
Discussion prompt
Before we open Rational Functions, Asymptotes, and Inequalities: without looking back, what was the main idea of Polynomial Functions, Division, and Zeros, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck covers degree and the four end-behavior cases, then zeros, multiplicity, and whether the graph crosses the axis or merely touches it. It works through long division with placeholder terms and synthetic division, the Remainder, Factor, and Rational Zero Theorems, and the full strategy for finding every zero of a cubic or quartic, including the Fundamental Theorem of Algebra and conjugate pairs. It targets synthetic division done with the wrong sign or a non-linear divisor, missing placeholders in long division, treating rational-zero candidates as though they were answers, and getting cross-versus-touch backwards.
Section
Part 1
Concept
A rational function is one polynomial divided by another. That is the whole definition.
\[ f(x) = \frac{P(x)}{Q(x)}, \qquad Q(x) \ne 0 \]
rational function — A function that can be written as a quotient of two polynomials, with the bottom polynomial not the zero polynomial. The word rational comes from ratio, not from being reasonable.
Counterexample
Discussion prompt
A rational function is one polynomial divided by another. That is the whole definition.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Concept
Both the top and the bottom must be polynomials: whole-number exponents, no variable under a radical, no variable in an exponent.
| Function | Rational? | Why |
|---|---|---|
| 1 over x | yes | constant over degree one |
| (2x + 1) over (x - 5) | yes | degree one over degree one |
| (x squared - 9) over 4 | yes | a polynomial over a constant |
| (square root of x) over (x + 1) | no | the top is not a polynomial |
| 2 to the x over x | no | a variable exponent is not a polynomial |
Discrimination
Sort into buckets
Sort these by Rational?, from memory, without looking back at Which functions qualify. Telling them apart on the spot is the skill; the table is only where the answer happens to be written down.
Picture it
Animation
Shows: Which part wins far out — a rendered Manim animation.
Rendered with Manim.
Takeaway: Top growing faster means the ratio grows without bound — no horizontal asymptote.
Intuition
Every interesting thing a rational function does traces back to its bottom.
Where the bottom is a normal nonzero number, the function behaves like any other curve. Where the bottom gets close to zero, the outputs explode. Where the bottom is exactly zero, there is no output at all.
So the first thing you ever do with a rational function is look at the denominator and ask: where is this zero?
Analogy
Discussion prompt
Explain The denominator is in charge by analogy to something with no College Algebra in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Every interesting thing a rational function does traces back to its bottom.
Concept
Division by zero is undefined, so every input that makes the bottom zero is removed from the domain. Everything else stays.
\[ \text{domain of } \frac{P(x)}{Q(x)} = \{\, x \in \mathbb{R} \;:\; Q(x) \ne 0 \,\} \]
Then write that set in interval notation by cutting the number line at each excluded value and joining the pieces with a union.
Explain it
Discussion prompt
Explain Domain: throw out the zeros of the denominator to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Division by zero is undefined, so every input that makes the bottom zero is removed from the domain. Everything else stays.
Ranking
Put in order
Put the moves of Worked example: a domain in interval notation into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Only a zero denominator can break the function.
Worked example
Find the domain of this function.
\[ f(x) = \frac{x + 3}{x^2 - 4x - 5} \]
Set the denominator equal to zero
Why: Only a zero denominator can break the function. The numerator is allowed to be anything, including zero.
Factor the denominator
Why: Two numbers multiplying to negative five and adding to negative four are negative five and positive one.
\[ x^2 - 4x - 5 = (x - 5)(x + 1) \]
Set each factor to zero
Why: A product is zero exactly when one of its factors is zero.
\[ x - 5 = 0 \;\Rightarrow\; x = 5 \qquad x + 1 = 0 \;\Rightarrow\; x = -1 \]
Cut the number line at the two excluded values
Why: Both excluded points get round brackets, because they are not in the domain.
\[ (-\infty,\, -1) \cup (-1,\, 5) \cup (5,\, \infty) \]
Verify by testing the excluded values and one allowed value
Why: At the excluded inputs the bottom really is zero, and an ordinary input really does produce a number.
\[ \begin{aligned} (-1)^2 - 4(-1) - 5 &= 1 + 4 - 5 = 0 \\ 5^2 - 4(5) - 5 &= 25 - 20 - 5 = 0 \\ f(0) &= \frac{3}{-5} = -\frac{3}{5} \end{aligned} \]
Picture it
Animation
Shows: Each line of the worked example "a domain in interval notation", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At the excluded inputs the bottom really is zero, and an ordinary input really does produce a number.
Concept
Figure (svg): Graph of the reciprocal function, two branches hugging the x-axis and the y-axis
The simplest rational function is the reciprocal function.
\[ f(x) = \frac{1}{x} \]
Its domain leaves out one input, and the graph splits into two separate branches because of it.
\[ (-\infty,\, 0) \cup (0,\, \infty) \]
The vertical line and the horizontal line the graph hugs but never reaches are the two asymptotes. Every rational function in this deck is a variation on this picture.
Intuition
Why does the graph run off the page near a denominator zero? Watch what one divided by a shrinking number does.
| input | output of 1 over x |
|---|---|
| 0.1 | 10 |
| 0.01 | 100 |
| 0.001 | 1000 |
| 0.0001 | 10000 |
The closer the bottom gets to zero, the bigger the output. It never reaches infinity and there is no output at zero itself, but the outputs grow without bound. That runaway is what a vertical asymptote records.
Approach from the negative side instead and the same thing happens downward: the outputs are large and negative.
Pattern
Step through it
Step through Dividing by a tiny number gives a huge answer one row at a time. What is driving the change, and what would the row after the last one be?
Concept
vertical asymptote — A vertical line that the graph shoots up or down along as the inputs approach a fixed value. The function is undefined at that value.
\[ x = a \text{ is a vertical asymptote if } |f(x)| \to \infty \text{ as } x \to a \]
It is a line you draw on your sketch as a guide. It is not part of the graph, so draw it dashed.
Definition probe
Sort into buckets
Every line below is part of the definition of rational function or of vertical asymptote — one or the other, never both. Put each where it belongs.
Picture it
Animation
Shows: A vertical asymptote is a forbidden input — a rendered Manim animation.
Rendered with Manim.
Takeaway: The denominator approaches zero while the numerator does not.
Concept
This one is absolute. A vertical asymptote sits at an input the function does not have, so there is no point there to cross with.
A function gives exactly one output per input. Crossing a vertical line would require a second point at that same input, or a point at an input that is excluded. Neither can happen.
Horizontal and slant asymptotes are different, and we will use that difference later. Only the vertical ones are uncrossable.
Section
Part 2
Concept
Before you say a single word about a rational function, factor the top and the bottom completely.
Unfactored, every denominator zero looks the same. Factored, you can see which ones cancel against the numerator and which ones survive. That one distinction decides hole versus vertical asymptote.
\[ \frac{x^2 - 9}{x^2 - 2x - 3} \;=\; \frac{(x-3)(x+3)}{(x-3)(x+1)} \]
Concept
If a factor appears on the top and the bottom, it cancels. The simplified function is fine at that input, but the original was not, so the point is punched out.
hole (removable discontinuity) — A single missing point on an otherwise ordinary curve, caused by a factor that cancels from top and bottom. The graph continues on both sides of it.
Cancelling never puts the input back in the domain. The domain is decided by the original denominator, before any simplification.
Step zero
Discussion prompt
Worked example: hole and vertical asymptote — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Factor top and bottom completely
Answer:
Worked example
Find every hole and vertical asymptote.
\[ f(x) = \frac{x^2 - 9}{x^2 - 2x - 3} \]
Factor top and bottom completely
Why: The top is a difference of squares. The bottom needs two numbers multiplying to negative three and adding to negative two: negative three and positive one.
\[ f(x) = \frac{(x-3)(x+3)}{(x-3)(x+1)} \]
Read the domain off the ORIGINAL denominator
Why: Both bottom factors are zero somewhere, so both inputs are excluded no matter what cancels later.
\[ x \ne 3, \qquad x \ne -1 \]
Cancel the shared factor
Why: The factor x minus three divides out of top and bottom, but only for inputs where it is not zero.
\[ f(x) = \frac{x+3}{x+1}, \qquad x \ne 3 \]
The surviving denominator zero is the vertical asymptote
Why: At this input the bottom is zero while the top is not, so the outputs blow up.
\[ x = -1 \text{ is a vertical asymptote} \]
The cancelled zero is a hole; get its height from the simplified form
Why: Plug the cancelled input into the simplified expression to find the y-value the graph is heading toward.
\[ y = \frac{3+3}{3+1} = \frac{6}{4} = \frac{3}{2} \;\Rightarrow\; \text{hole at } \left(3, \tfrac{3}{2}\right) \]
Verify with the original function on both sides of each break
Why: Near the hole the original outputs close in on three halves; near the asymptote they run away, which is exactly the difference we claimed.
| x | original f(x) | reading |
|---|---|---|
| 2.9 | about 1.513 | closing in on 1.5 |
| 3.1 | about 1.488 | closing in on 1.5 |
| -0.9 | 21 | running away upward |
| -1.1 | -19 | running away downward |
Picture it
Animation
Shows: Cancel first, then classify — a rendered Manim animation.
Rendered with Manim.
Takeaway: The uncancelled zeros are the walls; the cancelled ones are gaps.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Set the bottom equal to zero and call every answer a vertical asymptote.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The bottom really is zero at both inputs, so it feels like both should send the graph to infinity.
Factor first, then decide. Only a factor that survives the cancelling is an asymptote.
Why: The bottom really is zero at both inputs, so it feels like both should send the graph to infinity.
Trap
Set the bottom equal to zero and call every answer a vertical asymptote.
\[ f(x) = \frac{x^2 - 4}{x^2 - x - 2} \]
Solve the denominator equal to zero and claim two vertical asymptotes
Why: The bottom really is zero at both inputs, so it feels like both should send the graph to infinity.
\[ x^2 - x - 2 = 0 \;\Rightarrow\; x = 2 \text{ and } x = -1 \;\text{(both claimed as asymptotes)} \]
But the actual outputs near the first value do not blow up
Why: Evaluate the original function just to the left and just to the right of that input. Nothing runs away.
| x | f(x) | behavior |
|---|---|---|
| 1.9 | about 1.345 | settling near 1.333 |
| 2.1 | about 1.323 | settling near 1.333 |
Factor first, then decide. Only a factor that survives the cancelling is an asymptote.
\[ f(x) = \frac{(x-2)(x+2)}{(x-2)(x+1)} \]
Cancel the shared factor and record the exclusion
Why: The factor divides out of both parts, but the input it kills is still barred from the domain.
\[ f(x) = \frac{x+2}{x+1}, \qquad x \ne 2 \]
Surviving zero is the asymptote; the cancelled zero is a hole
Why: The height of the hole comes from the simplified expression, which agrees with those two table values.
\[ \text{VA: } x = -1, \qquad \text{hole at } \left(2, \tfrac{4}{3}\right) \]
Comparison
Comparison matrix
From Trap: calling a cancelled factor a vertical asymptote: refill the behavior column from what you know. The rest of the table is as it appeared.
| x | f(x) | behavior |
|---|---|---|
| 1.9 | about 1.345 | settling near 1.333 |
| 2.1 | about 1.323 | settling near 1.333 |
Concept
Figure (svg): A smooth rising curve with a single open circle punched out of it
A hole is not a dramatic feature. The curve walks right up to it from the left and from the right.
The only thing wrong is that one single point has been punched out. On a hand sketch you draw an open circle there to say the point is missing.
This is why a hole is also called a removable discontinuity: if you were allowed to fill that one point back in, the function would be perfectly continuous.
Estimation
Predict first
Find the domain, every hole, and every vertical asymptote.
Commit before you compute: what does Worked example: a hole on the left, an asymptote on the… come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify with values from the ORIGINAL function on both sides of each break
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Near the hole the outputs close in on five sixths, about 0.833.
Worked example
Find the domain, every hole, and every vertical asymptote.
\[ g(x) = \frac{x^2 + x - 6}{x^2 - 9} \]
Factor top and bottom completely
Why: The top needs two numbers multiplying to negative six and adding to positive one: positive three and negative two. The bottom is a difference of squares.
\[ g(x) = \frac{(x+3)(x-2)}{(x-3)(x+3)} \]
Read the domain off the original denominator
Why: Both bottom factors vanish somewhere, and cancelling later never restores a barred input.
\[ (-\infty,\, -3) \cup (-3,\, 3) \cup (3,\, \infty) \]
Cancel the shared factor and carry the exclusion along
Why: The factor of x plus three divides out, but only where it is not itself zero.
\[ g(x) = \frac{x-2}{x-3}, \qquad x \ne -3 \]
Name the vertical asymptote
Why: At this input the surviving bottom is zero while the top is one, not zero, so the quotient runs away.
\[ x = 3 \text{ is a vertical asymptote} \]
Name the hole and compute its height
Why: Substitute the cancelled input into the simplified expression to find the y-value the curve is aiming for.
\[ y = \frac{-3-2}{-3-3} = \frac{-5}{-6} = \frac{5}{6} \;\Rightarrow\; \text{hole at } \left(-3, \tfrac{5}{6}\right) \]
Verify with values from the ORIGINAL function on both sides of each break
Why: Near the hole the outputs close in on five sixths, about 0.833. Near the asymptote they blow up in opposite directions. That is exactly the claimed difference.
| x | original g(x) | reading |
|---|---|---|
| -3.1 | about 0.836 | closing in on 0.833 |
| -2.9 | about 0.831 | closing in on 0.833 |
| 2.9 | -9 | running away downward |
| 3.1 | 11 | running away upward |
Picture it
Animation
Shows: Behaviour on either side of a wall — a rendered Manim animation.
Rendered with Manim.
Takeaway: Both sides run up here; with an odd power they run opposite ways.
Ranking
Put in order
These are the steps of Recipe: holes and vertical asymptotes, scrambled. Put them back in order before the next slide shows you.
Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.
Pattern
Every hole-versus-asymptote question is the same five moves.
The domain does not change at step three. Cancelling changes the formula, never the set of legal inputs.
Edge cases
Discussion prompt
Recipe: holes and vertical asymptotes works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".
Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.
Answer:
Every hole-versus-asymptote question is the same five moves.
Elimination
Eliminate the wrong options
Which description of the graph is correct?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Factored, the function is (x-5)(x+5) over (x-5)(x+2). The factor x-5 cancels, so x = 5 is a hole; the surviving zero x = -2 is the vertical asymptote. The hole's height comes from the simplified form (x+5)/(x+2) at x = 5, which is 10/7.
Check
Factor both parts first, then decide. Work it on paper before you pick.
\[ h(x) = \frac{x^2 - 25}{x^2 - 3x - 10} \]
Check your understanding
Which description of the graph is correct?
Answer: A
Why: Factored, the function is (x-5)(x+5) over (x-5)(x+2). The factor x-5 cancels, so x = 5 is a hole; the surviving zero x = -2 is the vertical asymptote. The hole's height comes from the simplified form (x+5)/(x+2) at x = 5, which is 10/7.
Section
Part 3
Intuition
A vertical asymptote answers what happens near a bad input. A horizontal asymptote answers a different question: what happens when the input gets enormous?
\[ f(x) = \frac{3x^2 - 5}{2x^2 + x - 6} \]
| x | f(x) |
|---|---|
| 10 | about 1.4461 |
| 100 | about 1.4927 |
| 1000 | about 1.4993 |
| 10000 | about 1.4999 |
The outputs are marching toward a single number. Once the input is huge, the small terms are rounding error next to the squared terms, so the whole thing behaves like three of something divided by two of the same something.
Pattern
Step through it
Step through Far from the origin, only the biggest terms matter one row at a time. What is driving the change, and what would the row after the last one be?
Concept
horizontal asymptote — A horizontal line that the graph levels off toward as the inputs run far out to the right or far out to the left. It describes end behavior only.
\[ y = L \text{ is a horizontal asymptote if } f(x) \to L \text{ as } x \to \infty \text{ or as } x \to -\infty \]
A rational function has at most one horizontal asymptote. Which one it is depends on nothing but the two degrees and the two leading coefficients.
Picture it
Animation
Shows: The horizontal asymptote is end behaviour — a rendered Manim animation.
Rendered with Manim.
Takeaway: Far out, the leading coefficients decide everything.
Concept
If the denominator has the higher degree, the bottom grows faster than the top, so the fraction is squeezed toward zero.
\[ f(x) = \frac{3x + 1}{x^2 + 2} \;\Longrightarrow\; y = 0 \]
| x | f(x) |
|---|---|
| 10 | about 0.3039 |
| 100 | about 0.0301 |
| 1000 | about 0.0030 |
Pattern
Step through it
Step through Case 1: the bottom wins the degree race one row at a time. What is driving the change, and what would the row after the last one be?
Concept
If the two degrees are equal, neither side outgrows the other. The graph levels off at the ratio of the two leading coefficients.
\[ f(x) = \frac{a_n x^n + \cdots}{b_n x^n + \cdots} \;\Longrightarrow\; y = \frac{a_n}{b_n} \]
Leading coefficient means the number attached to the highest-power term, once each polynomial is written in standard form. Nothing else in either polynomial matters here.
Concept
If the numerator has the higher degree, the outputs grow without bound. There is no horizontal asymptote at all.
\[ f(x) = \frac{x^2 + 1}{x - 1} \]
| x | f(x) |
|---|---|
| 10 | about 11.222 |
| 100 | about 101.02 |
| 1000 | about 1001.0 |
Nothing is levelling off, so no horizontal line can describe it. When the top degree is exactly one more, a slanted line does the job instead. That is Part 4.
Pattern
Step through it
Step through Case 3: the top wins the degree race one row at a time. What is driving the change, and what would the row after the last one be?
Fill the middle
Fill in the blanks
From Worked example: a horizontal asymptote with tied degrees — finish the line. Write what belongs on the right of the equals sign before you look.
f(x) = \frac{3x^2 - 5}{2x^2 + x - 6}
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The highest power on top is two and the highest power on the bottom is two, so the degrees tie.
Worked example
Find the horizontal asymptote.
\[ f(x) = \frac{3x^2 - 5}{2x^2 + x - 6} \]
Compare the degrees, not the constants
Why: The highest power on top is two and the highest power on the bottom is two, so the degrees tie. That puts us in case two.
\[ \deg(\text{top}) = 2, \qquad \deg(\text{bottom}) = 2 \]
Take the ratio of the leading coefficients
Why: Only the highest-power terms survive the comparison when the inputs are huge; everything else is dwarfed.
\[ y = \frac{3}{2} \]
See why, by dividing every term by the highest power
Why: Each leftover fraction has a fixed number on top and a growing power on the bottom, so each one drains away to zero.
\[ \frac{3x^2 - 5}{2x^2 + x - 6} = \frac{3 - \dfrac{5}{x^2}}{2 + \dfrac{1}{x} - \dfrac{6}{x^2}} \;\longrightarrow\; \frac{3 - 0}{2 + 0 - 0} = \frac{3}{2} \]
Verify by evaluating far out on the axis
Why: At an input of one thousand the function returns about 1.4993, which is three halves to three decimal places. The claim checks out.
\[ f(1000) = \frac{2{,}999{,}995}{2{,}000{,}994} \approx 1.4993 \approx \frac{3}{2} \]
Picture it
Animation
Shows: Each line of the worked example "a horizontal asymptote with tied degrees", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At an input of one thousand the function returns about 1.4993, which is three halves to three decimal places. The claim checks out.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Grabbing the two numbers with no variable attached and dividing them.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: They are the easiest numbers to see, so they get grabbed first.
End behavior is decided by the leading terms, because those are the terms that get huge.
Why: They are the easiest numbers to see, so they get grabbed first. This is the single most common asymptote error.
Trap
Grabbing the two numbers with no variable attached and dividing them.
\[ f(x) = \frac{2x + 6}{x - 3} \]
Divide the constant terms
Why: They are the easiest numbers to see, so they get grabbed first. This is the single most common asymptote error.
\[ y = \frac{6}{-3} = -2 \quad \text{(claimed)} \]
Test it far out and watch it fail
Why: The real outputs are near positive two, nowhere near negative two. The claim is not just imprecise, it has the wrong sign.
| x | f(x) | claimed value |
|---|---|---|
| 100 | about 2.124 | -2 |
| 1000 | about 2.012 | -2 |
End behavior is decided by the leading terms, because those are the terms that get huge.
\[ f(x) = \frac{2x + 6}{x - 3} \]
Degrees tie at one, so divide the leading coefficients
Why: The constants six and negative three are fixed numbers; at an input of a thousand they are noise next to the terms carrying the variable.
\[ y = \frac{2}{1} = 2 \]
Test it far out and watch it hold
Why: The outputs squeeze down onto two from above, exactly as claimed.
| x | f(x) | claimed value |
|---|---|---|
| 100 | about 2.124 | 2 |
| 1000 | about 2.012 | 2 |
Trade off
Comparison matrix
From Trap: reading the horizontal asymptote off the constant…: every row here is a choice with a cost. Fill the claimed value column, then say which row you would actually pick and what you give up for it.
| x | f(x) | claimed value |
|---|---|---|
| 100 | about 2.124 | -2 |
| 1000 | about 2.012 | -2 |
Concept
This surprises people, so say it out loud: a horizontal asymptote is a statement about the far ends of the graph only.
In the middle of the picture the curve is free to wander across that line, once or even several times. It only has to settle down eventually.
Vertical asymptotes are the strict ones. A vertical asymptote sits at an input the function does not have, so there is no point available to cross with. Horizontal and slant asymptotes have no such rule.
Picture it
Animation
Shows: A graph may cross its horizontal asymptote — a rendered Manim animation.
Rendered with Manim.
Takeaway: It describes the far ends, not a barrier in the middle.
Missing information
Discussion prompt
Does this graph cross its horizontal asymptote, and if so where?
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
Both degrees are two and both leading coefficients are one, so the ratio is one.
Worked example
Does this graph cross its horizontal asymptote, and if so where?
\[ f(x) = \frac{x^2 + 2x}{x^2 + 1} \]
Find the horizontal asymptote first
Why: Both degrees are two and both leading coefficients are one, so the ratio is one.
\[ y = \frac{1}{1} = 1 \]
Set the function equal to that value
Why: A crossing point is an input where the output really equals the asymptote's height, not merely approaches it.
\[ \frac{x^2 + 2x}{x^2 + 1} = 1 \]
Multiply both sides by the denominator
Why: Here that is safe: a squared term plus one is always at least one, so it is never zero and never negative. This is an equation, not an inequality, so no direction can flip.
\[ x^2 + 2x = x^2 + 1 \]
Subtract the squared term from both sides and solve
Why: The squared terms are identical, so they cancel and a linear equation is left.
\[ 2x = 1 \;\Rightarrow\; x = \frac{1}{2} \]
Verify by substituting back into the original function
Why: The output really is one, so the point on the graph sits exactly on the asymptote. The graph crosses its horizontal asymptote once, at one half.
\[ f\!\left(\tfrac{1}{2}\right) = \frac{\tfrac{1}{4} + 1}{\tfrac{1}{4} + 1} = \frac{\tfrac{5}{4}}{\tfrac{5}{4}} = 1 \;\checkmark \]
Pattern
Write both polynomials in standard form, then compare only the degrees.
| Degrees | Horizontal asymptote | Reason |
|---|---|---|
| bottom degree is bigger | y = 0 | the bottom outgrows the top |
| degrees are equal | y = leading coefficient of top over leading coefficient of bottom | neither side outgrows the other |
| top degree is bigger | none | the outputs grow without bound |
Never look at the constant terms for this. Never look at whether factors cancel, either. Cancelling can change a degree, so simplify first, then compare.
Picture it
Animation
Shows: Three cases, decided by degree — a rendered Manim animation.
Rendered with Manim.
Takeaway: Compare degrees before doing any arithmetic.
Prediction
Predict first
What is the horizontal asymptote of this function?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: y = 2
Why: Both polynomials have degree three, so the degrees tie and the asymptote is the ratio of leading coefficients: four over two, which is two. Checking at an input of one thousand gives about 1.9950, closing in on 2.
Check
Compare the degrees first, then the leading coefficients.
\[ f(x) = \frac{4x^3 - x}{2x^3 + 5x^2 - 1} \]
Check your understanding
What is the horizontal asymptote of this function?
Answer: A
Why: Both polynomials have degree three, so the degrees tie and the asymptote is the ratio of leading coefficients: four over two, which is two. Checking at an input of one thousand gives about 1.9950, closing in on 2.
Section
Part 4
Concept
Case three said there is no horizontal asymptote when the top wins the degree race. But there is still a pattern to describe, and when the top wins by exactly one degree it is a slanted line.
slant (oblique) asymptote — A non-horizontal line that the graph levels off along at the far ends. It appears exactly when the numerator's degree is one more than the denominator's degree.
A rational function never has a horizontal asymptote and a slant asymptote at the same time. It is one or the other, or neither.
Intuition
Long division of numbers gives a quotient plus a remainder. Long division of polynomials does the same thing.
\[ \frac{P(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)} \]
When the top degree is one more, the quotient is a line and the leftover fraction has a bigger bottom than top.
That leftover fraction is exactly a case-one situation, so it drains to zero far out. Whatever is left is the line. That line is the slant asymptote.
Step zero
Discussion prompt
Worked example: a slant asymptote by long division — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Check the degrees before dividing
Answer:
Worked example
Find every asymptote.
\[ f(x) = \frac{x^2 + 3x - 4}{x - 2} \]
Check the degrees before dividing
Why: The top is degree two and the bottom is degree one, a difference of exactly one. So expect no horizontal asymptote and one slant asymptote.
\[ \deg(\text{top}) - \deg(\text{bottom}) = 2 - 1 = 1 \]
Name the vertical asymptote
Why: The bottom is zero at two, and the top there is four plus six minus four, which is six, not zero. Nothing cancels, so the graph blows up there.
\[ x = 2 \text{ is a vertical asymptote} \]
Divide: first term of the quotient
Why: The leading term of the top divided by the leading term of the bottom gives the first piece of the quotient. Multiply back and subtract.
\[ \frac{x^2}{x} = x, \qquad x(x-2) = x^2 - 2x, \qquad (x^2 + 3x) - (x^2 - 2x) = 5x \]
Divide: second term of the quotient
Why: Bring down the constant and repeat. The remainder that is left has degree lower than the divisor, so the division stops.
\[ \frac{5x}{x} = 5, \qquad 5(x-2) = 5x - 10, \qquad (5x - 4) - (5x - 10) = 6 \]
Write the function as quotient plus remainder over divisor
Why: The leftover fraction has a constant on top and a growing bottom, so it drains to zero at the far ends. What remains is the line.
\[ f(x) = x + 5 + \frac{6}{x-2} \;\Longrightarrow\; \text{slant asymptote } y = x + 5 \]
Verify by multiplying the quotient back and by testing far out
Why: The rebuilt numerator matches the original exactly, and at an input of one hundred the function sits about six hundredths above the line, just as the leftover fraction predicts.
\[ \begin{aligned} (x+5)(x-2) + 6 &= x^2 + 3x - 10 + 6 = x^2 + 3x - 4 \;\checkmark \\ f(100) &= \frac{10296}{98} \approx 105.061, \qquad 100 + 5 = 105 \;\checkmark \end{aligned} \]
Picture it
Animation
Shows: A slant asymptote from long division — a rendered Manim animation.
Rendered with Manim.
Takeaway: Top-heavy by exactly one degree leaves a line plus a fading remainder.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Reaching for the leading-coefficient ratio without checking that the degrees actually tie.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Both leading coefficients happen to be one, so the ratio looks reassuringly clean.
Compare the degrees before touching the coefficients. Here the top wins by one, so divide instead.
Why: Both leading coefficients happen to be one, so the ratio looks reassuringly clean. But that rule only applies when the degrees are equal.
Trap
Reaching for the leading-coefficient ratio without checking that the degrees actually tie.
\[ f(x) = \frac{x^2 + 3x - 4}{x - 2} \]
Divide the leading coefficients and call it a horizontal asymptote
Why: Both leading coefficients happen to be one, so the ratio looks reassuringly clean. But that rule only applies when the degrees are equal.
\[ y = \frac{1}{1} = 1 \quad \text{(claimed)} \]
Test it far out and watch it collapse
Why: The outputs are in the thousands, not levelling off near one. Nothing about this graph settles onto a horizontal line.
| x | f(x) | claimed value |
|---|---|---|
| 100 | about 105.06 | 1 |
| 1000 | about 1005.0 | 1 |
Compare the degrees before touching the coefficients. Here the top wins by one, so divide instead.
\[ f(x) = \frac{x^2 + 3x - 4}{x - 2} = x + 5 + \frac{6}{x-2} \]
There is no horizontal asymptote; the slant asymptote is the quotient line
Why: The leftover fraction fades to zero far out, leaving the line to describe the end behavior.
\[ y = x + 5 \]
Test it far out and watch it hold
Why: The function tracks the line to within a few hundredths at large inputs, which is exactly what an asymptote should do.
| x | f(x) | line value |
|---|---|---|
| 100 | about 105.06 | 105 |
| 1000 | about 1005.0 | 1005 |
Comparison
Comparison matrix
From Trap: quoting a horizontal asymptote when the top degree is…: refill the claimed value column from what you know. The rest of the table is as it appeared.
| x | f(x) | claimed value |
|---|---|---|
| 100 | about 105.06 | 1 |
| 1000 | about 1005.0 | 1 |
Pattern
Simplify first, then subtract the degrees: top degree minus bottom degree.
| Top degree minus bottom degree | End-behavior asymptote |
|---|---|
| negative (bottom is bigger) | horizontal, at y = 0 |
| zero (degrees tie) | horizontal, at the ratio of leading coefficients |
| exactly one | slant, found by long division |
| two or more | neither; the ends curve away |
Vertical asymptotes are a separate question with a separate answer. Compute those from the surviving denominator factors, no matter which row of this table you land in.
Comparison
Comparison matrix
From Recipe: which asymptote am I even looking for?: refill the End-behavior asymptote column from what you know. The rest of the table is as it appeared.
| Top degree minus bottom degree | End-behavior asymptote |
|---|---|
| negative (bottom is bigger) | horizontal, at y = 0 |
| zero (degrees tie) | horizontal, at the ratio of leading coefficients |
| exactly one | slant, found by long division |
| two or more | neither; the ends curve away |
Prediction
Predict first
What is the slant asymptote of this function?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: y = 2x - 5
Why: Long division gives a quotient of 2x - 5 with a remainder of 6, so the function equals 2x - 5 plus 6 over (x + 1). Multiplying back confirms it: (2x - 5)(x + 1) + 6 equals 2x squared minus 3x plus 1.
Check
Confirm the degree gap, then divide. Do the long division on paper before you pick.
\[ f(x) = \frac{2x^2 - 3x + 1}{x + 1} \]
Check your understanding
What is the slant asymptote of this function?
Answer: A
Why: Long division gives a quotient of 2x - 5 with a remainder of 6, so the function equals 2x - 5 plus 6 over (x + 1). Multiplying back confirms it: (2x - 5)(x + 1) + 6 equals 2x squared minus 3x plus 1.
Section
Part 5
Concept
The y-intercept is where the graph meets the vertical axis, so the input is zero. Substitute zero and simplify.
\[ y\text{-intercept} = \bigl(0,\, f(0)\bigr) \]
A function has at most one y-intercept. If zero is not in the domain, meaning the denominator is zero there, then the graph simply has none.
Concept
A fraction equals zero exactly when its top is zero and its bottom is not. So set the numerator equal to zero.
\[ f(x) = 0 \iff P(x) = 0 \text{ and } Q(x) \ne 0 \]
Use the numerator of the simplified function. A numerator zero that cancelled away is a hole, not an intercept: there is no point on the graph there at all.
Fill the middle
Fill in the blanks
From Worked example: intercepts when a factor cancels — finish the line. Write what belongs on the right of the equals sign before you look.
f(x) = \frac{(x-3)(x+2)}{(x-2)(x+2)}
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The top needs two numbers multiplying to negative six and adding to negative one: negative three and positive two.
Worked example
Find the intercepts, the hole, and the vertical asymptote.
\[ f(x) = \frac{x^2 - x - 6}{x^2 - 4} \]
Factor both parts
Why: The top needs two numbers multiplying to negative six and adding to negative one: negative three and positive two. The bottom is a difference of squares.
\[ f(x) = \frac{(x-3)(x+2)}{(x-2)(x+2)} \]
Record the domain, then cancel
Why: Both original bottom zeros stay excluded. The shared factor divides out and leaves a much simpler expression.
\[ f(x) = \frac{x-3}{x-2}, \qquad x \ne 2,\; x \ne -2 \]
Sort the two excluded inputs
Why: The surviving factor makes the asymptote; the cancelled factor makes a hole whose height comes from the simplified form.
\[ \text{VA: } x = 2; \qquad \text{hole at } \left(-2,\; \frac{-2-3}{-2-2}\right) = \left(-2, \tfrac{5}{4}\right) \]
Find the y-intercept by substituting zero
Why: Zero is in the domain, so this point exists. Use either form of the function; they agree everywhere the original is defined.
\[ f(0) = \frac{0-3}{0-2} = \frac{3}{2} \;\Rightarrow\; \left(0, \tfrac{3}{2}\right) \]
Find the x-intercept from the simplified numerator
Why: The only surviving numerator zero is three, and three is a legal input. The other numerator zero, negative two, cancelled, so it is the hole rather than an intercept.
\[ x - 3 = 0 \;\Rightarrow\; x = 3 \;\Rightarrow\; (3, 0) \]
Verify both intercepts in the ORIGINAL function
Why: The original agrees with the simplified form at both points, which confirms the cancelling did not move anything.
\[ \begin{aligned} f(3) &= \frac{9 - 3 - 6}{9 - 4} = \frac{0}{5} = 0 \;\checkmark \\ f(0) &= \frac{0 - 0 - 6}{0 - 4} = \frac{-6}{-4} = \frac{3}{2} \;\checkmark \end{aligned} \]
Picture it
Animation
Shows: Each line of the worked example "intercepts when a factor cancels", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The original agrees with the simplified form at both points, which confirms the cancelling did not move anything.
Check
Factor, cancel, and only then read the intercepts.
\[ f(x) = \frac{x^2 - 4}{x^2 + x - 6} \]
Check your understanding
What are the x-intercepts of this graph?
Answer: A
Why: Factored, the function is (x-2)(x+2) over (x+3)(x-2). The factor x-2 cancels, leaving (x+2)/(x+3). Only x = -2 makes the surviving numerator zero, so the single x-intercept is (-2, 0); x = 2 is a hole with no point on the graph.
Concept
A rational function can only switch between positive and negative at two kinds of places: an x-intercept, where it passes through zero, or a vertical asymptote, where it jumps.
Between two consecutive such places the function keeps one sign the whole way. So test a single convenient input per interval and that answers the whole interval.
That is the entire idea of a sign chart, and it is the tool that turns a list of asymptotes into an actual picture.
Explain it
Discussion prompt
Explain A sign chart tells you which side of the axis each piece is on to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
A rational function can only switch between positive and negative at two kinds of places: an x-intercept, where it passes through zero, or a vertical asymptote, where it jumps.
Ranking
Put in order
Put the moves of Worked example: a complete sketch, start to finish into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Both parts are already fully factored and share nothing, so there are no holes to worry about.
Worked example
Figure (svg): Graph of (x+1)/(x-2): a dashed vertical asymptote at x equals 2, a dashed horizontal asymptote at y equals 1, a left branch crossing the axis at negative one and dropping, and a right branch coming down from above
Analyze and sketch this function completely.
\[ f(x) = \frac{x+1}{x-2} \]
Factor and look for cancelling
Why: Both parts are already fully factored and share nothing, so there are no holes to worry about.
\[ \text{no common factors} \;\Rightarrow\; \text{no holes} \]
Domain and vertical asymptote
Why: The bottom is zero at two and the top there is three, not zero, so this is a genuine blow-up rather than a hole.
\[ (-\infty, 2) \cup (2, \infty); \qquad \text{VA: } x = 2 \]
End behavior
Why: Both degrees are one and both leading coefficients are one, so the tied-degrees rule gives the ratio one.
\[ \text{HA: } y = 1 \]
Intercepts
Why: The numerator is zero at negative one, which is a legal input. Substituting zero gives the height where the curve meets the vertical axis.
\[ (-1, 0) \quad \text{and} \quad \left(0, -\tfrac{1}{2}\right) \]
Does it cross the horizontal asymptote?
Why: Setting the function equal to one gives a false statement, so no input works. This graph stays off its horizontal asymptote everywhere.
\[ \frac{x+1}{x-2} = 1 \;\Rightarrow\; x + 1 = x - 2 \;\Rightarrow\; 1 = -2 \;\text{(false)} \]
Build the sign chart from the two critical values
Why: The intercept and the asymptote cut the line into three intervals. One test value settles each interval completely.
| interval | test value | f(test) | sign |
|---|---|---|---|
| less than -1 | -2 | 1/4 | positive |
| between -1 and 2 | 0 | -1/2 | negative |
| greater than 2 | 3 | 4 | positive |
Verify the sketch against real function values at the ends
Why: Far to the right the outputs sit just above one; far to the left just below one. That matches the picture, where the right branch comes down onto the dashed line and the left branch rises up to it.
| x | f(x) | position relative to y = 1 |
|---|---|---|
| 100 | about 1.0306 | just above |
| -100 | about 0.9706 | just below |
| 1.9 | -29 | diving down the asymptote |
| 2.1 | 31 | climbing up the asymptote |
Picture it
Animation
Shows: Each line of the worked example "a complete sketch, start to finish", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Far to the right the outputs sit just above one; far to the left just below one. That matches the picture, where the right branch comes down onto the dashed line and the left branch rises up to it.
Hypothesis
Predict first
Worked example: a sketch with two vertical asymptotes is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Factor and check for cancelling
Why: The top is two times x times x and the bottom is a difference of squares. They share no factor, so there are no holes.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Figure (svg): Sign chart on a number line with critical values at negative three, zero, and three, marked plus, minus, minus, plus
Analyze this function completely.
\[ f(x) = \frac{2x^2}{x^2 - 9} \]
Factor and check for cancelling
Why: The top is two times x times x and the bottom is a difference of squares. They share no factor, so there are no holes.
\[ f(x) = \frac{2x \cdot x}{(x-3)(x+3)} \]
Two vertical asymptotes
Why: Both bottom factors survive, and at each of those inputs the top is eighteen, not zero.
\[ x = 3 \quad \text{and} \quad x = -3 \]
End behavior
Why: The degrees tie at two, so take the ratio of leading coefficients: two over one.
\[ \text{HA: } y = 2 \]
Intercepts, and a check for crossing
Why: The only numerator zero is zero itself, which serves as both intercepts. Setting the function equal to two gives a contradiction, so the graph never touches its horizontal asymptote.
\[ (0,0); \qquad \frac{2x^2}{x^2-9} = 2 \;\Rightarrow\; 2x^2 = 2x^2 - 18 \;\Rightarrow\; 0 = -18 \;\text{(false)} \]
Sign chart on the three critical values
Why: The zero at the origin is doubled, so the sign does not change as you pass it. The two asymptotes do flip the sign.
\[ \text{cut points: } x = -3,\; 0,\; 3 \]
Verify each region with a test value in the original function
Why: Every computed sign matches the chart, including the doubled zero at the origin where the graph touches the axis and turns back down instead of crossing.
| interval | test value | f(test) | sign |
|---|---|---|---|
| less than -3 | -4 | 32/7, about 4.571 | positive |
| between -3 and 0 | -1 | -1/4 | negative |
| between 0 and 3 | 1 | -1/4 | negative |
| greater than 3 | 4 | 32/7, about 4.571 | positive |
Pattern
Same seven moves every time, in this order. Do not skip step one.
Draw the asymptotes as dashed lines before you draw any curve. They are the scaffolding; getting them down first makes the branches almost draw themselves.
Picture it
Animation
Shows: Sign chart for a rational function — a rendered Manim animation.
Rendered with Manim.
Takeaway: Both sets of marks split the line, though only one set is in the domain.
Section
Part 6
Concept
Reading a rational function is a skill. Writing one to order is the same skill run backwards, and it is often an exam question.
| Feature you want | What to write |
|---|---|
| vertical asymptote at a | the factor (x - a) in the denominator ONLY |
| x-intercept at b | the factor (x - b) in the numerator ONLY |
| hole at c | the factor (x - c) in BOTH the numerator and the denominator |
| horizontal asymptote y = 0 | make the denominator's degree higher |
| horizontal asymptote y = k | match the degrees, then multiply the numerator so the leading coefficients divide to k |
The word only is doing all the work in the first two rows. A factor that shows up on both sides cancels, and a cancelled factor is a hole, not an asymptote and not an intercept.
Analogy
Discussion prompt
Explain Every feature you want is a factor you place by analogy to something with no College Algebra in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Reading a rational function is a skill. Writing one to order is the same skill run backwards, and it is often an exam question.
Step zero
Discussion prompt
Worked example: writing a function to order — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Place the vertical asymptote in the denominator
Answer:
Worked example
Write a rational function with a vertical asymptote at three, an x-intercept at negative one, and a horizontal asymptote at two.
Place the vertical asymptote in the denominator
Why: A vertical asymptote needs a bottom factor that is zero there and does not cancel, so it goes on the bottom and nowhere else.
\[ \text{denominator so far: } (x - 3) \]
Place the x-intercept in the numerator
Why: The function is zero exactly where the surviving top is zero, so the factor that vanishes at negative one goes on the top only.
\[ f(x) = \frac{x + 1}{x - 3} \]
Check the horizontal asymptote this draft already has
Why: Both degrees are one and both leading coefficients are one, so the draft levels off at one. We were asked for two.
\[ y = \frac{1}{1} = 1 \quad \text{(not yet what we want)} \]
Scale the numerator to fix the ratio
Why: Multiplying the whole top by two doubles the leading coefficient without moving any zero, so the intercept and the asymptote stay exactly where they were.
\[ f(x) = \frac{2(x + 1)}{x - 3} = \frac{2x + 2}{x - 3} \]
Verify all three requested features on the finished function
Why: The bottom is zero at three while the top there is eight, so that is a true asymptote and not a hole. The top is zero at negative one, which is a legal input. The degrees tie, so the ratio two over one is the horizontal asymptote, and a far-out value confirms it.
\[ \begin{aligned} \text{VA: } & x = 3 \;\checkmark \quad (\text{top at } 3 \text{ is } 8 \ne 0) \\ \text{x-int: } & 2(x+1) = 0 \Rightarrow x = -1 \;\checkmark \\ \text{HA: } & f(1000) = \frac{2002}{997} \approx 2.008 \;\checkmark \end{aligned} \]
Picture it
Animation
Shows: Each line of the worked example "writing a function to order", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Both degrees are one and both leading coefficients are one, so the draft levels off at one. We were asked for two.
Estimation
Predict first
Write a rational function with a vertical asymptote at negative four, a hole at two, an x-intercept at five, and a horizontal asymptote at three.
Commit before you compute: what does Worked example: writing one that also needs a hole come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify all four requested features
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Each one checks out on the finished function: the surviving bottom zero is the asymptote, the cancelled zero is the hole, the surviving top zero is the intercept, and the tied degrees give the ratio three.
Worked example
Write a rational function with a vertical asymptote at negative four, a hole at two, an x-intercept at five, and a horizontal asymptote at three.
Put the asymptote factor on the bottom and the intercept factor on the top
Why: Same two moves as before. Neither of these factors may appear on the other side, or it would cancel.
\[ \text{draft: } \frac{x - 5}{x + 4} \]
Put the hole factor on BOTH sides
Why: A hole is exactly a factor that cancels: it bars the input from the domain but does not make the graph blow up.
\[ \text{draft: } \frac{(x-5)(x-2)}{(x+4)(x-2)} \]
Fix the horizontal asymptote by scaling
Why: The degrees already tie at two, so the ratio of leading coefficients is currently one. Multiplying the top by three makes it three without disturbing any zero.
\[ f(x) = \frac{3(x-5)(x-2)}{(x+4)(x-2)} \]
Simplify to read the hole's height
Why: The cancelled factor's height comes from substituting two into the simplified expression.
\[ f(x) = \frac{3(x-5)}{x+4}, \; x \ne 2 \;\Rightarrow\; \text{hole at } \left(2, \frac{3(-3)}{6}\right) = \left(2, -\tfrac{3}{2}\right) \]
Verify all four requested features
Why: Each one checks out on the finished function: the surviving bottom zero is the asymptote, the cancelled zero is the hole, the surviving top zero is the intercept, and the tied degrees give the ratio three.
| requested | check on f | result |
|---|---|---|
| VA at -4 | bottom zero at -4, top there is 3(-9) = -27, not zero | confirmed |
| hole at 2 | factor (x - 2) cancels; height is -3/2 | confirmed |
| x-intercept at 5 | f(5) = 3(0)(3) over (9)(3) = 0 | confirmed |
| HA y = 3 | degrees tie at 2; leading coefficients 3 and 1 | confirmed |
Pattern
There is never exactly one right answer to these. Any function with the same factors and the same degrees works. Your job is to make sure yours really has every listed feature.
Picture it
Animation
Shows: Sketching a rational function — a rendered Manim animation.
Rendered with Manim.
Takeaway: In that order, the graph has almost no freedom left.
Section
Part 7
Intuition
Asking where an expression is positive is the same as asking where its graph sits above the horizontal axis. Negative means below.
A graph can only get from above to below by passing through the axis or by jumping across a vertical asymptote. There is no third way.
So the answer is always a union of whole intervals, and the endpoints of those intervals are exactly the zeros and the breaks. Find those, test one point in each interval, done.
Concept
critical value — An input where the expression is either zero or undefined. For a fraction, that means every zero of the numerator and every zero of the denominator.
Both kinds cut the number line, but they behave differently at the cut. A numerator zero makes the expression equal zero, so it may be included when the inequality allows equality.
A denominator zero makes the expression undefined, so it is never included, no matter which inequality symbol is used.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of vertical asymptote, hole (removable discontinuity), horizontal asymptote, slant (oblique) asymptote, critical value as Rational Functions, Asymptotes, and Inequalities uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Pattern
Predict first
The table runs: less than -3 | -4 | (-8)(-1) = 8 | positive · between -3 and 4 | 0 | (-4)(3) = -12 | negative
In Worked example: a quadratic inequality, given the rows so far: what is the next one — the row where interval is greater than 4?
Correct: greater than 4 | 5 | (1)(8) = 8 | positive
| interval | test value | product | sign |
|---|---|---|---|
| less than -3 | -4 | (-8)(-1) = 8 | positive |
| between -3 and 4 | 0 | (-4)(3) = -12 | negative |
| greater than 4 | 5 | (1)(8) = 8 | positive |
Why: The relationship between the columns, not the individual numbers, is what generates the next row. Sign charts only work on an expression compared to zero.
Worked example
Figure (svg): Sign chart on a number line with cut points at negative three and four, marked plus, minus, plus
Solve and write the answer in interval notation.
\[ x^2 - x - 12 \le 0 \]
Confirm one side is already zero
Why: Sign charts only work on an expression compared to zero. Here the right side is zero already, so nothing needs moving.
\[ x^2 - x - 12 \le 0 \]
Factor the left side
Why: Two numbers multiplying to negative twelve and adding to negative one are negative four and positive three. Factoring turns the sign question into a product of signs.
\[ (x - 4)(x + 3) \le 0 \]
List the critical values
Why: These are the only inputs where the product is zero, so they are the only places the sign can change.
\[ x = 4 \quad \text{and} \quad x = -3 \]
Test one value in each of the three intervals
Why: One test value settles an entire interval, because the sign cannot change without passing a critical value.
| interval | test value | product | sign |
|---|---|---|---|
| less than -3 | -4 | (-8)(-1) = 8 | positive |
| between -3 and 4 | 0 | (-4)(3) = -12 | negative |
| greater than 4 | 5 | (1)(8) = 8 | positive |
Read off the answer, including the endpoints
Why: We want the expression to be negative or zero. The middle interval is negative, and both critical values make it exactly zero, which the non-strict symbol allows. Polynomials are defined everywhere, so nothing is barred.
\[ [-3,\, 4] \]
Verify by testing a point in each region and both endpoints
Why: Inside the claimed interval the inequality is true; outside it is false on both sides; and the two endpoints give exactly zero, which the symbol permits. The answer is confirmed.
| x | value of the left side | is it at most zero? | expected |
|---|---|---|---|
| -4 (outside) | 16 + 4 - 12 = 8 | no | outside, correct |
| 0 (inside) | 0 - 0 - 12 = -12 | yes | inside, correct |
| 5 (outside) | 25 - 5 - 12 = 8 | no | outside, correct |
| -3 (endpoint) | 9 + 3 - 12 = 0 | yes | included, correct |
| 4 (endpoint) | 16 - 4 - 12 = 0 | yes | included, correct |
Picture it
Animation
Shows: Each line of the worked example "a quadratic inequality", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Inside the claimed interval the inequality is true; outside it is false on both sides; and the two endpoints give exactly zero, which the symbol permits. The answer is confirmed.
Estimation
Predict first
Solve and write the answer in interval notation.
Commit before you compute: what does Worked example: a cubic inequality with three cut points come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify by testing a point in each claimed region and each excluded one
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Both claimed regions test true, both rejected regions test false, and each endpoint gives exactly zero, which a strict inequality rejects.
Worked example
Figure (svg): Sign chart on a number line with cut points at negative two, zero, and two, marked minus, plus, minus, plus
Solve and write the answer in interval notation.
\[ x^3 - 4x > 0 \]
Factor completely
Why: Pull the common factor out first, then recognize the difference of squares. Never divide both sides by the variable factor: that would throw away information about where it is zero.
\[ x(x^2 - 4) = x(x-2)(x+2) > 0 \]
List the three critical values
Why: Each factor contributes one zero, and each zero is a place the product can change sign.
\[ x = -2,\; 0,\; 2 \]
Test one value in each of the four intervals
Why: Substitute into the original expression, not the factored one, so any algebra slip shows up immediately.
| interval | test value | value of the expression | sign |
|---|---|---|---|
| less than -2 | -3 | -27 + 12 = -15 | negative |
| between -2 and 0 | -1 | -1 + 4 = 3 | positive |
| between 0 and 2 | 1 | 1 - 4 = -3 | negative |
| greater than 2 | 3 | 27 - 12 = 15 | positive |
Collect the positive intervals with strict endpoints
Why: The symbol is strictly greater than, so the critical values themselves give exactly zero and are excluded. Every bracket is round.
\[ (-2,\, 0) \cup (2,\, \infty) \]
Verify by testing a point in each claimed region and each excluded one
Why: Both claimed regions test true, both rejected regions test false, and each endpoint gives exactly zero, which a strict inequality rejects. The interval answer holds.
| x | value | is it greater than zero? | expected |
|---|---|---|---|
| -1 (claimed) | 3 | yes | in the solution, correct |
| 3 (claimed) | 15 | yes | in the solution, correct |
| -3 (rejected) | -15 | no | out, correct |
| 1 (rejected) | -3 | no | out, correct |
| 0 (endpoint) | 0 | no | excluded, correct |
Concept
The sign-chart method needs a single expression compared to zero. So move everything to one side and combine it over one common denominator.
\[ \frac{A}{B} \le c \;\Longrightarrow\; \frac{A}{B} - c \le 0 \;\Longrightarrow\; \frac{A - cB}{B} \le 0 \]
Do not clear the denominator by multiplying. That is the one move that is legal for equations and illegal here, and the next slide shows exactly what it costs you.
Picture it
Animation
Shows: Stating the domain properly — a rendered Manim animation.
Rendered with Manim.
Takeaway: A hole is still excluded, even though the graph looks continuous there.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Clearing the fraction the way you would in an equation.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: It feels identical to clearing a fraction in an equation.
Move everything to one side, combine, and let a sign chart handle the unknown sign for you.
Why: It feels identical to clearing a fraction in an equation. But the sign of that denominator is unknown, and multiplying an inequality by a negative reverses the direction.
Trap
Clearing the fraction the way you would in an equation.
\[ \frac{x+1}{x-3} \le 2 \]
Multiply both sides by the denominator and solve
Why: It feels identical to clearing a fraction in an equation. But the sign of that denominator is unknown, and multiplying an inequality by a negative reverses the direction.
\[ x + 1 \le 2(x - 3) \;\Rightarrow\; x + 1 \le 2x - 6 \;\Rightarrow\; 7 \le x \;\Rightarrow\; [7, \infty) \]
Test an input the wrong answer left out
Why: Zero satisfies the original inequality easily, yet the claimed answer excludes it. An entire interval of solutions was thrown away, because for inputs below three the denominator is negative.
\[ x = 0: \quad \frac{0+1}{0-3} = -\frac{1}{3} \le 2 \;\text{TRUE, but } 0 \notin [7,\infty) \]
Move everything to one side, combine, and let a sign chart handle the unknown sign for you.
\[ \frac{x+1}{x-3} \le 2 \]
Subtract and combine over one denominator
Why: Subtracting a number is always legal on an inequality. Now the question is just where a single fraction is negative or zero.
\[ \frac{x+1}{x-3} - 2 = \frac{x + 1 - 2(x-3)}{x-3} = \frac{7 - x}{x - 3} \le 0 \]
Sign chart on both critical values, then read the answer
Why: Three is excluded because the expression is undefined there; seven is included because it makes the fraction exactly zero. The lost interval is back.
\[ (-\infty,\, 3) \cup [7,\, \infty) \]
Notation
Annotate
From Trap: multiplying both sides by the denominator — read this one piece at a time. What is each part doing?
On: \( \frac{x+1}{x-3} \le 2 \)
Step zero
Discussion prompt
Worked example: a rational inequality with a constant on the right — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Subtract the constant from both sides
Answer:
Worked example
Solve and write the answer in interval notation.
\[ \frac{x+1}{x-3} \le 2 \]
Subtract the constant from both sides
Why: Adding or subtracting the same number never changes an inequality's direction, so this move is always safe. Multiplying by the denominator is not.
\[ \frac{x+1}{x-3} - 2 \le 0 \]
Combine over the common denominator
Why: Rewrite the two as a fraction with the same bottom, then subtract the numerators. Distribute the minus across the entire second numerator.
\[ \frac{x + 1 - 2(x - 3)}{x - 3} = \frac{x + 1 - 2x + 6}{x-3} = \frac{7 - x}{x - 3} \le 0 \]
List the critical values from top and bottom
Why: The numerator is zero at seven; the denominator is zero at three. Both cut the number line, for different reasons.
\[ x = 7 \;(\text{top}), \qquad x = 3 \;(\text{bottom}) \]
Test one value per interval
Why: Substitute into the combined fraction. Each result settles its whole interval.
| interval | test value | combined fraction | sign |
|---|---|---|---|
| less than 3 | 0 | 7 over -3 | negative |
| between 3 and 7 | 5 | 2 over 2 = 1 | positive |
| greater than 7 | 8 | -1 over 5 | negative |
Choose the negative intervals and set the brackets carefully
Why: Seven gets a square bracket because the fraction is exactly zero there and the symbol allows equality. Three gets a round bracket because the expression is undefined there, and undefined can never satisfy an inequality.
\[ (-\infty,\, 3) \cup [7,\, \infty) \]
Verify by testing a point in every region in the ORIGINAL inequality
Why: Both claimed regions test true, the rejected middle region tests false, the endpoint seven gives equality exactly, and three is undefined. Every bracket is justified.
| x | original left side | is it at most 2? | expected |
|---|---|---|---|
| 0 (claimed) | 1 over -3, about -0.333 | yes | in the solution, correct |
| 8 (claimed) | 9 over 5 = 1.8 | yes | in the solution, correct |
| 5 (rejected) | 6 over 2 = 3 | no | out, correct |
| 7 (endpoint) | 8 over 4 = 2 | yes, exactly equal | included, correct |
| 3 (endpoint) | undefined | no | excluded, correct |
Picture it
Animation
Shows: Average cost falls, then flattens — a rendered Manim animation.
Rendered with Manim.
Takeaway: The fixed cost is spread thinner, approaching the per-unit cost as a floor.
Concept
This is worth its own slide because it costs so many points. The symbol that allows equality does not buy you a denominator zero.
An expression that is undefined at an input has no value there at all. It is not big, not small, not zero. It cannot satisfy any comparison, so the input is out.
Practical rule for writing your interval answer: numerator zeros get square brackets when the symbol allows equality, and denominator zeros get round brackets always.
Counterexample
Discussion prompt
An expression that is undefined at an input has no value there at all. It is not big, not small, not zero. It cannot satisfy any comparison, so the input is out.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
Practical rule for writing your interval answer: numerator zeros get square brackets when the symbol allows equality, and denominator zeros get round brackets always.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Seeing a symbol that allows equality and closing every bracket.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The reasoning goes: the symbol allows equality, so include the endpoints.
Decide each bracket separately, by asking which polynomial produced that critical value.
Why: The reasoning goes: the symbol allows equality, so include the endpoints. That rule is right for the numerator zero and wrong for the denominator zero.
Trap
Seeing a symbol that allows equality and closing every bracket.
\[ \frac{x-5}{x-2} \le 0 \]
Find the critical values and close both brackets
Why: The reasoning goes: the symbol allows equality, so include the endpoints. That rule is right for the numerator zero and wrong for the denominator zero.
\[ [2,\, 5] \quad \text{(claimed)} \]
Test the left endpoint
Why: There is no output at all at that input, so it cannot be at most zero. Including it makes the answer wrong.
\[ \frac{2-5}{2-2} = \frac{-3}{0} \quad \text{undefined} \]
Decide each bracket separately, by asking which polynomial produced that critical value.
\[ \frac{x-5}{x-2} \le 0 \]
Sign chart the two critical values
Why: Testing zero gives a positive value, testing three gives negative two, and testing six gives one quarter. Only the middle interval works.
| interval | test value | value | sign |
|---|---|---|---|
| less than 2 | 0 | -5 over -2 = 2.5 | positive |
| between 2 and 5 | 3 | -2 over 1 = -2 | negative |
| greater than 5 | 6 | 1 over 4 = 0.25 | positive |
Square bracket on the numerator zero, round on the denominator zero
Why: At five the fraction is exactly zero, which the symbol allows, so it is in. At two the fraction is undefined, so it is out no matter what.
\[ (2,\, 5] \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Ranking
Put in order
Put the moves of Worked example: a rational inequality already compared to zero into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. It is, so there is nothing to move or combine.
Worked example
Solve and write the answer in interval notation.
\[ \frac{x-1}{x+2} \ge 0 \]
Check that one side is already zero and the fraction is a single quotient
Why: It is, so there is nothing to move or combine. Go straight to the critical values.
\[ \frac{x-1}{x+2} \ge 0 \]
List critical values from top and bottom
Why: The numerator is zero at one, and the denominator is zero at negative two. Both cut the line even though only one of them can be included.
\[ x = 1 \;(\text{top}), \qquad x = -2 \;(\text{bottom}) \]
Test one value per interval
Why: A negative divided by a negative is positive on the left; a negative over a positive is negative in the middle; positive over positive on the right.
| interval | test value | value | sign |
|---|---|---|---|
| less than -2 | -3 | -4 over -1 = 4 | positive |
| between -2 and 1 | 0 | -1 over 2 = -0.5 | negative |
| greater than 1 | 2 | 1 over 4 = 0.25 | positive |
Take the positive intervals and set the brackets
Why: One makes the fraction exactly zero, and the symbol allows equality, so it is included. Negative two makes the fraction undefined, so it is excluded.
\[ (-\infty,\, -2) \cup [1,\, \infty) \]
Verify by testing a point in each claimed region and each excluded one
Why: Both claimed regions test true, the rejected middle region tests false, the endpoint one gives exactly zero which the symbol allows, and negative two is undefined. The answer is confirmed.
| x | value | is it at least zero? | expected |
|---|---|---|---|
| -3 (claimed) | 4 | yes | in the solution, correct |
| 2 (claimed) | 0.25 | yes | in the solution, correct |
| 0 (rejected) | -0.5 | no | out, correct |
| 1 (endpoint) | 0 | yes, exactly zero | included, correct |
| -2 (endpoint) | undefined | no | excluded, correct |
Pattern
Polynomial inequalities are the same recipe with an empty step two: there is no denominator, so every critical value can be included when the symbol allows equality.
Picture it
Animation
Shows: Inverse variation — a rendered Manim animation.
Rendered with Manim.
Takeaway: Doubling the input halves the output — the defining shape of the reciprocal.
Elimination
Eliminate the wrong options
What is the solution set in interval notation?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The critical values are -4 from the numerator and 1 from the denominator. Testing gives positive at x = -5 (one sixth), negative at x = 0 (negative four), and positive at x = 2 (six). The numerator zero -4 is included because the fraction is exactly zero there, and the denominator zero 1 is excluded because the fraction is undefined there.
Check
Build the sign chart, then think hard about each bracket before you pick.
\[ \frac{x+4}{x-1} \ge 0 \]
Check your understanding
What is the solution set in interval notation?
Answer: A
Why: The critical values are -4 from the numerator and 1 from the denominator. Testing gives positive at x = -5 (one sixth), negative at x = 0 (negative four), and positive at x = 2 (six). The numerator zero -4 is included because the fraction is exactly zero there, and the denominator zero 1 is excluded because the fraction is undefined there.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — What a Rational Function Is · Holes vs Vertical Asymptotes · End Behavior and Horizontal Asymptotes · Slant (Oblique) Asymptotes · Intercepts and Full Sketches · Building a Function from a Description. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
Everything in this deck started with one move: factor the top and the bottom. From there each feature reads straight off the factors.
| Question | What you compare | Answer |
|---|---|---|
| vertical asymptote | surviving denominator factors | a vertical line at each zero |
| hole | factors that cancel | a point, height from the simplified form |
| horizontal asymptote | the two degrees and leading coefficients | y = 0, the ratio, or none |
| slant asymptote | top degree exactly one more | the quotient line from long division |
| x-intercepts | zeros of the SIMPLIFIED numerator | points on the horizontal axis |
| y-intercept | the function at zero | at most one point |
Next up: exponential functions, where a constant ratio replaces a constant rate, and the horizontal asymptote returns in a new disguise.
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