Introduces the first functions in the course whose graphs break. Factors numerator and denominator to separate holes from vertical asymptotes, compares degrees to find horizontal or slant asymptotes, locates intercepts, and assembles the whole into a sketch. Closes with rational inequalities, solved by sign analysis across the critical values.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 3 — Polynomial and Rational Functions
§3.7 Rational Functions, pp. 392-420
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §3.7 Rational Functions §3.7, pp. 392-420 — the pages these objectives are drawn from
Warm-up
Every function so far has been defined everywhere it mattered. This one is not.
Discussion prompt
Evaluate one over the quantity x minus 1 at 1.1, then 1.01, then 1.001. What is happening, and what happens at 1 itself?
Hint: Compute all three, and notice how fast they grow.
Answer:
The outputs are 10, then 100, then 1000 — growing without bound as the input approaches 1 from the right.
At 1 itself the rule is undefined: the denominator is zero and no output exists. The function does not have a large value there; it has no value at all.
So the graph shoots up towards a vertical line it never reaches. That line is a vertical asymptote, and it is a kind of behaviour no polynomial has ever shown — which is why rational functions need their own section.
Concept
At every input where the denominator vanishes, ask whether the numerator vanishes there too. If not, the graph has a vertical asymptote; if so, the factor cancels and the graph has a hole.
rational function — A quotient of two polynomials. It is undefined wherever the denominator is zero, and each such input produces either a vertical asymptote or a hole, according to whether the numerator vanishes there as well.
\[ f(x)=\frac{p(x)}{q(x)}, \qquad q(x)\ne 0 \]
The single most useful habit in this section is factoring both parts completely before doing anything else. Every question — domain, holes, asymptotes, intercepts, sign — is answered by looking at the factored form, and none of them can be answered reliably from the expanded one.
Figure (svg): Two graphs side by side: one with a vertical asymptote where the curve runs off to infinity, and one with a hole where the curve is simply missing a single point
OpenStax, Precalculus, §3.7 Rational Functions §3.7, pp. 392-399
Section
Section 1
Concept
Factor both parts. A denominator factor that cancels against a numerator factor produces a hole; one that does not produces a vertical asymptote.
The fourth point matters and is easy to lose. Cancelling a factor changes the formula but not the domain, exactly as in §1.2 and §1.4: the original was never defined at that input, so the hole stays even though the simplified rule would happily produce a value.
Figure (svg): Two graphs side by side: one with a vertical asymptote where the curve runs off to infinity, and one with a hole where the curve is simply missing a single point
OpenStax, Precalculus, §3.7 Rational Functions §3.7, pp. 392-401
Picture it
Same kind of rule, two very different behaviours, decided by one question.
Figure (svg): Two graphs side by side: one with a vertical asymptote where the curve runs off to infinity, and one with a hole where the curve is simply missing a single point
Both inputs are outside the domain. The difference is what the graph does nearby: run off to infinity, or carry on as though nothing happened apart from one missing point.
Worked example
Factor both, then compare.
\[ \text{Analyse } f(x)=\frac{x^2-4}{x^2-x-6}. \]
Factor the numerator
Why: A difference of squares.
\[ (x - 2) (x + 2) \]
Factor the denominator
Why: Two numbers multiplying to -6, adding to -1.
\[ (x - 3) (x + 2) \]
Find the shared factor
Why: Both contain x plus 2.
Find the unshared one
Why: The denominator's x minus 3 is alone.
\[ \text{asymptote at } 3 \]
Figure (svg): A completed analysis card for a rational function, listing its zeros, holes, vertical asymptotes and horizontal asymptote in the order they should be found
\[ \text{hole at } x=-2, \quad \text{asymptote at } x=3 \]
Verify: find the hole's height
Why: After cancelling, the rule is x minus 2 over x minus 3. At negative 2 that gives negative 4 over negative 5, which is four fifths. So the hole sits at the point with coordinates negative 2 and four fifths — the graph looks perfectly ordinary there apart from that single missing point.
OpenStax, Precalculus, §3.7 Rational Functions §3.7, pp. 394-397
Sorting
Compare the factors of the numerator and the denominator.
Sort into buckets
Sort each denominator zero.
Worked example
Cancelling once may not remove the factor entirely.
\[ \text{Analyse } f(x)=\frac{x-1}{(x-1)^2(x+4)}. \]
Cancel one copy of the shared factor
Why: The numerator has one, the denominator two.
\[ \frac{1}{(x - 1) (x + 4)} \]
Check what remains
Why: One copy of x minus 1 survives below.
Classify that input
Why: The denominator still vanishes there.
\[ \text{asymptote at } 1 \]
Classify the other
Why: Unshared from the start.
\[ \text{asymptote at } -4 \]
Figure (svg): The solution to Worked example a repeated denominator factor shown as a ladder of expressions, one row per legal move
\[ \text{asymptotes at } x=1 \text{ and } x=-4 \]
Verify: check why there is no hole
Why: Cancelling removed one copy of the factor but the denominator started with two, so one is left. A hole requires the factor to vanish from the denominator completely, which needs the numerator to contain it at least as many times. Comparing the multiplicities, not just the presence of the factor, is what decides this case.
OpenStax, Precalculus, §3.7 Rational Functions §3.7, pp. 398-401
Trap
\[ f(x)=\frac{x^2-4}{x+2} = x-2 \;\Longrightarrow\; \text{a line, defined everywhere} \]
Factor and cancel
Why: The numerator factors and the common factor cancels cleanly.
The function is described as the line x minus 2, with no exclusions.
The original was never defined at negative 2, and cancelling does not create a value there. The graph is the line with a hole punched out at that point.
This is the §1.2 trap in its natural habitat: the domain belongs to the rule you were given, not to the one you simplified to.
Record every excluded input before cancelling. The cancellation is legitimate and useful — it tells you the behaviour is a hole rather than an asymptote — but it erases the evidence, so the exclusion must be noted first.
Prediction
A rational function has a hole at 2 and a vertical asymptote at 5.
Predict first
What is its domain?
Correct: All reals except 2 and 5.
Why: Both inputs make the original denominator zero, so both are excluded, whatever the graph does nearby. A hole is just as undefined as an asymptote — the difference is visual rather than a difference in the domain, which is why cancelling never enlarges it.
Faded example
After cancelling, a function simplifies to x plus 3 over x minus 1, with a hole at x equal to 4.
Fill in the blanks
\text7 = \frac7/3___ = \frac___}___ = ___
Why: Substituting the hole's input into the SIMPLIFIED rule gives its height, which is seven thirds. The original rule cannot be used, since it is undefined there — that is what makes it a hole. The point is drawn as an open circle at that height.
Socratic
A hole removes one point from the graph.
Discussion prompt
Why is a hole so easy to miss on a calculator plot, and what does that imply about how to find them?
Hint: How many pixels wide is a single point?
Answer:
A hole removes exactly one point, which has no width. A plotter samples the function at finitely many inputs, and the chance of sampling precisely the excluded one is essentially zero — so the plot looks continuous.
The graph on either side is entirely normal, so there is no visual cue at all that anything is missing. Unlike an asymptote, which is unmistakable, a hole is invisible to a plot.
So holes must be found algebraically, by factoring and looking for shared factors. This is one of the clearest cases in the course where the picture is genuinely insufficient and the algebra is doing something the graph cannot.
Section
Section 2
Concept
Far from the origin only the leading terms matter, so the end behaviour of a rational function is decided by comparing the numerator's degree with the denominator's.
The reasoning is §3.3's dominance argument applied to a quotient. Far out, the function behaves like the leading term of the top divided by the leading term of the bottom, and that quotient is a power function whose behaviour is immediate.
Figure (svg): The three cases of end behaviour for a rational function, decided by comparing the degree of the numerator with that of the denominator
OpenStax, Precalculus, §3.7 Rational Functions §3.7, pp. 401-409
Picture it
One comparison of degrees decides all of it.
Figure (svg): The three cases of end behaviour for a rational function, decided by comparing the degree of the numerator with that of the denominator
The middle case is the one where the coefficients matter; in the other two the degrees alone settle the answer.
Worked example
The ratio of the leading coefficients, not of the constants.
\[ \text{Find the horizontal asymptote of } f(x)=\frac{3x^2-1}{2x^2+5x}. \]
Compare the degrees
Why: Both are 2.
Identify the leading coefficients
Why: Three on top, two below.
\[ 3\text{ and } 2 \]
Take their ratio
Why: Top over bottom.
\[ \frac{3}{2} \]
State the asymptote
Why: A horizontal line at that height.
\[ y = \frac{3}{2} \]
Figure (svg): The three cases of end behaviour for a rational function, decided by comparing the degree of the numerator with that of the denominator
\[ y=\tfrac{3}{2} \]
Verify: test a large input
Why: At x equal to 1000 the top is about three million and the bottom about two million, and the ratio is about 1.4988 — close to 1.5 and approaching it. Note that the ratio uses the LEADING coefficients: taking negative 1 over 0 from the constants would be meaningless, and the lower-degree terms have no effect on the end behaviour at all.
OpenStax, Precalculus, §3.7 Rational Functions §3.7, pp. 403-405
Sorting
Compare the two degrees.
Sort into buckets
Sort each rational function by its end behaviour.
Worked example
One degree higher on top means dividing.
\[ \text{Find the slant asymptote of } f(x)=\frac{x^2+1}{x-1}. \]
Compare the degrees
Why: Two over one, a difference of one.
Divide
Why: Long or synthetic division.
\[ \text{quotient } x + 1,\text{ remainder } 2 \]
Write the result
Why: Quotient plus remainder over divisor.
\[ x + 1 + \frac{2}{x - 1} \]
Discard the vanishing part
Why: The fraction shrinks to zero far out.
\[ y = x + 1 \]
Figure (svg): A rational function whose numerator degree exceeds the denominator's by one, drawn with its slant asymptote as a dashed line that the curve approaches at both ends
\[ y=x+1 \]
Verify: check the remainder term dies
Why: At x equal to 100 the remainder term is 2 over 99, about 0.02 — negligible against the quotient's 101. That is why the quotient alone describes the end behaviour. Note that the function approaches the slant line from above on the right and from below on the left, according to the sign of the remainder term.
OpenStax, Precalculus, §3.7 Rational Functions §3.7, pp. 406-409
Error analysis
A student finds a horizontal asymptote for a quotient of quadratics.
Annotate
On: \( \frac{3x^2-1}{2x^2+5x} \;\Longrightarrow\; y = \frac{-1}{0} \)
Far from the origin the leading terms win, exactly as in §3.3. Constants matter for the y-intercept and for nothing about the ends.
Prediction
A rational function has a horizontal asymptote at y equal to 2.
Predict first
Can its graph cross that line?
Correct: Yes, though not far out.
Why: A horizontal asymptote describes behaviour only at the extremes; nothing forbids the graph crossing it in the middle. Setting the function equal to the asymptote's value and solving finds any crossings. A vertical asymptote is different — the function is undefined there, so it genuinely cannot be crossed.
Faded example
For the quotient of 7x cubed minus x by 2x cubed plus 5.
Fill in the blanks
\text7 y = \frac2}___}
Why: Both degrees are 3, so the asymptote is the ratio of the leading coefficients, seven halves. The lower-degree terms contribute nothing to the end behaviour, which is why they can be ignored entirely once the degrees have been compared.
Explain it to yourself
The rules look like three separate facts.
Discussion prompt
Explain all three from the single observation that only leading terms matter far out.
Hint: What does the leading term of the top over the leading term of the bottom look like?
Answer:
Far out the function behaves like one power divided by another, which simplifies to a constant times a power of x. Everything follows from what that power is.
If the bottom degree is larger, the power is negative and the quotient shrinks to zero. If the degrees are equal, the powers cancel and a constant remains — the ratio of the coefficients. If the top is larger, the power is positive and the quotient grows without bound.
The slant case is the sub-case where that positive power is exactly one, so the growth is linear and a line describes it. Three cases, one argument — and it is the same dominance argument that gave polynomials their end behaviour.
Section
Section 3
Concept
A rational function is zero exactly where its numerator is zero and its denominator is not. The y-intercept is found by substituting zero, provided that input is in the domain.
The second bullet is the one to watch. A factor that cancels contributes neither an intercept nor an asymptote; it contributes a hole. Reading the intercepts off the original numerator without cancelling first produces a phantom intercept at a point where the graph does not exist.
Figure (svg): A completed analysis card for a rational function, listing its zeros, holes, vertical asymptotes and horizontal asymptote in the order they should be found
OpenStax, Precalculus, §3.7 Rational Functions §3.7, pp. 409-412
Picture it
Each row of this card is one of the section's questions.
Figure (svg): A completed analysis card for a rational function, listing its zeros, holes, vertical asymptotes and horizontal asymptote in the order they should be found
The zeros row records both what cancelled and what survived, because they lead to entirely different features.
Worked example
Cancel first, then read.
\[ \text{Find the intercepts of } f(x)=\frac{x^2-4}{x^2-x-6}. \]
Use the factored form
Why: From the earlier analysis.
\[ \frac{x - 2}{x - 3}\text{ after cancelling} \]
Set the simplified numerator to zero
Why: The surviving factor.
\[ x = 2 \]
Note what cancelled
Why: The factor x plus 2 gave a hole, not an intercept.
\[ -2\text{ is } a\text{ hole} \]
Substitute zero for the y-intercept
Why: Using the simplified form.
\[ \frac{-2}{-3} = \frac{2}{3} \]
Figure (svg): A completed analysis card for a rational function, listing its zeros, holes, vertical asymptotes and horizontal asymptote in the order they should be found
\[ (2,\,0) \text{ and } (0,\tfrac{2}{3}) \]
Verify: check against the original
Why: The original at zero gives negative 4 over negative 6, which is two thirds — matching. And at 2 the original numerator is zero while the denominator is negative 4, so the output really is zero. The input negative 2 makes both zero, which is a hole rather than an intercept, exactly as the cancellation indicated.
OpenStax, Precalculus, §3.7 Rational Functions §3.7, pp. 410-411
Sorting
Compare what vanishes at each input.
Sort into buckets
Sort each input for a rational function.
Worked example
The vertical axis may be outside the domain.
\[ \text{Find the intercepts of } f(x)=\frac{x-3}{x}. \]
Set the numerator to zero
Why: It does not share a factor with the bottom.
\[ x = 3 \]
Attempt the y-intercept
Why: Substitute zero.
Interpret
Why: Zero is not in the domain.
Note what is there instead
Why: The denominator vanishes with the numerator nonzero.
\[ \text{asymptote at } 0 \]
Figure (svg): The solution to Worked example no y-intercept shown as a ladder of expressions, one row per legal move
\[ (3,0); \text{ no } y\text{-intercept} \]
Verify: confirm from the shape
Why: The vertical asymptote sits exactly on the vertical axis, so the graph runs alongside that axis without touching it — there is nowhere for a y-intercept to be. Any function with a vertical asymptote or a hole at zero has no y-intercept, which is worth checking before attempting the substitution.
OpenStax, Precalculus, §3.7 Rational Functions §3.7, pp. 411-412
Trap
\[ \frac{(x-2)(x+2)}{(x-3)(x+2)}: \quad x\text{-intercepts at } 2 \text{ and } -2 \]
Set the original numerator to zero
Why: Both of its factors vanish, at 2 and at negative 2.
Two x-intercepts are reported.
The factor at negative 2 cancels, so that input is not in the domain at all. There is a hole there, not an intercept — the graph does not exist at that point.
The only x-intercept is at 2, where the numerator vanishes and the denominator does not.
Cancel before reading intercepts. A numerator zero is an intercept only if the denominator is nonzero there, and the shared factors are exactly the ones where it is not.
Prediction
A rational function is being analysed.
Predict first
When does it have no y-intercept?
Correct: When zero is not in its domain.
Why: The y-intercept is the output at zero, so it exists exactly when zero is a legal input. A vertical asymptote elsewhere is irrelevant — only one at zero itself matters, and a hole at zero has the same effect. Many functions have no y-intercept, so the last option is false in general.
Faded example
For the function that divides x plus 6 by x minus 2.
Fill in the blanks
f(0) = \frac-2-3 = \frac______} = ___
Why: Substituting zero gives 6 over negative 2, which is negative 3. Zero is in the domain because the denominator vanishes at 2 rather than at zero, so the intercept exists. Checking that the input is legal before substituting is the habit worth building.
Prediction
A rational function's surviving numerator factor is squared.
Predict first
What does the graph do at that intercept?
Correct: Touches the axis and turns back.
Why: Multiplicity works exactly as it did for polynomials in §3.4: an even multiplicity means the output does not change sign, so the graph bounces. The denominator plays no part in this, since it is nonzero at an intercept and therefore does not change sign there either.
Section
Section 4
Concept
The vertical asymptotes divide the domain into intervals. On each interval the sign is constant, so the graph's position relative to the axis is determined by one test point.
The sign only changes at a zero or at an undefined point, which is why those two together are called the critical values. Between them the graph stays on one side of the axis, so a single test point settles an entire interval.
Figure (svg): A rational function whose numerator degree exceeds the denominator's by one, drawn with its slant asymptote as a dashed line that the curve approaches at both ends
OpenStax, Precalculus, §3.7 Rational Functions §3.7, pp. 412-417
Picture it
A vertical asymptote splits the domain and a slant one governs both ends.
Figure (svg): A rational function whose numerator degree exceeds the denominator's by one, drawn with its slant asymptote as a dashed line that the curve approaches at both ends
Each branch is drawn separately, and each respects the vertical asymptote at one end and the slant one at the other.
Worked example
Asymptotes, intercepts, sign, then draw.
\[ \text{Sketch } f(x)=\frac{x-2}{x-3}. \]
Find the vertical asymptote
Why: The denominator vanishes at 3.
\[ x = 3 \]
Find the horizontal asymptote
Why: Equal degrees, coefficients both 1.
\[ y = 1 \]
Find the intercepts
Why: Numerator zero at 2; output at zero is two thirds.
\[ (2, 0)\text{ and } (0, \frac{2}{3}) \]
Test the sign on each interval
Why: Below 2, between 2 and 3, and above 3.
\[ +, -, + \]
Figure (svg): The solution to Worked example sketch a rational function shown as a ladder of expressions, one row per legal move
\[ \text{VA } x=3, \; \text{HA } y=1, \; (2,0), \; (0,\tfrac{2}{3}) \]
Verify: check one sign by substitution
Why: At x equal to 2.5 the numerator is 0.5 and the denominator negative 0.5, giving negative 1 — negative, as predicted for that middle interval. Testing one point per interval confirms the whole sign pattern, and any disagreement means a critical value was missed.
OpenStax, Precalculus, §3.7 Rational Functions §3.7, pp. 413-415
Ranking
The order that makes a rational sketch come out right.
Put in order
Why: Factoring comes first because every later step reads off the factored form. The asymptotes frame the picture, the intercepts add the crossings, and the sign test fills in which side of the axis each branch is on — which only makes sense once all the critical values are known.
Worked example
Set the function equal to the asymptote's value.
\[ \text{Does } f(x)=\frac{x-2}{x-3} \text{ cross } y=1? \]
Set the function equal to the value
Why: The asymptote's height.
\[ \frac{x - 2}{x - 3} = 1 \]
Multiply up
Why: Clearing the denominator.
\[ x - 2 = x - 3 \]
Simplify
Why: The x terms cancel.
\[ -2 = -3 \]
Interpret
Why: A false statement.
Figure (svg): The solution to Worked example find where a graph crosses its horizontal asymptote shown as a ladder of expressions, one row per legal move
\[ \text{Never crosses } y=1. \]
Verify: contrast with a function that does cross
Why: Had the algebra produced a solution, that input would be a genuine crossing. Many rational functions do cross their horizontal asymptotes — the crossing simply has to happen at a finite input, and the asymptote governs only the far behaviour. Here the two branches approach the line from opposite sides without meeting it.
OpenStax, Precalculus, §3.7 Rational Functions §3.7, pp. 415-417
Error analysis
A student sketches a curve meeting its vertical asymptote.
Annotate
On: \( \text{a branch drawn reaching the line } x=3 \)
The two kinds of asymptote are not symmetric. A vertical asymptote sits at an input outside the domain and is unreachable; a horizontal one describes far-out behaviour and constrains nothing in the middle.
Discrimination
The two kinds behave differently.
Sort into buckets
Sort each statement.
Faded example
Does the function that divides 2x by x plus 1 cross its horizontal asymptote at y equal to 2?
Fill in the blanks
\frac2___ = 2 \;\Longrightarrow\; 2x = 2x + ___ \;\Longrightarrow\; 0 = ___
Why: Multiplying up gives 2x equal to 2x plus 2, and cancelling leaves 0 equal to 2, which is false. So there is no crossing. Had the equation produced a solution, that input would be a genuine point where the graph meets the line before eventually approaching it.
Explain it to yourself
Testing one point per interval settles the whole interval.
Discussion prompt
Explain why the sign of a rational function cannot change except at a zero or an undefined point.
Hint: What would have to happen for the sign to change?
Answer:
To change from positive to negative the function must either pass through zero or jump. Passing through zero happens only where the numerator vanishes; jumping happens only where the function is undefined.
Between consecutive critical values the function is continuous and never zero, so it can do neither. Its sign is therefore constant on each such interval.
That is why one test point decides an entire interval, and it is what makes the sign-chart method work. The same argument underlies solving inequalities in the next section, where the critical values are exactly the boundaries of the answer.
Section
Section 5
Concept
To solve a rational inequality, get everything on one side, find where the expression is zero or undefined, and test the sign on each interval those points create.
The first bullet is the one that must not be skipped. Multiplying both sides by a denominator whose sign is unknown is invalid, because multiplying an inequality by a negative reverses it — and a variable denominator is negative on some intervals and positive on others.
Figure (svg): A completed analysis card for a rational function, listing its zeros, holes, vertical asymptotes and horizontal asymptote in the order they should be found
OpenStax, Precalculus, §3.7 Rational Functions §3.7, pp. 417-420
Picture it
The zeros and the undefined points are exactly the critical values.
Figure (svg): A completed analysis card for a rational function, listing its zeros, holes, vertical asymptotes and horizontal asymptote in the order they should be found
Every row of this card contributes a boundary. Between them the sign is constant, which is what makes the interval test valid.
Worked example
One side, one fraction, then test.
\[ \text{Solve } \frac{x-1}{x+2} > 0. \]
Find the critical values
Why: Numerator zero and denominator zero.
\[ x = 1\text{ and } x = -2 \]
Split the line into intervals
Why: Three of them.
Test one point in each
Why: Try -3, then 0, then 2.
\[ +, -, + \]
Collect the intervals where it is positive
Why: Strict inequality excludes both boundaries.
\[ (-\infty, -2) U(1, \infty) \]
Figure (svg): The solution to Worked example solve a rational inequality shown as a ladder of expressions, one row per legal move
\[ (-\infty,-2)\cup(1,\infty) \]
Verify: check both boundaries are correctly excluded
Why: At 1 the expression is zero, which is not greater than zero, so 1 is out. At negative 2 it is undefined, so that is out too — and it would be out even for a non-strict inequality, since an undefined point can never satisfy anything. The two exclusions have different reasons, which matters when the inequality allows equality.
OpenStax, Precalculus, §3.7 Rational Functions §3.7, pp. 418-419
Sorting
Numerator zeros may be included; denominator zeros never are.
Sort into buckets
For a non-strict inequality, sort each critical value.
Worked example
Never cross-multiply an inequality with a variable denominator.
\[ \text{Solve } \frac{3}{x-1} \le 1. \]
Move everything to one side
Why: Subtract 1 from both sides.
\[ \frac{3}{x - 1} - 1 \le 0 \]
Combine into one fraction
Why: Common denominator.
\[ \frac{3 - (x - 1)}{x - 1} \le 0 \]
Simplify the numerator
Why: Three minus x plus one.
\[ \frac{4 - x}{x - 1} \le 0 \]
Find critical values and test
Why: Zeros at 4 and 1.
\[ (-\infty, 1) U [4, \infty] \]
Figure (svg): The solution to Worked example combine into one fraction first shown as a ladder of expressions, one row per legal move
\[ (-\infty,1)\cup[4,\infty) \]
Verify: check the two boundaries differ
Why: The value 4 makes the numerator zero, so the expression is zero and satisfies the non-strict inequality — hence a square bracket. The value 1 makes it undefined, so it is excluded regardless — hence a round bracket. Cross-multiplying at the start would have given x at or above 4 only, losing the entire left-hand interval, because it assumed the denominator was positive.
OpenStax, Precalculus, §3.7 Rational Functions §3.7, pp. 419-420
Trap
\[ \frac{3}{x-1} \le 1 \;\Longrightarrow\; 3 \le x-1 \;\Longrightarrow\; x \ge 4 \]
Multiply both sides by the denominator
Why: Clearing the fraction as one would for an equation.
The answer is given as x at or above 4.
The denominator's sign is unknown. When x is below 1 it is negative, and multiplying an inequality by a negative reverses it — so the step is invalid on that whole interval.
The complete answer includes everything below 1 as well, which the cross-multiplication silently discarded. Testing x equal to 0 confirms it: 3 over negative 1 is negative 3, which is at most 1.
Move everything to one side and combine instead. Comparing a single fraction against zero avoids multiplying by anything of unknown sign, and it is the only reliable method.
Prediction
A rational inequality has two numerator zeros and one denominator zero, all distinct.
Predict first
How many intervals must be tested?
Correct: Four.
Why: Three critical values cut the real line into four intervals, one below the smallest, two between consecutive pairs, and one above the largest. Each needs one test point. The general rule is one more interval than the number of critical values, which is a useful check that none has been missed.
Faded example
Rearrange the inequality comparing 2 over x with 5.
Fill in the blanks
\frac5___ - 5 \le 0 \;\Longrightarrow\; \frac___}x}___ \le 0
Why: Writing 5 as 5x over x and subtracting gives 2 minus 5x over x. The critical values are then two fifths from the numerator and zero from the denominator. Combining into a single fraction is what makes the sign test valid, since the sign of one expression is what is being tracked.
Real world
Any condition on a rate, a ratio or a concentration is one.
Discussion prompt
A concentration is modelled by a rational function of time, and a safety rule requires it to stay below a threshold. What kind of problem is that?
Hint: What do you get when you compare a rational function against a constant?
Answer:
It is exactly a rational inequality: the rational model compared against a constant threshold, with the answer being the set of times at which the rule is satisfied.
The method is this section's. Move the threshold across, combine into one fraction, find the critical values and test. The critical values have real meaning: the numerator zeros are the times the concentration equals the threshold, and the denominator zeros are times the model breaks down.
The answer will typically be a union of intervals, which reads naturally as 'safe before this time and after that one'. And the §2.3 warning applies: the model's own domain may exclude some of what the algebra returns, so the mathematical answer has to be intersected with the times that physically exist.
Comparison
Fill the blanks from memory. Both are excluded inputs, and they look nothing alike.
Comparison matrix
| a hole | a vertical asymptote | |
|---|---|---|
| cause | the factor cancels completely | a denominator copy survives |
| in the domain | no | no |
| what the graph does | one point is missing | runs off to infinity |
| visible on a plot | no, essentially never | yes, unmistakably |
| how to find it | factor and look for shared factors | factor and look for unshared denominator factors |
The fourth row is why the algebra cannot be skipped. An asymptote announces itself; a hole has to be deduced.
Pattern
Seven steps, and the first one makes all the others readable.
Step 2 must come before step 3. Cancelling erases the evidence of the excluded input, so recording the domain first is what preserves it.
OpenStax Algebra and Trigonometry 2e, §5.6 Rational Functions §5.6
Check
Does the factor cancel?
Check your understanding
The function (x-1)(x+3) over (x+3)(x-4) has what at x = -3?
Answer: A
Why: The factor x plus 3 appears once in each, so it cancels completely and leaves a hole at negative 3. The input is still outside the domain, but the graph is otherwise ordinary there.
Check
Compare the degrees.
Check your understanding
What is the horizontal asymptote of (4x^2 + 1) over (2x^2 - 3)?
Answer: A
Why: The degrees are equal, so the asymptote is the ratio of the leading coefficients, 4 over 2, which is 2. The lower-degree terms have no bearing on the end behaviour.
Check
Do not cross-multiply.
Check your understanding
Which is the correct first move for the inequality comparing 5 over x minus 2 against 3?
Answer: A
Why: Getting everything on one side and combining lets the sign of a single expression be analysed. Multiplying by the denominator is invalid because its sign is unknown, and reversing the inequality depends on that sign.
Real world
Rational functions describe anything that approaches a limit rather than growing without bound.
Discussion prompt
A drug's concentration after a dose is modelled by a rational function that tends to zero far out. What do the horizontal asymptote and any vertical one mean?
Hint: What is happening physically at each?
Answer:
The horizontal asymptote at zero says the drug clears from the body: as time grows, concentration approaches zero without ever quite reaching it, which is exactly how elimination works.
A vertical asymptote would be physically impossible — an infinite concentration — so its presence would signal that the model has broken down at that time, usually because the input is outside the range the model was built for.
This is a good example of the §2.3 point in a new setting. The mathematics permits a vertical asymptote and the situation does not, so the model's domain must exclude it. Reading the asymptotes is therefore a way of checking a model as much as of describing it.
Commit first
State your confidence along with your answer.
Predict first
A rational function has a horizontal asymptote at y equal to 4. What can its graph do?
Correct: Cross that line at a finite input, but approach it far out.
Why: A horizontal asymptote constrains the far behaviour only, so crossings in the middle are permitted and common. Setting the function equal to 4 and solving finds them. A rational function can cross only finitely often, since the resulting equation is polynomial, and it may approach from either side or from opposite sides at the two ends.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate the difference between a hole and a vertical asymptote, and why both mean the same thing for the domain.
Hint: What does the denominator do at each, and what does the numerator do?
Answer:
At both, the denominator is zero, so the function is undefined and the input is out of the domain. That part is identical, which is why cancelling never restores the input.
The difference is the numerator. If it also vanishes there, the shared factor cancels and the function is well behaved on either side — the graph has one missing point. If it does not vanish, you are dividing something nonzero by something approaching zero, and the outputs grow without bound.
A good explanation includes the practical consequence: an asymptote is impossible to miss on a graph and a hole is impossible to see. So the algebra is not an alternative to the picture here — for holes it is the only route.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The first is the section's organising question and everything else builds on it. The fourth causes wrong answers that look entirely reasonable, because cross-multiplying produces a clean-looking interval that is simply missing half the solution set.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Write a rational function whose numerator and denominator share exactly one factor and whose denominator has one more factor besides. Factor both, and list in order: the domain, the hole with its height, the vertical asymptote, the horizontal asymptote from the degree comparison, and both intercepts. Then sketch it, drawing the asymptotes first and marking the hole as an open circle.
If your domain excludes both the hole and the asymptote, you have the point that cancelling most often erases.
Recap
Five things, and the first question is asked at every excluded input.
| if you remember one thing | it should be this |
|---|---|
| about excluded inputs | does the numerator vanish there too? that is the whole question |
| about end behaviour | compare the degrees; only the leading terms matter far out |
| about the domain | record it before cancelling, because cancelling erases the evidence |
| about inequalities | one side, one fraction, test intervals - never multiply by a variable |
Section 3.8 returns to §1.7's inverses, now that a function can have a restricted domain for interesting reasons, and works out how to invert the polynomial and radical functions of this chapter.
OpenStax, Precalculus, §3.7 Rational Functions §3.7, pp. 392-420 — everything on these slides traces back here
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