Rational functions and asymptotes

A rational rule is a quotient of polynomials on its stated domain

A rational function is represented by a quotient f(x) = P(x)/Q(x) of real polynomials, with Q not the zero polynomial. A polynomial itself is included: it can be written with denominator 1. Here the natural real domain of a specified quotient contains every real x for which its original denominator is nonzero. A problem may narrow that domain further. The word rational describes the quotient form; its permitted inputs and outputs need not be rational numbers.

Find all original denominator zeros before simplifying. For f(x) = (x² − 1)/(x − 1), first exclude x = 1. Factoring and cancelling gives x + 1 only on x ≠ 1. The original function and the polynomial x + 1 on all real inputs are different functions because their domains differ. Keep the recorded exclusions when using the reduced formula to evaluate, find intercepts or determine the range.

A zero numerator polynomial is allowed. The rule 0/(x − 1) has value zero whenever x ≠ 1, but it is undefined at 1. Zero divided by a nonzero denominator is zero; zero divided by zero is undefined. This case must be checked directly rather than assigning a degree to its zero numerator or treating every denominator zero as a pole.

Cancellation can leave a hole or leave a genuine pole

A hole is an excluded point whose nearby outputs approach one finite height from both sides. For (x² − 1)/(x − 1), the reduced rule x + 1 approaches 2 as x approaches 1, while the original remains undefined there. Its graph has the hole (1, 2). Giving the function a new value 2 at input 1 would fill that hole, but it would change the original definition.

Cancel common factors completely before classifying a denominator zero. If the remaining denominator is nonzero at the excluded input a, evaluate the remaining quotient there to obtain the finite hole height. If the remaining denominator still vanishes there and the remaining numerator does not, the outputs become unbounded near a: this is a pole and the line x = a is a vertical asymptote. For (x − 1)/(x − 1)², cancellation leaves 1/(x − 1), so x = 1 is still a pole despite a common factor having been cancelled.

More generally, suppose the numerator and denominator have factors (x − a)^m and (x − a)^n, with their other factors nonzero at a and n > 0. If m ≥ n, all copies in the denominator cancel and a is a hole. If m < n, a denominator power remains and there is a pole. To justify the unbounded behaviour, the remaining nonzero quotient stays near a nonzero finite value, while the denominator power approaches zero; its reciprocal magnitude can therefore exceed any fixed bound. These finite-factor multiplicities apply to a nonzero numerator; the zero-numerator case instead has constant zero output on its original domain.

A numerator zero is an intercept only when the input is permitted

A defined fraction is zero exactly when its numerator is zero and its denominator is nonzero. Solve the numerator equation, then filter through every original exclusion and the stated domain. The real zero a gives the x-intercept (a, 0). The y-intercept is (0, f(0)) only if input 0 belongs to the original domain; an algebraic reduced formula alone cannot create that intercept.

For (x² − 1)/(x − 1), numerator zeros −1 and 1 are candidates, but only −1 is permitted. Thus the only x-intercept is (−1, 0); the missing point at 1 is the hole (1, 2), not an intercept. At input 0 the original fraction equals 1, giving y-intercept (0, 1). By contrast, x²/x equals x on x ≠ 0 but has no x-intercept and no y-intercept: the origin is its excluded hole, not a point of its original graph.

Approaching an input is different from evaluating at it

Saying that f(x) approaches a height L near input a describes outputs at permitted inputs increasingly close to a; it does not assign a value at a. The notation x → a⁻ means approach from inputs below a, and x → a⁺ means approach from inputs above a. A finite two-sided limit requires the same approached height from both sides. A few table entries illustrate an approach, while the simplified rule or a bound must justify what all sufficiently nearby permitted inputs do.

For f(x) = 1/(x − 1), the denominator is a small negative number on the left of 1 and a small positive number on the right. Thus f(x) decreases without bound as x approaches 1 from the left and increases without bound from the right. Write the directions as f(x) → −∞ and f(x) → +∞ respectively. Infinity is not a real function value at the excluded input; it records unbounded behaviour.

For g(x) = 1/(x − 1)², the denominator is positive on either side and approaches zero, so both sides increase without bound. For h(x) = x³/(x² − 1), factor the denominator as (x − 1)(x + 1). Near −1 the numerator is negative, giving h(x) → −∞ from the left and +∞ from the right. Near 1 the numerator is positive, giving h(x) → −∞ from the left and +∞ from the right. Track the signs of all remaining factors, including any fixed negative coefficient; a squared pole need not point upward if its numerator is negative.

A horizontal asymptote describes the distant ends

The line y = L is a horizontal asymptote when the output difference from L approaches zero as x increases or decreases without bound. Check each relevant end. For rational rules on their natural real domains, the degree comparison gives the same horizontal height at both ends. This is a statement about unbounded inputs, not about arriving at the line at a finite input, and it is not a bound on the range.

For a nonzero numerator, compare its degree with the nonzero denominator's degree. A lower numerator degree gives height 0; equal degrees give the ratio of leading coefficients. Divide numerator and denominator by the highest relevant power of x to justify these results: terms with a remaining positive power of x in their denominator approach zero at the ends. A higher numerator degree gives no horizontal asymptote. A zero numerator gives constant zero output on the permitted domain, so its difference from y = 0 is identically zero without a degree comparison.

For (2x − 1)/(x − 1) = 2 + 1/(x − 1), the difference from 2 approaches zero at both ends, proving horizontal asymptote y = 2. That height is never attained because 1/(x − 1) cannot be zero. In contrast, x/(x² + 1) has horizontal asymptote y = 0 because (1/x)/(1 + 1/x²) approaches zero at either end, yet its permitted input 0 gives output 0. The output is negative for negative x and positive for positive x, so its graph crosses this horizontal asymptote at the origin.

Use the full polynomial quotient for an oblique line

An oblique, or slant, asymptote has equation y = mx + b with m ≠ 0. Its defining condition is that f(x) − (mx + b) approaches zero at the relevant unbounded end. This is the vertical separation between the graph and the line at the same input; for a fixed straight line, it also makes their perpendicular distance approach zero. Comparing only a leading-term ratio can identify a slope but miss the intercept b.

Polynomial division writes a rational rule as a polynomial quotient plus a remainder over the original denominator. The remainder has lower degree than that denominator, so this fractional difference approaches zero at both ends. If the numerator degree exceeds the denominator degree by one, the polynomial quotient is a nonconstant line and supplies the full oblique asymptote. If the excess is larger, the quotient has higher degree and gives a polynomial end comparison rather than a horizontal or oblique line.

For f(x) = x³/(x² − 1), division gives f(x) = x + x/(x² − 1) on x ≠ −1 and x ≠ 1. The difference from y = x is x/(x² − 1) = (1/x)/(1 − 1/x²) away from zero, which approaches zero at both ends. At input 0 the original rule equals the line's height 0. The difference is positive just left of 0 and negative just right of 0, so the graph crosses its oblique asymptote there. A horizontal or oblique asymptote can be crossed. Under this difference-to-zero definition, a rule already coinciding with a line on every permitted input also satisfies the end criterion; describe that coincidence explicitly rather than calling it a crossing.

A missing point does not automatically remove its height from the range

For f(x) = 2 + 1/(x − 1) on x ≠ 1, output 2 is impossible, and every real y ≠ 2 is attained at x = 1 + 1/(y − 2). That input differs from 1 and substitution returns y, proving range all real numbers except 2. For g(x) = 1/(x − 1)², all outputs are positive. Conversely, every y > 0 is attained at x = 1 + 1/√y, which is defined and differs from 1. Thus its exact range is y > 0, not y ≥ 0.

For (x² − 1)/(x − 1) = x + 1 on x ≠ 1, the height 2 would require the sole input x = 1 and is absent from the range. Every other real y is attained at x = y − 1, so the exact range excludes only 2. In 0/(x − 1) on x ≠ 1, however, the hole height 0 is attained at every permitted input, such as 0; its range is {0}. Both rules have holes, but the effect on their ranges differs.

Another example is (x² − 1)²/(x − 1)², which equals (x + 1)² on x ≠ 1 and has hole (1, 4). Output 4 remains attained at x = −3. In fact, its exact range is all y ≥ 0: squares are nonnegative, and any y ≥ 0 is produced by x = −1 − √y, which is at most −1 and therefore never the excluded input 1. Check all permitted preimages before deciding whether a removed graph point also removes an output height.

Use an algebraic witness or an existence argument for every proposed output

For f(x) = x/(x² + 1) on all real inputs, the positive denominator permits inequality comparisons. Since (x − 1)² ≥ 0, we have x² + 1 ≥ 2x and hence f(x) ≤ 1/2. Since (x + 1)² ≥ 0, we have x² + 1 ≥ −2x and hence f(x) ≥ −1/2. To attain any y with −1/2 ≤ y ≤ 1/2, solve yx² − x + y = 0. When y = 0, take x = 0. When y ≠ 0, the discriminant 1 − 4y² is nonnegative, and x = (1 + √(1 − 4y²))/(2y) is a real input satisfying that equation. Its denominator x² + 1 is positive, so rearranging gives f(x) = y. This proves the complete range, including both endpoints at inputs −1 and 1.

For g(x) = x³/(x² − 1) on x ≠ −1 and x ≠ 1, every real output is attained. To prove this without solving a cubic formula, fix an arbitrary real y and form the polynomial H(x) = x³ − yx² + y. It is continuous on [−1, 1], with H(−1) = −1 and H(1) = 1, independent of y. The intermediate value theorem says that a continuous function on an interval takes every height between its endpoint heights, so H has a zero a strictly between −1 and 1. These are values of the auxiliary polynomial H; the original rational function is undefined at the two endpoints.

At that interior a, a² − 1 is nonzero. The equation H(a) = 0 rearranges to a³ = y(a² − 1), so g(a) = y. Because y was arbitrary, the exact range is all real numbers. The argument proves existence of a permitted input for every y; it does not claim that a few controls find all preimages or that the graph is monotone. Use continuity on an interval where it is valid: a sign change across a pole, such as between −1 and 1 for 1/x, does not establish an intervening zero of that rational function.

A model's allowed inputs can change what its asymptotes mean

Suppose a stated order model charges a fixed 12-euro fee plus 2 euros per item. For a positive integer item count n, the total cost is T(n) = 12 + 2n euros and the average cost per item is A(n) = (12 + 2n)/n = 2 + 12/n euros per item. Division by n requires a nonzero count, and an average for an order of zero items is not defined. Fractional counts and negative counts are outside this model even though a corresponding real formula can be calculated at many of them.

If the model allows arbitrarily large positive integer counts, its average costs approach 2 euros per item from above as n grows; they never equal 2 at a finite permitted count. The continuous real extension has a vertical asymptote at n = 0, but positive integer counts cannot approach zero through smaller positive counts. That pole of the extension is not an accessible near-zero behaviour of the actual count model. If an order is limited to at most six items, its entire domain is {1, 2, 3, 4, 5, 6}; there is no unbounded-input end within that finite domain. Any end-asymptote statement then concerns an extension of the formula, not orders allowed by the stated model.

Keep excluded points, branches and end claims visible

A rational function is continuous wherever its denominator is nonzero, so a permitted finite input belongs to a continuous branch. An original excluded input can separate branches or leave a removable hole. A plotted curve must not join across a pole or fill a hole. A vertical asymptote is the excluded input's vertical line; a hole includes both its excluded input and its finite approached output. Neither infinity nor a reduced formula supplies a new original value there.

A fixed graph window can clip large outputs near poles, hide tiny holes and make a curve appear to lie exactly on an asymptote. Numerical curves illustrate the shape; they do not establish exact domain, range, intercepts or one-sided limits. Keep original factors and exact differences alongside the display. To review a rational rule, record the original domain, cancel completely, classify each exclusion, find permitted intercepts, establish side signs, justify end comparisons and prove any claimed full range. Then apply the problem's context and units.

Separate a selected value from the full domain, range and end behaviour

The explorer offers six original rules: (2x − 1)/(x − 1), 1/(x − 1)², (x² − 1)/(x − 1), 0/(x − 1), x/(x² + 1) and x³/(x² − 1). The first four exclude input 1; the fifth accepts every real input; the last excludes −1 and 1. Inspect poles and finite holes separately. The third rule coincides with y = x + 1 except for its hole, while the fourth coincides with y = 0 except for its hole; coincidence on the whole permitted domain is different from an isolated crossing.

Choose an original input and a proposed output independently. The input check determines whether the original expression is defined and, if so, its exact value. The output check asks whether at least one permitted input produces the proposed height; it need not be your selected input. Read the complete range proof as well as any displayed preimages. Compare exact differences from horizontal or oblique reference lines, and distinguish equality at the selected input from a statement about the unbounded ends. The sampled graph has a limited window and approximate curves; the algebra states the complete domain and behaviour. The next lesson compares function families when choosing a model for a relationship.

Worked examples

Try it yourself

1. For f(x) = (x² − 9)/(x − 3), state the original domain, reduced formula, hole and permitted intercepts. Does the excluded input give a vertical asymptote?

   Show the answer and reasoning

   Domain x ≠ 3; reduced formula x + 3 on that domain; hole (3, 6). Intercepts (−3, 0) and (0, 3). There is no vertical asymptote.

   Factor the numerator as (x − 3)(x + 3), but retain the original exclusion. The reduced rule approaches the finite height 6 at input 3, giving a hole instead of a pole. Numerator candidate 3 is excluded, while −3 is permitted; input 0 is also permitted.

2. Classify the exclusion at 2 in f(x) = (x − 2)/(x − 2)². Find the horizontal asymptote and the directions of unbounded behaviour on the two sides of 2. Explain why cancelling a common factor does not make a hole here.

   Show the answer and reasoning

   Input 2 is excluded and gives a genuine pole, with vertical asymptote x = 2. The left side decreases without bound and the right side increases without bound. The horizontal asymptote is y = 0; there is no hole.

   Cancellation leaves 1/(x − 2) on x ≠ 2: the denominator had an extra copy of the same factor. The remaining numerator is 1, so the reciprocal grows without bound in magnitude near 2 with the denominator's sign. At both unbounded ends the reciprocal approaches zero.

3. For f(x) = −3/(x + 2)², give the domain, vertical and horizontal asymptotes, both one-sided pole directions and complete range. Prove that every claimed output is attained.

   Show the answer and reasoning

   Domain x ≠ −2; vertical asymptote x = −2, with both sides decreasing without bound; horizontal asymptote y = 0. The exact range is all real y < 0.

   The squared denominator is positive whenever defined, so every output is negative and grows negatively without bound as the denominator approaches zero. For any y < 0, choose x = −2 + √(−3/y). The radicand is positive and the root is nonzero, so x ≠ −2 and substitution gives f(x) = y. Zero is never attained, while the difference from zero approaches zero at both ends.

4. For f(x) = x/(x² + 4), find the domain, horizontal asymptote and crossing point. Prove its complete range, including endpoint values.

   Show the answer and reasoning

   Domain all real numbers; horizontal asymptote y = 0, crossed at (0, 0). The complete range is −1/4 ≤ y ≤ 1/4, attained at endpoint inputs −2 and 2.

   The denominator is positive. The inequalities (x − 2)² ≥ 0 and (x + 2)² ≥ 0 give the upper and lower bounds ±1/4. For an arbitrary y in those bounds, y = 0 is attained at x = 0; for y ≠ 0, solve yx² − x + 4y = 0, whose discriminant 1 − 16y² is nonnegative. A quadratic root is a permitted real input giving the proposed y, proving there are no gaps. Dividing by x² at the ends shows the quotient approaches zero; negative and positive inputs give negative and positive outputs, so it crosses at zero.

5. For f(x) = (x³ + x²)/(x² + 1), find the original domain and full oblique asymptote. Prove the difference approaches zero at both ends, and find all permitted inputs where the graph meets that line.

   Show the answer and reasoning

   Domain all real numbers; oblique asymptote y = x + 1. The difference is −(x + 1)/(x² + 1), so the sole intersection is at x = −1, giving point (−1, 0); the graph crosses there.

   The denominator is always positive. Division gives f(x) = x + 1 − (x + 1)/(x² + 1), verified by multiplying back. The remainder quotient has lower numerator degree and approaches zero at both ends. Equality with the line requires x + 1 = 0, so x = −1 is the only input. Since the denominator stays positive, the difference changes from positive to negative across −1, proving a crossing rather than just an intersection.

6. For the original rule f(x) = (x² − 4)²/(x − 2)², find the hole and complete range. Is the hole height missing from that range? Supply a permitted input for every claimed output.

   Show the answer and reasoning

   Domain x ≠ 2; reduced rule (x + 2)²; hole (2, 16). Its complete range is y ≥ 0, and hole height 16 is still attained at x = −6.

   Cancellation preserves the exclusion. Every permitted output is a square and therefore nonnegative. For any y ≥ 0, choose x = −2 − √y, which is at most −2 and never equals the excluded input 2; then (x + 2)² = y. At y = 16 this construction gives −6, so removing the point (2, 16) does not remove height 16.

7. A stated model charges a fixed 15-euro fee plus 3 euros per item and allows only integer counts n = 1, 2, 3, 4, 5. Find the average cost per item and its complete range. Is 3 euros per item an attained output or an end within this finite domain?

   Show the answer and reasoning

   A(n) = 3 + 15/n euros per item. The range is {6, 27/4, 8, 21/2, 18} euros per item. Three euros per item is neither attained nor an unbounded-input end of this finite model.

   Dividing the total 15 + 3n euros by the permitted nonzero count gives the average. The five counts give 18, 21/2, 8, 27/4 and 6 euros per item. A hypothetical extension to unbounded positive counts would approach 3 from above, but the actual domain stops at 5; its finite range and restrictions must remain visible.

What to carry forward

• A rational rule is a quotient of real polynomials with denominator not identically zero. Record its original denominator exclusions and any chosen context restrictions before cancelling.
• A fully cancelled exclusion with a finite approached value is a hole. A remaining denominator zero with nonzero remaining numerator gives a pole; cancellation alone does not settle the distinction.
• A numerator zero is an attained zero only when the input is permitted in the original function. A reduced expression cannot create a missing intercept.
• One-sided approach describes nearby permitted inputs, not a function value at the excluded point. Track all factor signs to distinguish +∞ and −∞; infinity is not a real output there.
• Horizontal and oblique asymptotes require the difference from their line to approach zero at an unbounded end. They may be crossed; use the full division quotient to find an oblique intercept.
• Prove a complete range by restricting every output and supplying or establishing a permitted input for every proposed output. A hole removes an output height only when no other permitted preimage remains.
• Graph windows, approximate curves and mathematical extensions have limits. Preserve exact branch exclusions, distinguish selected equality from end behaviour, and interpret a model only on its stated domain with its units.