Loav · VCE Specialist Mathematics · Units 3 & 4
S34-F02 · Recommended sequence: Unit 3
1. A quotient is a rule together with a domain
A rational function is a quotient of polynomials, f(x) = P(x)/Q(x), with Q(x) ≠ 0. Its behaviour comes from both the algebraic expression and the excluded inputs. Factoring, cancelling, dividing and decomposing into partial fractions can reveal different features of the same graph. None of those operations silently restores an input excluded by the original rule.
Begin a graph investigation by writing the original domain. Next factor numerator and denominator where practical. An uncancelled denominator zero usually produces a vertical asymptote for the low-degree rational functions considered here; a cancelled factor can instead leave a removable hole. The decisive question is what the function approaches near the excluded input, rather than whether a denominator appeared in the first line.
An x-intercept requires the numerator to be zero at an allowed input. A y-intercept is f(0) only if zero belongs to the domain. A point that would lie on a simplified graph may still be missing from the actual function. Use an open circle for a hole and state its coordinates; an empty point is not an asymptote.
2. Division explains end behaviour
Polynomial division expresses an improper rational function as a polynomial plus a proper fraction. If the remaining fraction tends to zero as x becomes large in magnitude, the polynomial identifies the graph's end behaviour. In this guide, the important cases produce a constant horizontal asymptote or a linear oblique asymptote. Give an asymptote as an equation such as y = 2x − 1, rather than listing a slope or intercept separately.
When the numerator degree is lower than the denominator degree, the horizontal asymptote is y = 0. When their degrees are equal, it is the ratio of their leading coefficients. When the numerator degree is one higher, divide to obtain the oblique line. These shortcuts describe large-|x| behaviour; they do not determine vertical asymptotes, holes or turning points.
A graph may cross a horizontal or oblique asymptote. An asymptote constrains a limiting difference, not all finite points. For y = x/(x² + 1), the graph crosses y = 0 at the origin while approaching that same line at both ends. Conversely, y = x + 4/x never crosses y = x because their difference 4/x cannot equal zero. Use the actual difference to settle a crossing question.
One-sided behaviour near a vertical asymptote needs sign analysis. For 3/(x − 2), a small positive denominator gives a large positive value, while a small negative denominator gives a large negative value. For 3/(x − 2)², both sides approach positive infinity. Multiplicity and the other factors determine the direction; a dotted vertical line alone is not a complete graph.
Worked example 1: Cancellation leaves a missing point
Analyse f(x) = (x² − 1)/(x² − x − 2). Factor the denominator as (x − 2)(x + 1), so the original domain excludes −1 and 2. Cancelling x + 1 gives f(x) = (x − 1)/(x − 2) = 1 + 1/(x − 2), still for x ≠ −1, 2.
The simplified expression tends to 2/3 at x = −1, so there is a hole at (−1, 2/3). The vertical asymptote is x = 2 and the horizontal asymptote is y = 1. The x-intercept is (1, 0); the y-intercept is (0, 1/2). Although the original numerator also vanished at −1, that input was excluded and is not an intercept.
Finally f′(x) = −1/(x − 2)² is negative on every allowed interval, so there are no stationary points. The graph decreases on (−∞, −1), (−1, 2) and (2, ∞). The hole must remain visible even if a CAS simplifies it away.
3. Partial fractions expose the separate components
A proper rational expression with factorised denominator can often be written as a sum of simpler quotients. For example, (3x + 5)/[(x + 1)(x + 2)] = A/(x + 1) + B/(x + 2). Multiplying by the denominator gives the polynomial identity 3x + 5 = A(x + 2) + B(x + 1). Equating coefficients or choosing x = −1 and −2 gives A = 2 and B = 1.
The coefficient identity can be evaluated at −1 and −2 because it is an identity between polynomials. The original fractions remain undefined at those inputs. This distinction explains why substitution into the cleared identity is legitimate without claiming the original function has acquired values at its poles.
A repeated linear factor needs each power: a denominator (x − a)² uses terms A/(x − a) + B/(x − a)². An irreducible quadratic may require a linear numerator. Divide first if the expression is improper. For graph analysis, these forms expose shifts and local signs; for integration using partial fractions, they expose standard antiderivatives. Always recombine your proposed fractions as an algebraic check.
4. Join algebra to calculus
Stationary points satisfy f′(x) = 0 at allowed inputs. Solve that equation, obtain the corresponding coordinates and use the sign of f′ to determine increasing and decreasing intervals. A derivative zero does not by itself distinguish a maximum, minimum or stationary inflection. A point where the function is undefined cannot become stationary simply because a denominator was cleared.
The second derivative describes concavity. Positive f″ means increasing gradient, or concave up; negative f″ means decreasing gradient, or concave down. An inflection point requires a change of concavity at a point on the curve. A vertical asymptote can separate regions with opposite concavity without creating an inflection point, because there is no point of the function there.
Check symmetry from the rule and domain. An even function satisfies f(−x) = f(x) on a domain symmetric about zero. An odd function satisfies f(−x) = −f(x). A translated reciprocal graph may instead have rotational symmetry about the intersection of its asymptotes. Symmetry is useful evidence, but it cannot replace labelling intercepts, asymptotes and critical points precisely.
Worked example 2: A complete rational-function sketch
Let g(x) = x + 4/x for x ≠ 0. Division is already visible: x = 0 is a vertical asymptote and y = x is oblique. There are no axial intercepts: zero is excluded, and x + 4/x = 0 would require x² = −4. Since g is odd, the two branches are related by a half-turn about the origin.
Differentiate: g′(x) = 1 − 4/x² and g″(x) = 8/x³. Stationary inputs are x = −2 and 2, giving (−2, −4) and (2, 4). The derivative is positive when |x| > 2 and negative when 0 < |x| < 2. Thus the first point is a local maximum and the second a local minimum.
The graph is concave down for x < 0 and concave up for x > 0, but has no inflection point: zero is outside its domain. The range is (−∞, −4] ∪ [4, ∞). This follows from the extrema and the branch limits, not from the oblique asymptote alone.
5. Simple non-rational quotients need the same discipline
A quotient containing a logarithm, exponential or circular function need not be rational. The graphing workflow still works, but numerator-degree shortcuts do not. First intersect every domain condition: a logarithm needs a positive argument, an inverse circular function has its own permitted inputs, and a denominator cannot be zero. Then use appropriate limiting or numerical reasoning for the ends.
For instance, ln x/x exists only for x > 0. The positive denominator never vanishes inside that domain, but it approaches zero at the boundary. A graphing window that starts at x = 0.1 can hide the rapid fall near zero. A sensible sketch records the boundary behaviour explicitly and does not draw an unsupported left-hand branch.
Worked example 3: Different critical points in a quotient graph
For h(x) = ln x/x, x > 0, the quotient rule gives h′(x) = (1 − ln x)/x². Thus h increases on (0, e), decreases on (e, ∞) and has a maximum at (e, 1/e). Its only axial intercept is (1, 0). As x approaches zero from the right, h tends to negative infinity; as x tends to infinity, h tends to zero from above.
Differentiating again gives h″(x) = (2ln x − 3)/x³. Its sign changes from negative to positive at x = e3/2, giving a nonstationary inflection at (e3/2, 3/[2e3/2]). The maximum and inflection are different points. A CAS can confirm the derivatives, but the sign argument is what justifies the classifications.
6. Investigation and technology checks
Consider the family x + a/x. For a > 0, stationary inputs are ±√a, and the two branches have separated ranges. For a < 0, the derivative 1 − a/x² is always positive and there are no stationary points. At a = 0, the original expression still excludes zero unless the function is deliberately redefined. A parameter can change graph structure as well as numerical coordinates.
With technology, store the original rule and domain before simplifying. Request exact roots and derivatives, then test the sign on each interval cut by critical and excluded inputs. Plot at more than one scale. A missing hole, a false vertical join or a turning point outside the window is a display limitation, not a new mathematical property.
For a technology-free response, a feature list and sign chart provide a robust route to a smooth sketch. Label the requested equations and coordinates, draw asymptotes distinctly and show a hole clearly. State exact values such as e or √a unless approximations are requested. The graph and the written reasoning should tell the same story.
Practise, then check
- Describe the graph of (x² − 9)/(x − 3), retaining its original domain.
- Find all asymptotes and intercepts of (2x + 1)/(x − 4).
- Decompose (5x + 7)/[(x + 1)(x + 2)] into partial fractions.
- Find the stationary points and range of x + 9/x.
- For p(x) = x/(x² + 1), determine symmetry, stationary points and horizontal asymptote.
- Why does y = 1/x have no inflection point, despite its second derivative changing sign across zero?
- For q(x) = ex/x on x > 0, find its minimum and decide whether it has an inflection point.
Fully explained answers
- Factor and cancel to obtain y = x + 3 for x ≠ 3. The graph is that line with a hole at (3, 6). There is no vertical asymptote because the nearby values approach the finite value 6. The cancellation does not define the missing input.
- Write the rule as 2 + 9/(x − 4). The asymptotes are x = 4 and y = 2. Setting the numerator to zero gives (−1/2, 0), and substituting zero gives (0, −1/4). Both inputs are allowed.
- The identity 5x + 7 = A(x + 2) + B(x + 1) gives A = 2 and B = 3. Hence the answer is 2/(x + 1) + 3/(x + 2), for x ≠ −1, −2. Recombining produces the original numerator.
- The derivative is 1 − 9/x², so stationary inputs are ±3. Values are −6 and 6; the first is a maximum and the second a minimum by derivative sign changes. The range is (−∞, −6] ∪ [6, ∞), with zero excluded from the domain.
- The denominator is always positive and p is odd. Its derivative is (1 − x²)/(x² + 1)², giving a minimum at (−1, −1/2) and maximum at (1, 1/2). The horizontal asymptote is y = 0, which the graph crosses at the origin.
- The derivative calculation gives y″ = 2/x³, negative left of zero and positive right of zero. However, an inflection point must be on the curve. Since x = 0 is excluded, the two disconnected branches do not meet in an inflection.
- Here q′ = ex(x − 1)/x², so the derivative changes from negative to positive at x = 1, giving minimum (1, e). Also q″ = ex(x² − 2x + 2)/x³, positive for x > 0 because x² − 2x + 2 = (x − 1)² + 1. There is no inflection point.
Curriculum and sources
VCE Specialist Mathematics Units 3 and 4, Areas of Study 2 as applicable, with Outcomes 1–3. Scope follows the VCAA Mathematics Study Design 2023–2027 and the current Specialist Mathematics FAQ. Curriculum checked 15 October 2026. Teaching reference: Cambridge Senior Mathematics VCE Specialist Mathematics 3&4, second edition (2023), sections 8G (pp354–362);9G (pp407–414). All teaching prose, examples, practice questions and diagrams are original. Independent teaching material; not a VCAA publication.