1The idea in one paragraph

A rational function is one polynomial divided by another. At HL you meet two new shapes. In (ax + b)/(cx² + dx + e) the bottom grows faster than the top, so for large x the fraction dies away to zero: the horizontal asymptote is y = 0. In (ax² + bx + c)/(dx + e) the top grows faster, so the fraction does not settle to a constant; divide it out and you get a straight line plus a remainder fraction that dies away, so the graph settles onto that line, an oblique asymptote. In both, the graph shoots off to infinity wherever the denominator is zero and the numerator is not, which gives the vertical asymptotes. Find the asymptotes and the intercepts, work out the sign of the function on each side, and the graph draws itself.

2Where the asymptotes come from

Two rules find every asymptote in this subtopic:

• Vertical asymptotes: x = k for every real k that makes the denominator zero and the numerator non-zero. Near such a k the fraction is a normal number divided by something tiny, so it is huge, positive or negative.
• The long-run asymptote comes from comparing degrees. Top degree lower: y = 0. Degrees equal (the SL case): y = a/c. Top degree one higher: an oblique line, found by division.

Why the degrees? For large x, only the leading terms matter. (ax + b)/(cx² + dx + e) behaves like ax/(cx²) = a/(cx), which tends to 0. (ax² + bx + c)/(dx + e) behaves like ax²/(dx) = (a/d)x, which grows without limit along a line of gradient a/d.

3Linear over quadratic: (ax + b)/(cx² + dx + e)

Worked example 1 · two vertical asymptotes (Paper 1). Sketch y = (x + 1)/(x² − x − 6).

Now the signs. The function can only change sign where the top or the bottom is zero, at −2, −1 and 3. Test one value in each interval, or count the negative factors:

That table is the sketch. On the far left y is negative and near 0, so the curve sits just below y = 0 and plunges to −∞ as it reaches x = −2. Just right of −2 it is positive, so it comes down from +∞, crosses the axis at −1, carries on negative through (0, −⅙) and plunges to −∞ at x = 3. Right of 3 it comes down from +∞ towards y = 0, staying above it. Figure 1 is the result.

Notice that the middle branch crosses the horizontal asymptote at (−1, 0). That is allowed: a horizontal asymptote describes what happens as x → ±∞, not near the middle. For this family the crossing happens exactly at the x-intercept, because the asymptote is y = 0.

When the denominator has no real roots. If the discriminant of cx² + dx + e is negative, the bottom is never zero and there is no vertical asymptote. The graph is one unbroken curve with a maximum and a minimum. Figure 2(a) shows y = (x + 1)/(x² + 3): x² + 3 ≥ 3 for all x. The intercepts are (−1, 0) and (0, ⅓), the asymptote is y = 0, and the turning points are (1, ½) and (−3, −⅙). You would find those with the GDC on Paper 2, or with calculus; section 6 finds the range without either.

When the denominator is a perfect square. For y = (x + 2)/(x − 1)² the only vertical asymptote is x = 1, but (x − 1)² is positive on both sides of it. Near x = 1 the numerator is about 3, so y → +∞ from both sides, as Figure 2(b) shows. A repeated factor in the denominator is the rational-function version of a double root: no change of sign.

The reciprocal function is a particular case. 1/x has a constant on top and a linear expression below, so it fits (ax² + bx + c)/(dx + e) with a = b = 0, c = 1, d = 1, e = 0: vertical asymptote x = 0, and the "oblique" asymptote is the line y = 0. Its reflection in y = x is itself, which is why 1/x is its own inverse (2.14). A reciprocal of a quadratic, such as 1/(x² − 4), is the other family with a = 0: asymptotes x = ±2 and y = 0.

A common factor is a hole, not an asymptote. If the top and bottom share a factor, as in (x − 2)/(x² − 4) = 1/(x + 2) for x ≠ 2, there is no asymptote at x = 2, only a missing point (a hole) at (2, ¼). That is why the rule says "denominator zero and numerator non-zero". Exam questions in this family are usually set so that it does not happen, but check.

4Quadratic over linear: (ax² + bx + c)/(dx + e) and the oblique asymptote

Here the top has the higher degree, so first divide. The aim is the form

f(x) = Ax + B + R/(dx + e)

where Ax + B is the quotient and R is a constant remainder. As x → ±∞, R/(dx + e) → 0, so the graph approaches the line y = Ax + B, the oblique asymptote. The vertical asymptote is x = −e/d as before.

Worked example 2 · finding the oblique asymptote (Paper 1). Let f(x) = (x² + 2x − 3)/(x − 2). Use comparing coefficients, as in 2.12:

Synthetic division gives the same with less writing: bring down 1, then 1 × 2 + 2 = 4, then 4 × 2 − 3 = 5. Quotient x + 4, remainder 5.

So the asymptotes are x = 2 and y = x + 4. The x-intercepts come from the numerator: x² + 2x − 3 = (x + 3)(x − 1), giving (−3, 0) and (1, 0). The y-intercept is f(0) = −3/−2, giving (0, 1.5).

Which side of the oblique asymptote? The gap between the curve and the line is exactly the remainder term, 5/(x − 2). For x > 2 it is positive, so the curve lies above y = x + 4; for x < 2 it is negative, so the curve lies below. Figure 3 shows both branches doing just that. It also shows why this curve can never cross its oblique asymptote: 5/(x − 2) is never zero.

On Paper 2, the GDC gives the turning points (−0.236, 1.53) and (4.24, 10.5); exactly, they are (2 − √5, 6 − 2√5) and (2 + √5, 6 + 2√5).

When the leading coefficient of the divisor is not 1. Divide 2x² − x + 3 by 2x + 1:

Oblique asymptote y = x − 1, vertical asymptote x = −½. Checking by multiplying back takes ten seconds: (2x + 1)(x − 1) + 4 = 2x² − x − 1 + 4 = 2x² − x + 3.

5Sketching: the full checklist

Figure 4 puts the asymptote rules on one card.

A sketch of a rational function earns its marks for these things, and only these: the correct shape of each branch, every asymptote drawn dashed and labelled with its equation, and every intercept labelled with its coordinates. Turning points are labelled only if the question asks.

1. Factorise the denominator; its real zeros give the vertical asymptotes. Check the numerator is not also zero there.
2. Compare degrees: y = 0, y = a/c, or divide for the oblique asymptote.
3. Intercepts: y-intercept f(0); x-intercepts from numerator = 0.
4. Signs: use the zeros of the top and bottom to split the x-axis, and find the sign in each interval. A single factor changes sign; a squared factor does not.
5. Near each vertical asymptote, the sign on each side tells you whether the curve goes to +∞ or −∞ there.
6. Far out, the sign of the remainder term (or of y itself, for the linear-over-quadratic family) says which side of the asymptote the curve approaches from.
7. Join up, without crossing any vertical asymptote and without the curve turning back across an oblique one.

6The range, and the gap in the middle

A function of the second family often misses a whole band of y-values. Figure 5 shows y = (x² + 3)/(x − 1) = x + 1 + 4/(x − 1). The left branch rises to a maximum at (−1, −2) and falls away; the right branch falls to a minimum at (3, 6) and rises. Nothing in between is ever reached: the range is y ≤ −2 or y ≥ 6.

On Paper 2 you would read the turning points off the GDC. On Paper 1 there is an algebraic method that needs no calculus, the discriminant method. It asks a clean question: for which values of y does the equation y = f(x) have a real solution x?

Worked example 3 · range by the discriminant (Paper 1). Find the range of f(x) = (x² + 3)/(x − 1).

The boundary values are where Δ = 0, a single repeated x, which is exactly the turning points: y = 6 gives x² − 6x + 9 = 0, so x = 3, and y = −2 gives x² + 2x + 1 = 0, so x = −1.

The same method works for the first family. For y = (x + 1)/(x² + 3), multiply up to get yx² − x + (3y − 1) = 0. For y ≠ 0 this is a quadratic in x, and real roots need 1 − 4y(3y − 1) ≥ 0, that is 12y² − 4y − 1 ≤ 0, or (6y + 1)(2y − 1) ≤ 0. So −⅙ ≤ y ≤ ½, and y = 0 is included too since x = −1 gives it. That is the range quoted in Figure 2(a).

Two cautions. The inequality from Δ is solved like any quadratic inequality (2.7), so sketch the parabola in y if the direction is unclear. And if the coefficient of x² in your quadratic can be zero for some y, check that value of y separately, as with y = 0 above.

7With technology

The guide expects graphs of these functions to be explored with dynamic graphing software, and Paper 2 expects the GDC. Graph the function, then graph each asymptote you found as a separate line and check the curve hugs it. Use the maximum and minimum tools for turning points and quote them to 3 s.f. Watch for two illusions: a connected-mode line drawn down a vertical asymptote, and a window too narrow to show that a curve is approaching an oblique line rather than bending away from it. Zoom out.