Rational functions and graphs

Further Maths · Further Pure Mathematics 1. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

A rational function is one polynomial divided by another, here with degree at most 2 on the top and the bottom. To sketch it, find the asymptotes, the points where the curve crosses the axes and any turning points.

The values of y that the function can take are found without calculus: rearrange y = f(x) into a quadratic in x and use the fact that real x needs a discriminant that is not negative.

You must also be able to turn the graph of y = f(x) into the graphs of y² = f(x), y = 1/f(x), y = |f(x)| and y = f(|x|), and use these sketches to solve equations and inequalities.

Rules to remember

Vertical asymptote: x = a where the denominator is zero at x = a (and the numerator is not)
Horizontal asymptote: y = 0 if the top has lower degree; y = (ratio of leading coefficients) if the degrees are equal
Oblique asymptote: if the top has degree one more than the bottom, divide; y = quotient is the asymptote
Set of values of y: rearrange to a quadratic in x and solve b² − 4ac ≥ 0 as an inequality in y
y = 1/f(x): zeros of f become vertical asymptotes, a maximum of f becomes a minimum, and the sign of y is unchanged
y² = f(x): exists only where f(x) ≥ 0, is y = ±√f(x), and is symmetrical about the x-axis
y = |f(x)|: reflect the part below the x-axis in the x-axis. y = f(|x|): keep the part for x ≥ 0 and reflect it in the y-axis

Common mistake

Students confuse y = |f(x)| with y = f(|x|). The first reflects in the x-axis the part of the graph that is below it; the second throws away the part for x < 0 and replaces it with the mirror image of the part for x > 0.

Worked example

Find the set of values taken by y = (x² + 3)/(x + 1) for real x ≠ −1.

Multiply up: y(x + 1) = x² + 3, so x² − yx + (3 − y) = 0.
For real x the discriminant is ≥ 0: y² − 4(3 − y) ≥ 0, so y² + 4y − 12 ≥ 0.
Factorise: (y + 6)(y − 2) ≥ 0, so y ≤ −6 or y ≥ 2.
Check: x = 1 gives y = 4/2 = 2 and x = −3 gives y = 12/(−2) = −6.

Answer: y ≤ −6 or y ≥ 2

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