Topic 2 of 8 · Form 5
Rational functions and curve sketching
For y = f(x) as a quotient, rearrange into a quadratic in x and use the discriminant ≥ 0 to find values y cannot take.
Notes for this topic 1 file, in order
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Asymptotes
- Vertical asymptotes where the denominator is zero (and the numerator is not).
- Horizontal asymptote: compare degrees – equal degrees give y = ratio of leading coefficients; lower numerator degree gives y = 0.
- Oblique asymptote when the numerator degree is one more than the denominator: use polynomial division.
Worked example: Find the asymptotes of y = (x² + 1)/(x − 1).
x² + 1 = (x − 1)(x + 1) + 2 → y = x + 1 + 2/(x − 1)
Vertical asymptote x = 1; oblique asymptote y = x + 1
Range and stationary points
For y = f(x) as a quotient, rearrange into a quadratic in x and use the discriminant ≥ 0 to find values y cannot take.
Worked example: Show that y = (x² + 1)/(x − 1) cannot take values between −0.83 and 4.83.
yx − y = x² + 1 → x² − yx + (1 + y) = 0 → y² − 4(1 + y) ≥ 0 → y² − 4y − 4 ≥ 0
y ≤ 2 − 2√2 ≈ −0.83 or y ≥ 2 + 2√2 ≈ 4.83
Related graphs
Sketch y = |f(x)|, y = f(|x|), y² = f(x) and y = 1/f(x) from y = f(x).
Inequalities
Solve rational inequalities by multiplying by the square of the denominator or by using sign diagrams – never multiply by an expression of unknown sign.
Check yourself
- Find the asymptotes of y = (2x + 3)/(x − 4).
- Solve (x + 1)/(x − 2) > 0.
- What happens to the graph of y = 1/f(x) where f(x) = 0?
Show answers
- x = 4 and y = 2
- x < −1 or x > 2
- It has vertical asymptotes there.
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