Paper 1 · Further Pure 1 ✦

Rational Functions & Graphs

Sketching quotients of polynomials — finding every asymptote, and pinning down the exact set of values the curve can reach.

Syllabus 1.2Curve sketchingAsymptotes & range

What the examiner expects you to do

Sketch simple rational functions, finding vertical, horizontal and oblique asymptotes (numerator/denominator degree ≤ 2).
Determine the set of values the function takes — often via a discriminant.
Use relationships between the graphs of $y = f (x)$ , $y^{2} = f (x)$ , $y = \frac{1}{f (x)}$ and $y = | f (x) |$ .

§1 Asymptotes & sketching

Three families of asymptote, decided by comparing the degrees of numerator and denominator.

◆ The three asymptote rules

Vertical: wherever the denominator is zero (and the numerator isn't) — the curve shoots to $\pm \infty$ .
Horizontal: if $\deg (num) < \deg (den)$ then $y \to 0$ ; if degrees are equal, $y \to$ ratio of leading coefficients.
Oblique (slant): if $\deg (num) = \deg (den) + 1$ , divide out — the quotient $y = m x + c$ is the asymptote.

✎ Worked example — an oblique asymptote

Sketch $y = \frac{x^{2} + 1}{x - 1}$ . Polynomial-divide the top by the bottom:

\frac{x^{2} + 1}{x - 1} = x + 1 + \frac{2}{x - 1} .

As $x \to \pm \infty$ the $\frac{2}{x - 1}$ term vanishes, so the oblique asymptote is $y = x + 1$ ; the vertical asymptote is $x = 1$ . The curve hugs $y = x + 1$ far out and blows up near $x = 1$ .

★ A* tip

To place the two branches, test the sign of $\frac{2}{x - 1}$ just either side of $x = 1$ : it decides whether each branch sits above or below the oblique asymptote.

§2 The set of values — via the discriminant

To find which $y$ -values are actually reached, set $y$ equal to the function, clear the fraction into a quadratic in $x$ , and demand a real solution: $discriminant \geq 0$ .

✎ Worked example

Find the range of $y = \frac{x}{x^{2} + 1}$ . Multiply up: $y (x^{2} + 1) = x$ , i.e.

y x^{2} - x + y = 0 .

For a real $x$ , the discriminant $(- 1)^{2} - 4 (y) (y) \geq 0$ :

1 - 4 y^{2} \geq 0 \Rightarrow y^{2} \leq \frac{1}{4} \Rightarrow - \frac{1}{2} \leq y \leq \frac{1}{2} .

So the curve only ever reaches $y$ -values in $[- \frac{1}{2}, \frac{1}{2}]$ .

⚠ Trap

When the leading coefficient is $y$ itself, the "quadratic" degenerates if $y = 0$ . Check that boundary case separately rather than trusting the discriminant blindly.

§4 Interactive: watch the asymptotes form

Explore $y = \frac{x^{2} + p}{x - a}$ . The dashed lines are the vertical asymptote $x = a$ and the oblique asymptote $y = x + a$ (from dividing out). Move the sliders and see the curve hug them.

Pole

a

1.0

p

1.0

drag the sliders ✎

Since $x^{2} + p = (x - a) (x + a) + (a^{2} + p)$ , the curve is $y = x + a + \frac{a^{2} + p}{x - a}$ .

§5 Examiner traps & A* checklist

⚠ The four most common mark-losers

Calling every rational function's horizontal asymptote $y = 0$ — check the degrees; equal degrees give a non-zero horizontal.
Missing the oblique asymptote when $\deg (num) = \deg (den) + 1$ — you must divide.
Forgetting to state coordinates of turning points and axis intercepts on the sketch.
In range questions, ignoring the degenerate case where the $x^{2}$ coefficient vanishes.

§6 Video explainers (curated)

▶Sketching rational functions
Curated search — vertical, horizontal & oblique asymptotes.↗ ▶Range via the discriminant
Curated search — the set of values a function takes.↗ ▶Related graphs (y², 1/f, |f|)
Curated search — transforming a known curve.↗

§7 Check yourself

Score 0 / 5

Q1 · vertical asymptotes

The vertical asymptotes of $y = \frac{x + 2}{x^{2} - 9}$ are:

Solution

Set the denominator to zero: $x^{2} - 9 = 0 \Rightarrow x = \pm 3$ (numerator non-zero there).

Q2 · behaviour at infinity

As $x \to \pm \infty$ , the curve $y = \frac{2 x^{2} + 1}{x^{2} - 4}$ approaches:

Solution

Equal degrees ⇒ horizontal asymptote at the ratio of leading coefficients, $y = \frac{2}{1} = 2$ .

Q3 · oblique asymptote

The oblique asymptote of $y = \frac{x^{2} + 1}{x - 1}$ is:

Solution

$\frac{x^{2} + 1}{x - 1} = x + 1 + \frac{2}{x - 1}$ ; the quotient $y = x + 1$ is the asymptote.

Q4 · range

The set of values taken by $y = \frac{x}{x^{2} + 1}$ is:

Solution

$y x^{2} - x + y = 0$ needs $1 - 4 y^{2} \geq 0$ , so $- \frac{1}{2} \leq y \leq \frac{1}{2}$ .

Q5 · full method

Find the set of values taken by $y = \frac{1}{x^{2} - 2 x + 2}$ .

Solution

Rearrange to a quadratic in $x$ : $y (x^{2} - 2 x + 2) = 1$ , i.e. $y x^{2} - 2 y x + (2 y - 1) = 0$ . For real $x$ ,

(- 2 y)^{2} - 4 y (2 y - 1) \geq 0 \Rightarrow 4 y - 4 y^{2} \geq 0 \Rightarrow 0 \leq y \leq 1 .

Since $x^{2} - 2 x + 2 = (x - 1)^{2} + 1 \geq 1 > 0$ , the value $y = 0$ is never reached, so the range is $0 < y \leq 1$ . $■$

FP1 · Topic 2 · Rational Functions & Graphs

Rational Functions & Graphs

What the examiner expects you to do

§1 Asymptotes & sketching

§2 The set of values — via the discriminant

§3 Related graphs from y=f(x)

§4 Interactive: watch the asymptotes form

§5 Examiner traps & A* checklist

§6 Video explainers (curated)

§7 Check yourself

§3 Related graphs from $y = f (x)$