Rational Functions and Their Graphs

A rational function is $y = \dfrac{p(x)}{q(x)}$y=q(x)p(x)​ for polynomials $p, q$p,q. Sketching one is a systematic act of detective work: the algebra of the function dictates every feature of the curve — where it blows up (vertical asymptotes), how it behaves far out (horizontal or oblique asymptotes), where it crosses the axes (intercepts), and the set of heights it can reach (its range, found by a beautiful discriminant argument). At Further-Maths level the prize feature is finding turning points and range without calculus, by demanding that "$y=k$y=k" has real solutions.

This lesson assembles a reliable sketching algorithm, derives the asymptote rules rather than just stating them, and develops the discriminant/range method that is the topic's signature exam technique. A correct sketch is worth several marks and underpins later work on polar curves, transformations and modelling.

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Where this sits in AQA 7367

This is the rational functions and graphs strand of the compulsory pure content examined on Paper 1 and Paper 2 (each 2 hours, 100 marks, 33⅓% of the A-Level). It extends A-Level Maths curve-sketching and the reciprocal/transformation graphs to general $p(x)/q(x)$p(x)/q(x). Identifying asymptotes and intercepts is AO1; deducing the range by a discriminant condition, and justifying the curve's behaviour in each region, are the AO2/AO3 reasoning marks. It links forward to further calculus (areas, the behaviour of integrands) and back to partial fractions (the previous lesson, which reveals near-asymptote behaviour).

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Core theory: the sketching algorithm

Every rational-function sketch is the same five-step procedure. Master the order and no feature is missed.

1. Vertical asymptotes — where the denominator vanishes

A vertical asymptote sits at $x=a$x=a when $q(a)=0$q(a)=0 but $p(a)\neq 0$p(a)=0: the denominator collapses while the numerator does not, forcing $y\to\pm\infty$y→±∞. Crucially you must determine the sign on each side. Near $x=a$x=a, write $y\approx \dfrac{p(a)}{q(x)}$y≈q(x)p(a)​ and ask whether $q(x)\to 0^+$q(x)→0+ or $0^-$0− as $x\to a^{\pm}$x→a±. (If $q(a)=0$q(a)=0 and $p(a)=0$p(a)=0, the factor cancels — there is a hole, not an asymptote.)

2. Horizontal / oblique asymptotes — behaviour as $x\to\pm\infty$x→±∞

The end behaviour is governed entirely by the degrees of $p$p and $q$q:

Condition	Asymptote as $x\to\pm\infty$x→±∞
$\deg p < \deg q$degp<degq	$y = 0$y=0
$\deg p = \deg q$degp=degq	$y = \dfrac{\text{leading coeff of }p}{\text{leading coeff of }q}$y=leading coeff of qleading coeff of p​
$\deg p = \deg q + 1$degp=degq+1	oblique (slant) $y = mx+c$y=mx+c, via division
$\deg p > \deg q + 1$degp>degq+1	no horizontal/oblique asymptote (a polynomial-like tail)

The reason is divide-through-by-the-highest-power: for $\deg p = \deg q$degp=degq, $\dfrac{a_nx^n+\cdots}{b_nx^n+\cdots} = \dfrac{a_n + \cdots/x^n}{b_n + \cdots/x^n} \to \dfrac{a_n}{b_n}$bn​xn+⋯an​xn+⋯​=bn​+⋯/xnan​+⋯/xn​→bn​an​​. When $\deg p = \deg q + 1$degp=degq+1, polynomial division gives $y = (mx+c) + \dfrac{\text{remainder}}{q(x)}$y=(mx+c)+q(x)remainder​; the remainder fraction $\to 0$→0, leaving the oblique asymptote $y = mx+c$y=mx+c.

3. Intercepts

• $y$y-intercept: set $x=0$x=0, giving $y = \dfrac{p(0)}{q(0)}$y=q(0)p(0)​ (if defined).
• $x$x-intercepts: set $y=0$y=0, i.e. solve $p(x)=0$p(x)=0 (provided $q\neq 0$q=0 there).

4. The range — the discriminant (or "set $y=k$y=k") method

This is the Further-Maths signature. To find which heights $y$y the curve attains, set $y=k$y=k, clear the denominator to get a polynomial in $x$x with $k$k as a parameter, and demand that this polynomial has real solutions. For a function that becomes a quadratic in $x$x, the condition is the discriminant $\ge 0$≥0; solving the resulting inequality in $k$k gives the range. The boundary values of $k$k (where the discriminant is exactly $0$0, i.e. a repeated root) are precisely the $y$y-coordinates of the turning points — so this one method delivers range and stationary values without differentiating.

The logic deserves to be spelled out, because it is the conceptual heart of the topic. A height $y=k$y=k is attained by the curve if and only if there is some $x$x with $\tfrac{p(x)}{q(x)} = k$q(x)p(x)​=k — that is, if and only if the equation $p(x) = k\,q(x)$p(x)=kq(x) has a real solution. So the range is the set of $k$k-values for which this equation is solvable. When the equation is a quadratic in $x$x, "solvable" means "discriminant $\ge 0$≥0", which is itself an inequality in $k$k. The repeated-root case (discriminant exactly $0$0) is the borderline where the curve just touches the height $k$k once — geometrically a turning point, which is why the boundaries of the range are the stationary values. One health warning: when the equation reduces from quadratic to linear (the $x^2$x2 coefficient vanishing for some special $k$k), that $k$k must be checked separately, since the discriminant argument no longer applies. Example 4 below shows exactly this subtlety.

5. Behaviour in each region and the sketch

Combine the above: draw asymptotes as dashed lines, mark intercepts with coordinates, mark turning points, and join with smooth branches respecting the sign analysis from step 1 and the approach to asymptotes from step 2. The vertical asymptotes partition the $x$x-axis into regions; within each region the curve is continuous, so once you know its sign just inside each boundary and its behaviour at the region's outer edge, the branch is essentially determined. A useful sanity check is that a rational function changes sign only at an $x$x-intercept (numerator zero) or across a vertical asymptote of odd multiplicity; across an asymptote from an even-power factor the curve returns on the same side. Building the sketch region by region, rather than trying to draw it all at once, is the reliable route to a correct picture and full sketch marks.

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Worked examples (with mark scheme)

Example 1 — a full sketch with the range

Sketch $y = \dfrac{x^2 - 4}{x^2 - 1}$y=x2−1x2−4​, stating all asymptotes, intercepts, and the range.

Factorise: $y = \dfrac{(x-2)(x+2)}{(x-1)(x+1)}$y=(x−1)(x+1)(x−2)(x+2)​. No common factor, so no holes. (M1: factorise)

Vertical asymptotes: $q=0$q=0 at $x=1$x=1 and $x=-1$x=−1; numerator nonzero there. (A1)

Horizontal asymptote: $\deg p = \deg q = 2$degp=degq=2, leading coefficients both $1$1, so $y=1$y=1. (A1)

Intercepts: $x$x-intercepts at $x=\pm 2$x=±2; $y$y-intercept $y = \dfrac{-4}{-1} = 4$y=−1−4​=4. (A1)

Sign near $x=1$x=1: as $x\to 1^+$x→1+, numerator $\to 1-4 = -3$→1−4=−3 and denominator $\to 0^+$→0+, so $y\to-\infty$y→−∞; as $x\to 1^-$x→1−, denominator $\to 0^-$→0−, so $y\to+\infty$y→+∞. (M1: sign analysis)

Range (set $y=k$y=k): $k(x^2-1) = x^2 - 4 \Rightarrow x^2(k-1) = k-4 \Rightarrow x^2 = \dfrac{k-4}{k-1}$k(x2−1)=x2−4⇒x2(k−1)=k−4⇒x2=k−1k−4​. For real $x$x we need $\dfrac{k-4}{k-1}\ge 0$k−1k−4​≥0. (M1: rearrange) A sign diagram on $k$k (critical values $1, 4$1,4) gives $k<1$k<1 or $k\ge 4$k≥4. (A1)

$\therefore\;\text{range: } y < 1 \text{ or } y \ge 4.$∴range: y<1 or y≥4.

The value $k=4$k=4 (where $x^2 = 0$x2=0, i.e. a repeated root at $x=0$x=0) is the turning point $(0,4)$(0,4); the curve never takes a value in $[1,4)$[1,4). Here is the resulting sketch:

(M1 factorise; A1 VAs; A1 HA; A1 intercepts; M1 sign analysis; M1 set $y=k$y=k and rearrange; A1 range. The outer branches approach $y=1$y=1 from below and dip to cross the $x$x-axis at $\pm 2$±2; the central branch sits above $y=4$y=4 at its minimum $(0,4)$(0,4).)

Example 2 — an oblique asymptote

Find the oblique asymptote of $y = \dfrac{x^2+1}{x-1}$y=x−1x2+1​ and the coordinates of any stationary points.

$\deg p = 2 = \deg q + 1$degp=2=degq+1, so divide:

$\frac{x^2+1}{x-1} = x + 1 + \frac{2}{x-1}. \quad (\text{M1 divide; A1})$x−1x2+1​=x+1+x−12​.(M1 divide; A1)

As $x\to\pm\infty$x→±∞, $\dfrac{2}{x-1}\to 0$x−12​→0, so the oblique asymptote is $y = x+1$y=x+1; there is a vertical asymptote at $x=1$x=1.

For stationary points, set $y=k$y=k: $x^2+1 = k(x-1) \Rightarrow x^2 - kx + (k+1) = 0$x2+1=k(x−1)⇒x2−kx+(k+1)=0. Repeated root when the discriminant is zero: $k^2 - 4(k+1) = 0 \Rightarrow k^2 - 4k - 4 = 0 \Rightarrow k = 2 \pm 2\sqrt2$k2−4(k+1)=0⇒k2−4k−4=0⇒k=2±22​. (M1 set $y=k$y=k; M1 discriminant; A1) So the stationary values are $y = 2+2\sqrt2$y=2+22​ and $y = 2-2\sqrt2$y=2−22​; the curve takes no value strictly between them.

(M1 divide; A1 asymptote; M1/M1 the $y=k$y=k-discriminant method; A1 the two stationary values. The discriminant method handles the turning points without a single derivative — a major time-saver.)

Example 3 — range of a bounded rational function

Find the range of $y = \dfrac{x^2}{x^2+1}$y=x2+1x2​ for real $x$x.

$y(x^2+1) = x^2 \;\Rightarrow\; x^2(1-y) = y \;\Rightarrow\; x^2 = \frac{y}{1-y}. \quad (\text{M1 rearrange})$y(x2+1)=x2⇒x2(1−y)=y⇒x2=1−yy​.(M1 rearrange)

For real $x$x we need $x^2 \ge 0$x2≥0, i.e. $\dfrac{y}{1-y} \ge 0$1−yy​≥0. A sign diagram (critical values $0$0 and $1$1) gives $0 \le y < 1$0≤y<1. (M1 inequality; A1)

$\therefore\; \text{range } [0,1).$∴range [0,1).

(M1 rearrange to $x^2 = \cdots$x2=⋯; M1 demand $\ge 0$≥0 and solve; A1 range $[0,1)$[0,1). The endpoint $y=1$y=1 is excluded (the horizontal asymptote — approached, never reached); $y=0$y=0 is included (attained at $x=0$x=0). Getting the open/closed brackets right is where the final A-mark lives.)

Example 4 — a quadratic-over-quadratic with a forbidden band

Find the range of $y = \dfrac{x^2 + x + 1}{x^2 + 1}$y=x2+1x2+x+1​.

Set $y = k$y=k and clear the denominator:

$k(x^2 + 1) = x^2 + x + 1 \;\Rightarrow\; (k-1)x^2 - x + (k-1) = 0. \quad (\text{M1 rearrange to a quadratic in } x)$k(x2+1)=x2+x+1⇒(k−1)x2−x+(k−1)=0.(M1 rearrange to a quadratic in x)

For real $x$x, if $k\neq1$k=1 this quadratic needs discriminant $\ge 0$≥0:

$(-1)^2 - 4(k-1)(k-1) \ge 0 \;\Rightarrow\; 1 - 4(k-1)^2 \ge 0 \;\Rightarrow\; (k-1)^2 \le \tfrac14. \quad (\text{M1 discriminant})$(−1)2−4(k−1)(k−1)≥0⇒1−4(k−1)2≥0⇒(k−1)2≤41​.(M1 discriminant)

So $|k-1| \le \tfrac12$∣k−1∣≤21​, i.e. $\tfrac12 \le k \le \tfrac32$21​≤k≤23​. (A1) The case $k=1$k=1 gives the linear equation $-x = 0$−x=0, i.e. $x=0$x=0, which yields $y=1$y=1 — already inside the interval. Hence

$\text{range: } \tfrac12 \le y \le \tfrac32.$range: 21​≤y≤23​.

The stationary values are the boundary heights $y = \tfrac12$y=21​ and $y = \tfrac32$y=23​ (where the discriminant is zero), so the curve oscillates between a minimum of $\tfrac12$21​ and a maximum of $\tfrac32$23​, approaching its horizontal asymptote $y=1$y=1 at both ends.

(M1 rearrange to a quadratic in $x$x; M1 apply the discriminant condition; A1 the closed interval. The $k=1$k=1 case must be checked separately because the $x^2$x2 coefficient $(k-1)$(k−1) vanishes there — a subtlety that distinguishes a complete answer. Both endpoints are attained (turning points), so the interval is closed at both ends.)

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Specimen-style exam question

(specimen-style — not from any past paper)

The curve $C$C has equation $y = \dfrac{x^2 - 3}{x - 2}$y=x−2x2−3​. (a) Find the equations of the asymptotes of $C$C. (b) Show that $C$C takes no value with $2 < y < 6$2<y<6, and write down the $y$y-coordinates of its stationary points.

(a) Divide: $\dfrac{x^2-3}{x-2} = x + 2 + \dfrac{1}{x-2}$x−2x2−3​=x+2+x−21​. Vertical asymptote $x=2$x=2; oblique asymptote $y = x+2$y=x+2.