Rational Functions & Graphs | 9231 Further Pure P1

1. Rational Functions — Definition and Scope

A rational function is any function of the form y = P(x)/Q(x) where P and Q are polynomials and Q(x) ≠ 0. For Cambridge 9231, both P and Q have degree at most 2.

📐 The Four Types in This Syllabus

Type 1y = (ax+b)/(cx+d)linear / linear → horizontal asymptote

Type 2y = (ax²+bx+c)/(dx+e)quadratic / linear → oblique asymptote

Type 3y = (ax+b)/(cx²+dx+e)linear / quadratic → horizontal asymptote y=0

Type 4y = (ax²+bx+c)/(dx²+ex+f)quadratic / quadratic → horizontal asymptote

2. Asymptotes — Three Types

Type 1

Vertical Asymptote

Occurs where the denominator is zero and the numerator is non-zero. The curve shoots to ±∞ on each side.

Find: set Q(x) = 0
Equation: x = a
Check: P(a) ≠ 0 (else hole)

Type 2

Horizontal Asymptote

Exists when deg(P) ≤ deg(Q). Found by examining the limit as x → ±∞.

deg(P) < deg(Q): y = 0
deg(P) = deg(Q): y = lead(P)/lead(Q)
deg(P) > deg(Q): none

Type 3

Oblique Asymptote

Exists when deg(P) = deg(Q) + 1. Found by algebraic long division of P by Q.

Divide: P(x) = Q(x)·(mx+c) + r
As x→±∞, remainder/Q → 0
Asymptote: y = mx + c

Finding the Oblique Asymptote — Long Division

For y = (x² − 2x + 5)/(x − 1), divide numerator by denominator:

x² − 2x + 5 ÷ (x − 1)

Step 1:x² ÷ x = x → multiply: x(x−1) = x²−x

Step 2:Subtract: (x²−2x+5) − (x²−x) = −x + 5

Step 3:−x ÷ x = −1 → multiply: −1(x−1) = −x+1

Step 4:Subtract: (−x+5) − (−x+1) = 4 (remainder)

Result: y = (x−1) + 4/(x−1) → oblique asymptote y = x − 1

⭐ Key Insight

As x→±∞, the term 4/(x−1) → 0, so the curve approaches y = x − 1. The remainder tells you which side of the asymptote the curve lies: positive remainder → curve above asymptote for large positive x.

3. Sketching — The 6-Step Method

Cambridge expects sketches showing all significant features. Never plot points mechanically — follow this systematic approach.

1Domain: Find where the function is undefined (denominator = 0). These give vertical asymptotes.
2Asymptotes: Find all vertical, horizontal or oblique asymptotes. Draw them as dashed lines first.
3Intercepts: Set x=0 for y-intercept; set numerator=0 for x-intercepts. Label these points.
4Turning points: Differentiate (quotient rule), set dy/dx = 0, solve. Find coordinates. Determine nature (max/min).
5Behaviour near asymptotes: Check whether the curve approaches from above or below on each side of each vertical asymptote.
6Sketch: Draw the curve through all plotted features, approaching asymptotes correctly from both sides.

⚠ Behaviour Near Vertical Asymptotes — Must Show

To determine the sign of y near x = a (vertical asymptote), test a value slightly to the left and right. For example, near x = 1 in y = (x+2)/(x−1):

x → 1⁺(1⁺+2)/(1⁺−1) = (+3)/(+ε) → +∞

x → 1⁻(1⁻+2)/(1⁻−1) = (+3)/(−ε) → −∞

4. Range — Discriminant Method

To find the set of values that a rational function can take, set y = f(x) and rearrange to get a quadratic in x. For the function to have real solutions, the discriminant must be ≥ 0.

⭐ Finding the Range

For y = (ax²+bx+c)/(dx²+ex+f), set y = k and rearrange to:

Step 1Rearrange: Ax² + Bx + C = 0 (where A,B,C involve k)

Step 2For real x: discriminant B² − 4AC ≥ 0

Step 3Solve the inequality in k → this gives the range

The y-values of turning points are exactly where the discriminant equals zero.

5. Related Curves — Five Transformations

Given the graph of y = f(x), Cambridge expects you to sketch four related curves. Each has a systematic construction rule.

y = f(x) (original)

Reference curve Mark turning points P, Q and asymptotes. All other curves are derived from this.

y² = f(x)

Only exists where f(x) ≥ 0 • Where f(x) > 0: two branches y = ±√f(x)
• Where f(x) < 0: no curve
• Where f(x) = 0: x-axis crossings
• Turning points of f become x-axis points

y = 1/f(x)

Reciprocal — inverts the function • Zeros of f → vertical asymptotes
• Vertical asymptotes of f → zeros
• Where f↑, 1/f ↓ (and vice versa)
• Turning points: reflected, same x-coord
• Horizontal asymptote y=0 where |f|→∞

y = |f(x)|

Reflect negatives in x-axis • Positive parts stay unchanged
• Negative parts reflected in x-axis (y → −y)
• x-intercepts become "bounce" points
• The curve never goes below x-axis

y = f(|x|)

Reflect right half in y-axis • Keep the part for x ≥ 0
• Reflect it in the y-axis for x < 0
• Result is always an even function
• y-axis becomes an axis of symmetry

💡 Exam Tip — Transformations

For y² = f(x): first shade where f(x) < 0 on the original — no curve can exist there.
For y = 1/f(x): turning points of f become turning points of 1/f, but with reciprocal y-values (max → min if positive).
For y = |f(x)|: the graph is always above or on the x-axis — mark this explicitly.
For y = f(|x|): the right half of the original is copied — do not use the left half.

Example 1 Linear/Linear — Full Sketch ★☆☆ Standard

Sketch the graph of y = (2x + 3)/(x − 1), showing all asymptotes and intercepts clearly. State the range of the function.

1

Asymptotes

Verticalx − 1 = 0 → x = 1

Horizontaldeg equal: y = 2/1 = 2ratio of leading coefficients

2

Intercepts

y-interceptx=0: y = 3/(−1) = −3 → (0, −3)

x-intercept2x+3=0 → x = −3/2 → (−3/2, 0)

3

Behaviour near vertical asymptote x = 1

x → 1⁺Numerator ≈ 5 (positive), denominator small positive → y → +∞

x → 1⁻Numerator ≈ 5 (positive), denominator small negative → y → −∞

4

Range using discriminant

Set y = k: k(x−1) = 2x+3 → kx − k = 2x + 3 → x(k−2) = k+3 → x = (k+3)/(k−2).

This has a solution for all k ≠ 2. So the range is all real numbers except y = 2 (the horizontal asymptote).

✓ Range

y ∈ ℝ, y ≠ 2

Example 2 Quadratic/Linear — Oblique Asymptote + Turning Points ★★☆ Challenging

Sketch the graph of y = (x² + x − 2)/(x − 1), finding all asymptotes, intercepts and turning points.

1

Check for cancellation first

Numerator: x² + x − 2 = (x+2)(x−1). So y = (x+2)(x−1)/(x−1) = x + 2 for x ≠ 1.

This is a straight line with a hole at x = 1, y = 3! No asymptotes — just a removed point.

This demonstrates: always factorise before assuming asymptotes exist.

Now sketch y = (x² − 2x + 5)/(x − 1). Find asymptotes, turning points, and range.

2

Long division → oblique asymptote

x² − 2x + 5 = (x−1)(x−1) + 4

y = (x−1) + 4/(x−1)oblique: y = x−1

Verticalx = 1

Obliquey = x − 1

3

Intercepts

y-interceptx=0: y = 5/(−1) = −5 → (0, −5)

x-interceptsx² − 2x + 5 = 0: discriminant = 4−20 = −16 < 0

No real x-intercepts

4

Turning points — differentiate using quotient rule

dy/dx= [(2x−2)(x−1) − (x²−2x+5)(1)] / (x−1)²

= [2x²−4x+2 − x²+2x−5] / (x−1)²

= (x² − 2x − 3) / (x−1)²

= (x−3)(x+1) / (x−1)²

dy/dx = 0 when x = 3 or x = −1:

x = 3:y = (9−6+5)/(3−1) = 8/2 = 4 → (3, 4) minimum

x = −1:y = (1+2+5)/(−2) = 8/(−2) = −4 → (−1, −4) maximum

Check: (x−1)² > 0 always. Sign of dy/dx depends on (x−3)(x+1): negative for −1 < x < 3, so y is decreasing between turning points — confirming (−1,−4) is a local max and (3,4) is a local min.

Example 3 Range by Discriminant — Quadratic/Quadratic ★★☆ Challenging

Find the range of the function y = (x² + 2x + 3)/(x² + 1).

1

Set y = k and rearrange to quadratic in x

k(x² + 1) = x² + 2x + 3

kx² + k = x² + 2x + 3

(k−1)x² − 2x + (k−3) = 0

2

Apply discriminant ≥ 0 condition

For real x (with k ≠ 1 — check k=1 separately): discriminant ≥ 0:

Δ= (−2)² − 4(k−1)(k−3) ≥ 0

4 − 4(k²−4k+3) ≥ 0

4 − 4k² + 16k − 12 ≥ 0

−4k² + 16k − 8 ≥ 0

k² − 4k + 2 ≤ 0

3

Solve the quadratic inequality in k

k² − 4k + 2 = 0k = (4 ± √8)/2 = 2 ± √2

k² − 4k + 2 ≤ 0when 2−√2 ≤ k ≤ 2+√2

When k = 1: the equation becomes −2x + (−2) = 0 → x = −1 ✓ (real), so k = 1 is in the range. Check: 1 is between 2−√2 ≈ 0.586 and 2+√2 ≈ 3.414 ✓.

✓ Range

2 − √2 ≤ y ≤ 2 + √2

Example 4 Related Curves — y = 1/f(x) and y = |f(x)| ★★★ A* Level

The curve y = f(x) where f(x) = (x−1)/(x+2) has a vertical asymptote at x=−2, horizontal asymptote y=1, x-intercept at x=1 and y-intercept at y=−1/2. Describe how to sketch y = 1/f(x) and y = |f(x)|, stating their key features.

1

y = 1/f(x) = (x+2)/(x−1) — key features

New vert. asymptotex = 1 (was x-intercept of f)

New x-interceptx = −2 (was vertical asymptote of f)

New horiz. asymptotey = 1/1 = 1 (was y = 1 → reciprocal = 1)

New y-interceptx=0: y = 2/(−1) = −2 (was −1/2 → reciprocal = −2)

2

y = |f(x)| = |(x−1)/(x+2)|

For x > 1:f(x) > 0, so |f(x)| = f(x) (unchanged)

For −2 < x < 1:f(x) < 0, so |f(x)| = −f(x) (reflect in x-axis)

For x < −2:f(x) > 0, so |f(x)| = f(x) (unchanged)

The x-intercept at x=1 becomes a "touch" point. The vertical asymptote at x=−2 remains. The graph never goes below the x-axis.

Interactive Rational Function Plotter

Enter numerator and denominator coefficients. Switch between y=f(x), y=1/f(x), y=|f(x)| and y=f(|x|). Asymptotes are drawn automatically.

📈 y = (ax² + bx + c) / (dx² + ex + f)

Numerator: ax² + bx + c

a

b

c

Denominator: dx² + ex + f

d

e

f

Practice Questions

Question 1 — Asymptotes and Intercepts

For the function y = (3x − 2)/(x + 4), find: (a) the equations of all asymptotes, (b) the coordinates of the intercepts with the axes, (c) the range of the function. Sketch the curve.

Vertical asymptote: x+4=0. Horizontal: degrees equal, so y = 3/1. Intercepts: set x=0 and then set numerator=0. Range: set y=k, rearrange to get x in terms of k — find values of k with no solution.

✓ Solution

(a) Vert.x = −4 Horiz: y = 3

(b) y-intx=0: y = −2/4 = −1/2 → (0, −½)

x-int3x−2=0 → x = 2/3 → (2/3, 0)

(c) Set y=k:k(x+4)=3x−2 → x(k−3)=−2−4k → x=(−2−4k)/(k−3)

Range:y ∈ ℝ, y ≠ 3 (no solution when k=3)

Question 2 — Oblique Asymptote

Find the asymptotes of y = (x² + 3x − 4)/(x + 2) and sketch the curve, showing all significant features including turning points.

First check: does numerator factorise? x²+3x−4 = (x+4)(x−1). Substitute x=−2: denominator 0, numerator = (−2+4)(−2−1) = 2×(−3) ≠ 0, so x=−2 is a genuine VA. Do long division to find oblique asymptote. Differentiate to find TPs.

✓ Solution

Numerator doesn't cancel with denominator. Long division:

x² + 3x − 4 = (x+2)(x+1) − 6

y = (x+1) − 6/(x+2)

Verticalx = −2

Obliquey = x + 1

y-intx=0: y = −4/2 = −2

x-int(x+4)(x−1)=0: x=−4 and x=1

dy/dx = [(x+2)(2x+3) − (x²+3x−4)(1)]/(x+2)² = (x²+4x+10)/(x+2)². Discriminant of numerator = 16−40 = −24 < 0, so dy/dx > 0 always — no turning points.

Question 3 — Range by Discriminant

Find the set of values taken by the function y = (x² − x + 1)/(x² + x + 1) for real x. Hence state the range and find the coordinates of the turning points.

Set y=k and cross-multiply. Rearrange to quadratic in x: (k−1)x² + (k+1)x + (k−1) = 0. For real x, discriminant ≥ 0. Note: if k=1, the equation becomes linear — check separately. The discriminant gives a quadratic inequality in k.

✓ Solution

k(x²+x+1) = x²−x+1 → (k−1)x² + (k+1)x + (k−1) = 0

k=1:2x = 0 → x=0. Check: y(0) = 1 ✓. So k=1 is in range.

Δ ≥ 0:(k+1)² − 4(k−1)² ≥ 0

k²+2k+1 − 4k²+8k−4 ≥ 0

−3k²+10k−3 ≥ 0 → 3k²−10k+3 ≤ 0

(3k−1)(k−3) ≤ 0 → 1/3 ≤ y ≤ 3

Turning ptsy=1/3: x²+x+1=3(x²−x+1) → 2x²−4x+2=0 → x=1. So (1, 1/3)

y=3: x²+x+1=1/3(x²−x+1) → 2x²+4x+2=0 → x=−1. So (−1, 3)

Question 4 — Related Curves (A* Level)

The diagram (sketch) shows the curve y = f(x) where f(x) = (x−2)/(x²−1). It has vertical asymptotes at x=±1, a horizontal asymptote y=0, and passes through (0, 2) and (2, 0). On separate diagrams, sketch: (a) y = 1/f(x), (b) y = |f(x)|, (c) y = f(|x|), stating the key features of each.

(a) 1/f(x) = (x²−1)/(x−2): VA moves to x=2, zeros at x=±1, oblique asymptote (do long division). (b) |f(x)|: reflect any negative portion in the x-axis. (c) f(|x|): keep the right half (x≥0) and reflect in y-axis for x<0 — check whether x=0 to ∞ portion is what we keep.

✓ Solution

(a) y = 1/f(x) = (x²−1)/(x−2):

New VAx = 2 (was x-intercept of f)

New x-intx = ±1 (were VAs of f)

Long div:(x²−1)÷(x−2) = x+2 + 3/(x−2) → oblique asymptote y = x+2

New y-intx=0: y = −1/(−2) = 1/2

(b) y = |f(x)|: f(x) < 0 when x is between the roots in a way to make the fraction negative. Note f(x) = (x−2)/((x−1)(x+1)). For −1 < x < 1: denominator is (−)(−)= + if x > 0, wait — analyse carefully. The negative portions of f are reflected to give a graph always ≥ 0.

(c) y = f(|x|) = (|x|−2)/(|x|²−1) = (|x|−2)/(x²−1): Keep the graph of f for x ≥ 0, then reflect this in the y-axis. VAs remain at x = ±1. The result is an even function symmetric about the y-axis.

Formula Reference Sheet

Complete reference for Rational Functions and Graphs — Cambridge 9231 P1, Section 1.2.

Asymptote Decision Table

deg(P) < deg(Q)H.A.: y = 0

deg(P) = deg(Q)H.A.: y = lead(P)/lead(Q)

deg(P) = deg(Q)+1Oblique: use long div.

deg(P) > deg(Q)+1No H.A., no O.A.

Q(a) = 0, P(a) ≠ 0V.A.: x = a

Sketching Checklist

1. Domain (V.A.s)

2. Asymptotes (all types)

3. Intercepts (x=0 and y=0)

4. Turning points (dy/dx=0)

5. Behaviour near V.A.s

6. Sketch through features

Range — Discriminant Method

Set y = k

Rearrange → quadratic in x

Δ = B² − 4AC ≥ 0

Solve inequality in k

Check linear case separately

y² = f(x)

Only where f(x) ≥ 0

Two branches: ±√f(x)

Zeros of f → x-axis touch

Symmetric about x-axis

y = 1/f(x)

Zeros of f → new V.A.s

V.A.s of f → new zeros

Max of f → min of 1/f

H.A. y=L → H.A. y=1/L

y = |f(x)| and y = f(|x|)

|f(x)|: reflect −ve in x-axis

Always y ≥ 0

f(|x|): keep x≥0, reflect

Result: even function

📋 Cambridge Exam Strategy — Rational Functions

Always factorise first — if P and Q share a common factor, it's a hole (removable discontinuity), not an asymptote.
For oblique asymptotes, show your long division working explicitly — Cambridge awards method marks for this.
When sketching, draw asymptotes as dashed lines first, then build the curve around them. Never let the curve cross a vertical asymptote.
For the discriminant method: when you get the quadratic inequality in k, factorise it and read off the range from the parabola's sign.
For transformation sketches: state the key features of the new curve — asymptotes, intercepts — don't just draw a vague shape.
Check that your sketch has the correct number of branches — a function with two vertical asymptotes typically has three separate branches.

Rational Functions& Graphs

1. Rational Functions — Definition and Scope

2. Asymptotes — Three Types

Vertical Asymptote

Horizontal Asymptote

Oblique Asymptote

Finding the Oblique Asymptote — Long Division

3. Sketching — The 6-Step Method

4. Range — Discriminant Method

5. Related Curves — Five Transformations

Interactive Rational Function Plotter

Practice Questions

Formula Reference Sheet

Rational Functions
& Graphs