Cubic, Reciprocal & Exponential Graphs

SPEC A12 • ALGEBRA — GRAPH FAMILIES

Three new graph families — each with a shape you can recognise instantly once you know what to look for.

Cubic Graphs: y = x³

A cubic has an x³ term as its highest power. The simplest cubic is y = x³.

Key features

Passes through the origin (0, 0).

Comes from the bottom-left and heads to the top-right (for y = +x³).

Has a point of inflection at the origin — the curve flattens momentarily then carries on.

No turning points — no maximum or minimum.

Cubics With Other Terms

Adding an x or x² term can give the cubic a local maximum and minimum — a "hump" shape. Example: y = x³ − 3x.

Shape summary

y = ax³ (pure cubic): S-shape, no turning points, through origin.

y = ax³ + bx + c (general): can have a local max and min — the classic "hump and dip" cubic shape.

If a > 0: bottom-left to top-right. If a < 0: top-left to bottom-right (flipped).

Reciprocal Graphs: y = k/x

A reciprocal has x in the denominator. The graph has two separate branches that never touch — and never touch the axes.

Key features

Undefined at x = 0 — the curve never exists there. x = 0 is a vertical asymptote.

Approaches but never touches the x-axis. y = 0 is a horizontal asymptote.

Two branches in opposite quadrants (1st and 3rd for y = 1/x).

Reciprocal: Sign of k Changes the Quadrants

k > 0: branches in 1st and 3rd quadrants

Examples: y = 1/x, y = 2/x, y = 5/x. Top-right and bottom-left branches.

k < 0: branches in 2nd and 4th quadrants

Examples: y = −1/x, y = −3/x. Top-left and bottom-right branches.

Asymptote

A line the curve gets arbitrarily close to but never touches. For y = k/x, the two asymptotes are the x-axis and the y-axis. The curve shoots off to ±∞ near x = 0, and flattens toward y = 0 as |x| gets large.

Why two branches?

You can't divide by zero, so y is undefined at x = 0. That creates a gap the curve has to jump over — producing two disconnected branches on either side of the y-axis.

Exponential Graphs: y = kˣ HIGHER

An exponential has x in the exponent. Here k is a positive constant (k > 0, k ≠ 1).

k > 1: growth

Curve rises steeply to the right. Approaches y = 0 to the left (asymptote).

0 < k < 1: decay

Curve falls toward 0 as x increases. Approaches y = 0 to the right (asymptote).

Exponential Key Features HIGHER

Every exponential y = kˣ

Passes through (0, 1) — because k⁰ = 1 for any k > 0.

Always positive — the curve sits entirely above the x-axis.

Never crosses or touches the x-axis — y = 0 is a horizontal asymptote.

No maximum or minimum — the curve is always increasing (if k > 1) or always decreasing (if 0 < k < 1).

⚠️ k⁰ = 1, not 0. Any positive number raised to the power zero equals 1. That's why every exponential curve crosses the y-axis at exactly y = 1 — regardless of k.

Quick check

If you sketch an exponential and it doesn't pass through (0, 1), something's wrong. That's the universal anchor point for every y = kˣ curve.

Worked Example — Sketch y = x³01

Sketch the graph of y = x³ for −2 ≤ x ≤ 2, showing at least 5 points. [3 marks]

M

Method: build a table of values by substituting x-values; plot; join with a smooth curve.

C

Table:

x = −2 → y = −8. x = −1 → y = −1. x = 0 → y = 0. x = 1 → y = 1. x = 2 → y = 8.

Plot (−2, −8), (−1, −1), (0, 0), (1, 1), (2, 8). Draw a smooth S-shape through the points — no turning points, flattens briefly at the origin.

E

Evaluate: correct S-shape, through origin, with points plotted. B1 table, B1 correct points plotted, B1 smooth curve. Do not join with straight lines — cubic curves are smoothly curved.

THE SKETCH

Worked Example — Sketch y = 4/x02

Sketch the graph of y = 4/x. [3 marks]

M

Method: recognise as reciprocal with k = 4 > 0 (so 1st and 3rd quadrants). Plot a few points on each branch, noting the asymptotes x = 0 and y = 0.

C

Table (x > 0 branch): x = 1 → y = 4; x = 2 → y = 2; x = 4 → y = 1; x = 8 → y = 0.5.

Table (x < 0 branch): x = −1 → y = −4; x = −2 → y = −2; x = −4 → y = −1.

Sketch: two branches, curves never touching axes. Leave a gap at x = 0 — asymptote.

E

Evaluate: B1 correct shape in 1st and 3rd quadrants. B1 both branches plotted. B1 asymptotic behaviour (no crossing axes). Always leave a visible gap at x = 0 to signal the asymptote.

THE SKETCH

Worked Example — Sketch y = 2ˣ

HIGHER

Sketch the graph of y = 2ˣ for −3 ≤ x ≤ 3. [3 marks]

M

Method: k = 2 > 1, so exponential growth. Plot points including the anchor (0, 1). Note horizontal asymptote at y = 0 on the left.

C

Table:

x = −3 → y = 1/8 = 0.125; x = −2 → y = 0.25; x = −1 → y = 0.5; x = 0 → y = 1; x = 1 → y = 2; x = 2 → y = 4; x = 3 → y = 8.

Plot points. Draw a smooth curve — never crossing the x-axis, passing through (0, 1), rising steeply to the right.

E

Evaluate: curve must (i) pass through (0, 1), (ii) stay above x-axis, (iii) be smoothly increasing. B1 correct shape, B1 through (0, 1), B1 asymptote at y = 0 on the left.

THE SKETCH

Recognising Each Graph Family

Given a sketch with no equation, use the shape alone to identify the family.

FamilyEquationShape ClueTier

Lineary = mx + cStraight lineBasic F+H

Quadraticy = ax² + bx + cParabola (U or ∩)Basic F+H

Cubicy = ax³ + bx + cS-shape; may have hump and dipF+H

Reciprocaly = k/xTwo separate branches, asymptotes on both axesF+H

Exponential Hy = kˣSingle curve through (0, 1), always positive, horizontal asymptoteHigher

Fast identification

Three tell-tale questions: Does it cross zero through origin/point of inflection? → cubic. Does it have two disconnected branches? → reciprocal. Does it pass through (0, 1) and stay positive? → exponential.

Examiner Alert: Reciprocal Asymptotes

⚠️ EXAMINER ALERT: A real AQA Jun23 Paper 1 question (Q9) asked students to criticise a sketch of y = 1/x. The mark scheme accepted things like "the graph touches the y-axis" or "one branch should be moved to the right" — but rejected vague statements like "the graph is wrong" or "it should be symmetrical".

Valid criticism (B1)

"The graph touches the y-axis" (should have vertical asymptote)

"The graph should not cross the x-axis"

"One branch is in the wrong quadrant"

Not accepted (B0)

"The graph is wrong"

"It doesn't look symmetrical"

"There are no numbers on the axes"

Be specific

When criticising a graph, name precisely what is wrong — which feature, which axis, which quadrant. Generic words like "wrong" or "symmetrical" score zero marks.

Examiner Alert: k⁰ = 1 HIGHER

⚠️ EXAMINER ALERT: Students often think 2⁰ = 0, or that y = kˣ passes through the origin. Both are wrong. Any positive k raised to the power 0 equals 1. So every exponential y = kˣ passes through (0, 1) — not (0, 0).

ExpressionValueWhat students often write (wrong)

2⁰12 × 0 = 0 (wrong — it's not multiplication)

5⁰10 or 5 (wrong)

(1/3)⁰11/3 or 0

kˣ at x = 01 for any k > 0"It depends on k" (no — always 1)

Quick mnemonic

"Anything to the zero is one." Exponentials therefore always cross the y-axis at y = 1, regardless of the value of k.

The MCE Framework for Graph Sketching

Whether you're asked to sketch a cubic, reciprocal, or exponential, the three-step pattern is the same.

MMethod: Identify the family (cubic, reciprocal, exponential). Note key features — intercepts, asymptotes, direction.

CCalculate: Build a small table of values. Plot the points. Mark asymptotes where relevant.

EEvaluate: Draw a smooth curve. Check: does shape match the family? Does the curve cross the right axes? Does it pass through the anchor point?

Three anchor points to remember

Cubic y = x³: through (0, 0), point of inflection there.

Reciprocal y = k/x: never touches axes — leave visible gaps.

Exponential y = kˣ H: through (0, 1), asymptote at y = 0.

Practice Question 1

Here are three graphs labelled A, B and C.

Match each graph to its equation. [3 marks]

(i) y = 1/x (ii) y = x³ (iii) y = x² − 3

M

Method: match each graph to its family based on shape — cubic (S-shape through origin), reciprocal (two branches, asymptotes), quadratic (U-shape).

C

Graph A: S-shape through origin, no turning points → cubic → (ii) y = x³.

Graph B: two separate branches in opposite quadrants, never touching axes → reciprocal → (i) y = 1/x.

Graph C: U-shaped parabola crossing x-axis twice → quadratic → (iii) y = x² − 3.

E

Evaluate: A→(ii), B→(i), C→(iii). B1 each correct match. Shape recognition alone is enough — no plotting needed.

Practice Question 2

Amara sketches the graph of y = 1/x like this:

Make two specific criticisms of Amara's sketch. [2 marks]

M

Method: compare Amara's sketch to the correct features of y = 1/x. Identify each specific error with precise language.

C

Correct features of y = 1/x: (i) undefined at x = 0, so there's a vertical asymptote; (ii) never crosses or touches either axis; (iii) branches lie in 1st and 3rd quadrants.

Amara's errors:

Her curves pass through the origin (0, 0) — but y = 1/x is undefined at x = 0.

Her curves touch both axes — but the x-axis and y-axis are asymptotes; the curves should approach but never meet them.

E

Evaluate: any two of the above score B1 + B1 = B2. Generic criticisms like "the graph is wrong" or "it doesn't look right" score B0 — you must be specific about which feature is wrong.

Practice Question 3 HIGHER

The graph of y = kˣ passes through the point (4, 81).

k is a positive constant.

(a) Write down the y-intercept of the graph. [1 mark]

(b) Work out the value of k. [2 marks]

M

Method: (a) every exponential passes through (0, 1). (b) substitute the given point into y = kˣ to form an equation, then solve for k.

C

(a) At x = 0: y = k⁰ = 1. So y-intercept is (0, 1).

(b) Substitute (4, 81) into y = kˣ: 81 = k⁴.

Recognise that 81 = 3⁴ (since 3 × 3 × 3 × 3 = 81). So k⁴ = 3⁴ → k = 3.

E

Evaluate: (a) B1. (b) M1 form k⁴ = 81, A1 k = 3. Sanity check — at x = 4, 3⁴ = 81 ✓. Since k > 1, the curve is growing, which is consistent with the point (4, 81) being far above the y-intercept.

Summary

Three new graph families

Cubic y = ax³ : S-shape through origin, no turning points. With other terms (y = x³ − 3x), may have local max and min.

Reciprocal y = k/x : two separate branches, undefined at x = 0. Axes are asymptotes (curve never touches them). k > 0 → 1st and 3rd quadrants; k < 0 → 2nd and 4th.

Exponential y = kˣ H : always positive, passes through (0, 1), horizontal asymptote at y = 0. k > 1 → growth; 0 < k < 1 → decay.

Anchor points

Cubic: (0, 0).

Exponential: (0, 1).

Reciprocal: no anchor — avoid both axes.

Common errors to avoid

Reciprocal touching axes.

Exponential through (0, 0) instead of (0, 1).

Joining cubic points with straight lines.

Vague criticism ("it's wrong") — be specific.

Exam Tip

Before sketching, say the family name out loud: "This is a reciprocal graph with k > 0 — so two branches in the 1st and 3rd quadrants, with asymptotes on both axes." The rest follows automatically.