Ad Banner Placeholder

Cambridge IGCSE Mathematics — 0580 Extended

Topic 2.11: Algebra and graphs — Sketching Curves

This topic is recognition and sketching, not accurate plots (that is E2.10). A sketch needs the right shape, labelled intercepts, turning points and asymptotes — not a table of values.

Forms you must sketch: \(ax + by = c\) (linear); \(y = ax^2 + bx + c\) (quadratic); \(y = ax^3 + b\) and \(y = ax^3 + bx^2 + cx\) (cubic); \(y = \dfrac{a}{x} + b\) (reciprocal); \(y = ar^x + b\) with \(r > 0\) rational (exponential).

Linear graphs \(ax + by = c\)

A linear graph is a straight line. The fastest sketch uses intercepts.

Method

  1. Set \(y = 0\) to find the \(x\)-intercept.
  2. Set \(x = 0\) to find the \(y\)-intercept.
  3. Plot those two points and join them with a rulered line.

Sketch \(2x + 3y = 12\).

Straight line through x-intercept 6, 0 and y-intercept 0, 4
\(y = 0 \Rightarrow 2x = 12 \Rightarrow x = 6\). \(x = 0 \Rightarrow 3y = 12 \Rightarrow y = 4\).

Quadratic sketches

For \(y = ax^2 + bx + c\) you need: the roots (if they exist), the turning point, and whether it is a minimum (\(a > 0\)) or maximum (\(a < 0\)). The line of symmetry is \(x = -\dfrac{b}{2a}\). Completing the square gives the turning point without a formula sheet.

Method

  1. Factor (or use the quadratic formula) to find roots, if they are required.
  2. Complete the square to write \(y = a(x-h)^2 + k\). The turning point is \((h, k)\).
  3. Sketch: roots on the \(x\)-axis, turning point marked, and the correct way up.

Sketch \(y = x^2 - 4x + 3\).

\(y = (x-1)(x-3) = (x-2)^2 - 4 + 3 = (x-2)^2 - 1\). Roots \(1\) and \(3\); vertex \((2, -1)\); minimum because the coefficient of \(x^2\) is positive.

Parabola with roots at 1 and 3 and minimum turning point at 2, negative 1
Minimum at \((2, -1)\); line of symmetry \(x = 2\).

Find the turning point of \(y = x^2 + 6x + 5\) by completing the square.

Completing the square: y equals (x plus 3) squared minus 4, minimum at negative 3, negative 4
Half of \(6\) is \(3\); \(3^2 = 9\). Then \(y = (x+3)^2 - 4\).

Paper 2 (non-calculator)

Complete the square mentally for \(x^2 + 6x\): half of \(6\) is \(3\), and \(3^2 = 9\), so \(x^2 + 6x = (x+3)^2 - 9\). Keep the vertex as exact coordinates, not a decimal estimate.

Cubic sketches

\(y = ax^3 + b\) is a cube graph translated up or down by \(b\). \(y = ax^3 + bx^2 + cx\) can have up to two turning points. Factorising shows the roots; a sketch only needs those roots and the right end-behaviour (if \(a > 0\), the graph goes down to the left and up to the right).

Compare \(y = x^3\) with \(y = x(x-1)(x+1) = x^3 - x\).

y equals x cubed through the origin beside y equals x cubed minus x with roots at negative 1, 0 and 1
\(y = x^3 - x\) crosses at \(-1\), \(0\) and \(1\). \(y = x^3\) has a single root at \(0\).

Reciprocal graphs

For \(y = \dfrac{a}{x} + b\): vertical asymptote \(x = 0\), horizontal asymptote \(y = b\). The two branches never cross the asymptotes.

Sketch \(y = \dfrac{1}{x} + 2\).

Reciprocal graph translated up by 2 with vertical asymptote x equals 0 and horizontal asymptote y equals 2
This is \(y = 1/x\) shifted up by \(2\). It is not \(y = 1/(x+2)\).

Exponential sketches

\(y = ar^x + b\) always has a horizontal asymptote \(y = b\). When \(b = 0\) and \(a = 1\), \(y = 2^x\) passes through \((0, 1)\) and approaches \(y = 0\) as \(x \to -\infty\). Adding \(1\) lifts every point by \(1\): \(y = 2^x + 1\) passes through \((0, 2)\) with asymptote \(y = 1\).

Sketch \(y = 2^x\) and \(y = 2^x + 1\) on the same axes.

Exponential graphs y equals 2 to the x through (0, 1) and y equals 2 to the x plus 1 through (0, 2)
Translation up by \(1\) moves the asymptote from \(y = 0\) to \(y = 1\).

Try this

By completing the square, find the vertex of \(y = x^2 - 2x - 3\).

Show answer
Answer

  1. Half of \(-2\) is \(-1\); \((-1)^2 = 1\)

    \[ x^2 - 2x = (x-1)^2 - 1 \]

  2. Replace the constant

    \[ y = (x-1)^2 - 1 - 3 = (x-1)^2 - 4 \]

  3. Vertex (a minimum)

    \[ (1,\ -4) \]

Exam Traps

  • \(y = \dfrac{1}{x} + 2\) is not the same as \(y = \dfrac{1}{x+2}\). The first has asymptotes \(x = 0\) and \(y = 2\); the second has \(x = -2\) and \(y = 0\).

0/10

Ad Banner Placeholder