E2.11 Sketching curves

Learning objectives

E2.11.1Recognise, sketch and interpret graphs of the following functions: (a)• Recognise, sketch and interpret graphs of the following functions: (a) linear (b) quadratic (c) cubic (d) reciprocal (e) exponential. Functions will be equivalent to: • ax + by = c • y = ax2 + bx + c • y = ax3 + b • y = ax3 + bx2 + cx • y x a b= + • y = arx + b where a, b and c are rational numbers and r is a rational, positive number. Knowledge of turning points, roots and symmetry is required. Knowledge of vertical and horizontal asymptotes is required. Finding turning points of quadratics by completing the square is required.

A sketch records a function’s essential structure rather than plotting a dense table: family shape, intercepts, turning points, symmetry, asymptotes and end behaviour must agree with its equation.

Family and syllabus form	Essential sketch features
linear: $ax+by=c$	straight line; $x$ - and $y$ -intercepts
quadratic: $y=ax^2+bx+c$	parabola; axis of symmetry; one maximum or minimum; roots
cubic: $y=ax^3+b$ or $y=ax^3+bx^2+cx$	cubic end direction; roots; up to two turning points
reciprocal: $y=a/x+b$	two branches; vertical asymptote $x=0$ ; horizontal asymptote $y=b$
exponential: $y=ar^x+b$	$y$ -intercept $a+b$ ; horizontal asymptote $y=b$ ; growth or decay direction

Identify the family and leading sign; find exact intercepts where possible; find required symmetry, turning points and asymptotes; place and label these features; draw a smooth shape with correct end behaviour that approaches but does not cross a vertical asymptote.

$y=x^2+10x+14=(x+5)^2-11$

The completed-square form gives the minimum $(-5,-11)$ and axis of symmetry $x=-5$ . The positive squared coefficient makes the parabola open upwards; roots, if required, are symmetric about $x=-5$ .

Factorisation reveals intercept behaviour. In $y=(x+1)(x-3)^2$ , the graph crosses at $x=-1$ but touches and turns at the repeated root $x=3$ ; $y=9$ when $x=0$ .

For $y=2/x-1$ , the graph is undefined at $x=0$ , so $x=0$ is vertical; as $|x|$ grows, $2/x$ approaches $0$ , so $y=-1$ is horizontal.

Do not use differentiation to find cubic turning points in this objective. A sketch is still constrained: labelled roots, repeated-root behaviour, symmetry, turning points and asymptotes must match the equation.

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