POP

E2.11 Sketching Curves: Linear, Quadratic, Cubic, Reciprocal & Exponential

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Sketching is different from plotting: instead of a precise table of coordinates, a sketch must show the correct overall shape, the key features (roots, intercepts, turning points, symmetry), and what happens as $x$ becomes very large or very small — without necessarily being to scale.

How do I know what overall shape a curve should have?

Check the highest power of $x$ in the function.

Function	Shape	Key Features
$y=mx+c$	Straight line	Gradient $m$ , $y$ -intercept $c$
$y=ax^2+bx+c$	Parabola	Roots, line of symmetry, one turning point
$y=ax^3+bx^2+cx+d$	S-shaped curve	Up to 3 roots, up to 2 turning points
$y=\dfrac{a}{x}$	Two-branch curve	Asymptotic to both axes, symmetric about the origin
$y=ab^x$	Growth/decay curve	$y$ -intercept at $(0,a)$ , asymptotic to $y=0$

Exam Tip

Drawing a single symmetric U-shape for a cubic function is one of the most common sketching errors — always check the highest power first: power $2$ gives a parabola, power $3$ gives an S-shape with up to two turning points.

How do I find where a curve crosses the axes?

Set $y=0$ to find the $x$ -intercepts (easiest when the function is already factorised — the roots can be read off directly), and set $x=0$ to find the single $y$ -intercept.

For $y=(x+1)(x-2)(x-4)$ : the $x$ -intercepts are $x=-1,\ 2,\ 4$ directly from the factors, and the $y$ -intercept is $y=(1)(-2)(-4)=8$ .

How do I work out what happens to a curve as $x$ becomes very large (positive or negative)?

Look only at the term with the highest power — for large $|x|$ , it dominates every other term in the expression. For $y=(x+1)(x-2)(x-4)$ , expanding shows the leading term is $+x^3$ , so as $x\to+\infty$ , $y\to+\infty$ , and as $x\to-\infty$ , $y\to-\infty$ .

How do I sketch a reciprocal or exponential curve?

For $y=\dfrac{a}{x}$ : the curve never touches either axis (both are asymptotes), and it sits in opposite quadrants depending on the sign of $a$ . For $y=ab^x$ : the curve crosses the $y$ -axis at $(0,a)$ and flattens towards — but never reaches — the line $y=0$ as $x$ moves in one direction.

Worked Example

Sketch the curve $y=(x+1)(x-2)(x-4)$ , showing the coordinates of any points where the curve crosses the axes.

$x$ -intercepts, straight from the factors: $x=-1,\ x=2,\ x=4$ .
$y$ -intercept, substituting $x=0$ : $y=(1)(-2)(-4)=8$ .
End behaviour: the leading term is $+x^3$ , so the curve rises on the right and falls on the left.
Sketch: an S-shaped curve rising from the bottom-left, crossing at $x=-1$ , curving up to a local maximum, dipping down through $x=2$ , curving to a local minimum, then rising back up through $x=4$ and continuing upward — passing through $(0,8)$ on the way.