Cubic graphs and graphs of powers of x

AS · P1 · 11 min

Paper 1 assumes you know the shapes of the graphs $y = kx^n$ and can sketch a cubic from its factors. These sketches are rarely a whole question on their own, but they are used constantly: to count the solutions of an equation, to see which region an area integral covers, to check the nature of stationary points, and as the "given graph" in transformation questions. This note collects the shapes you need and the method for sketching any factorised cubic.

Graphs of powers of x

The syllabus lists, as assumed knowledge, the shapes of $y = kx^n$ where $n$ is a positive or negative integer, or $\pm\tfrac{1}{2}$ . They fall into a few families.

Key result

Power	Shape for $k > 0$	Examples
$n$ even and positive	$\cup$ -like, through the origin, symmetric in the $y$ -axis	$x^2$ , $x^4$
$n$ odd and positive	rises from bottom left to top right through the origin, with rotational symmetry about $O$	$x$ , $x^3$ , $x^5$
$n$ odd and negative	two branches in quadrants 1 and 3; asymptotes $x = 0$ and $y = 0$	$\dfrac{1}{x}$ , $\dfrac{1}{x^3}$
$n$ even and negative	two branches in quadrants 1 and 2, both above the axis; asymptotes $x = 0$ and $y = 0$	$\dfrac{1}{x^2}$
$n = \tfrac{1}{2}$	half a sideways parabola, from the origin, for $x \ge 0$	$\sqrt{x}$
$n = -\tfrac{1}{2}$	for $x > 0$ only, falling towards the $x$ -axis; asymptotes $x = 0$ and $y = 0$	$\dfrac{1}{\sqrt{x}}$

If $k < 0$ , reflect the graph in the $x$ -axis.

y = x^2 y = x^3 y = 1/x y = 1/x^2

All four of these pass through $(1, 1)$ . For $0 < x < 1$ higher powers are smaller ( $x^3 < x^2$ ), and for $x > 1$ they are larger. This is a useful check when sketching two powers on the same axes.

y = sqrt(x) y = 1/sqrt(x)

$y = \sqrt{x}$ and $y = \dfrac{1}{\sqrt{x}}$ exist only for $x \ge 0$ and $x > 0$ respectively, and meet at $(1, 1)$ .

The shape of a cubic

A cubic is $y = ax^3 + bx^2 + cx + d$ with $a \neq 0$ . For large $|x|$ the $ax^3$ term dominates, so the ends of the graph behave like $y = ax^3$ :

$a > 0$ : the curve comes up from the bottom left and goes off to the top right;
$a < 0$ : it comes down from the top left and goes off to the bottom right.

In between, a cubic has either two turning points (a "hump and a dip") or none. It meets the $x$ -axis at least once and at most three times.

What the factors tell you

If the cubic is factorised, each factor gives a root, and the power of the factor tells you how the curve behaves there:

Key result

A single factor $(x - \alpha)$ : the curve crosses the $x$ -axis at $x = \alpha$ .
A squared factor $(x - \alpha)^2$ : the curve touches the $x$ -axis at $x = \alpha$ and turns back (a turning point on the axis).
A cubed factor $(x - \alpha)^3$ : the curve flattens and crosses at $x = \alpha$ , like $y = x^3$ at the origin.

Sketching a factorised cubic

Find the sign of the $x^3$ coefficient (multiply the leading coefficients of the factors) to fix the ends.
Mark the roots from the factors, noting which are single (cross) and which are squared (touch).
Find the $y$ -intercept by putting $x = 0$ .
Starting from the correct end, draw a smooth curve through the roots, crossing or touching as required.
Label all intercepts with coordinates.

You do not need the exact positions of the turning points unless the question asks. Finding them uses calculus: see Stationary points.

Three distinct roots

Sketch $y = (x + 2)(x - 1)(x - 3)$ .

Solution

The $x^3$ coefficient is $1 > 0$ : bottom left to top right.
Roots $x = -2, 1, 3$ , all single, so the curve crosses at each.
$y$ -intercept: $(2)(-1)(-3) = 6$ , so $(0, 6)$ .

y = (x + 2)(x - 1)(x - 3) (-2, 0) (1, 0) (3, 0) (0, 6)

Coming up from the bottom left, the curve crosses at $-2$ , rises through $(0, 6)$ , turns, crosses at $1$ , dips, and crosses back up at $3$ .

A repeated root and a negative leading coefficient

Sketch $y = -x(x - 2)^2$ .

Solution

The $x^3$ coefficient is $-1 < 0$ : top left to bottom right.
Single root at $x = 0$ (crosses); repeated root at $x = 2$ (touches).
$y$ -intercept: $0$ .

y = -x(x - 2)^2 (0, 0) (2, 0)

From the top left the curve comes down and crosses at the origin, dips below the axis, rises to touch the axis at $(2, 0)$ , and turns back down.

Factorise first

Sketch $y = x^3 - 4x$ .

Solution

Factorise: $x^3 - 4x = x\left(x^2 - 4\right) = x(x - 2)(x + 2)$ .

$a = 1 > 0$ .
Roots $-2, 0, 2$ , all crossings. The $y$ -intercept is the root at the origin.

y = x^3 - 4x (-2, 0) (0, 0) (2, 0)

The curve has rotational symmetry about the origin, because $x^3 - 4x$ contains only odd powers.

Finding the equation of a cubic from its graph

Write down the factors from the roots, include an unknown constant $a$ , and use one more point.

Touch and cross

A cubic curve touches the $x$ -axis at $x = -1$ , crosses it at $x = 2$ , and passes through $(0, 4)$ . Find its equation.

Solution

Touching at $-1$ means a factor $(x + 1)^2$ ; crossing at $2$ means a factor $(x - 2)$ :

y = a(x + 1)^2(x - 2)

At $(0, 4)$ : $4 = a(1)(-2)$ , so $a = -2$ .

y = -2(x + 1)^2(x - 2)

Transformed cubics

Transformations apply exactly as for any graph (see Transformations of graphs).

A translated cubic

Describe the transformation that maps $y = x^3$ onto $y = (x - 2)^3 + 1$ , and find where the new curve meets the axes.

Solution

Translation by $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ . The point of symmetry moves from $(0, 0)$ to $(2, 1)$ .

$y$ -intercept: $(-2)^3 + 1 = -7$ , so $(0, -7)$ .

$x$ -intercept: $(x - 2)^3 = -1$ , so $x - 2 = -1$ and $x = 1$ , giving $(1, 0)$ . (A cube root has only one real value, so there is only one intercept.)

y = x^3 y = (x - 2)^3 + 1 (1, 0) (0, -7) (2, 1)

Using sketches to count solutions

The number of real solutions of $f(x) = g(x)$ is the number of points where $y = f(x)$ and $y = g(x)$ meet. A sketch tells you how many to look for; algebra then finds them.

A cubic and a line

On the same diagram sketch $y = x(x - 3)^2$ and $y = 2x$ . Hence state the number of real solutions of $x(x - 3)^2 = 2x$ , and find them exactly.

Solution

$y = x(x - 3)^2$ : $a > 0$ , crosses at $0$ , touches at $3$ . $y = 2x$ : a line through the origin.

y = x(x - 3)^2 y = 2x

The line meets the curve three times: at the origin, once while the curve comes down towards $(3, 0)$ , and once after it rises again. So there are three solutions.

Algebra: do not divide by $x$ (that would lose $x = 0$ ).

x(x - 3)^2 - 2x = 0 \quad\Rightarrow\quad x\left[(x - 3)^2 - 2\right] = 0

So $x = 0$ or $(x - 3)^2 = 2$ , giving $x = 3 \pm \sqrt{2}$ .

Solutions: $x = 0$ , $x = 3 - \sqrt{2}$ , $x = 3 + \sqrt{2}$ .

Watch out

Wrong ends. The sign of the $x^3$ term decides the ends. In $y = (2 - x)(x + 1)(x - 4)$ the $x^3$ coefficient is $-1$ , because of the $-x$ in the first factor.

Crossing at a repeated root. A squared factor means the curve touches the axis and turns back; drawing it crossing loses the mark.

Turning points drawn at the roots. Between two single roots the turning point is somewhere between them, not at either root.

Dividing by $x$ when counting solutions loses the root $x = 0$ .

Forgetting where power graphs exist. $y = \sqrt{x}$ has no part for $x < 0$ ; $y = \dfrac{1}{x^2}$ is never below the $x$ -axis.

Exam tip

"Sketch" means the correct shape with intercepts labelled. Turning points need only be in sensible positions, unless coordinates are asked for.
Expect to use these shapes inside other questions: "By sketching suitable graphs, show that the equation ... has exactly one root" requires a clear sketch of both graphs and a sentence stating the number of intersections.
When a question gives a graph "with given features" (a few labelled points) and asks for its transformed image, track each labelled point and draw the same shape through the images.

Summary

$y = kx^n$ : even positive $n$ is $\cup$ -like; odd positive $n$ rises through the origin; negative $n$ has asymptotes $x = 0$ and $y = 0$ ; $\sqrt{x}$ exists only for $x \ge 0$ .
A cubic with $a > 0$ goes from bottom left to top right; with $a < 0$ , top left to bottom right.
Single factor: cross. Squared factor: touch. Cubed factor: flatten and cross.
Find the $y$ -intercept by putting $x = 0$ , and label every intercept.
Equation from a graph: build the factors, then use one point to find $a$ .
The number of intersections of two graphs is the number of solutions of the equation they form.

Practice questions

Question

Sketch $y = x(x + 3)(x - 2)$ .
Sketch $y = (x - 1)^2(x + 4)$ .
Sketch $y = (2 - x)(x + 1)(x - 4)$ .
Factorise $x^3 - 6x^2 + 9x$ and hence sketch $y = x^3 - 6x^2 + 9x$ .
The curve $y = ax(x + 3)(x - 5)$ passes through $(1, -8)$ . Find $a$ .
A cubic crosses the $x$ -axis at $-2$ , $1$ and $3$ and the $y$ -axis at $(0, -12)$ . Find its equation.
Describe a sequence of transformations mapping $y = x^3$ onto $y = 2(x + 1)^3 - 3$ , and find the coordinates of the points where the new curve meets the axes (the $x$ -intercept to 3 significant figures).
On the same axes sketch $y = x^2$ and $y = \dfrac{1}{x}$ . Show that they meet at exactly one point and find it.
Find the coordinates of the points where the curve $y = x(x - 4)^2$ meets the line $y = 9x$ .
Show that, for every value of the constant $k$ , the curve $y = (x - k)\left(x^2 - 2x + 5\right)$ meets the $x$ -axis at exactly one point.

Answers

$a > 0$ ; crosses at $-3$ , $0$ , $2$ ; $y$ -intercept $0$ . Bottom left to top right.
$a > 0$ ; crosses at $-4$ , touches at $1$ ; $y$ -intercept $(1)(4) = 4$ , so $(0, 4)$ .
$x^3$ coefficient $-1 < 0$ : top left to bottom right. Crosses at $-1$ , $2$ , $4$ . $y$ -intercept $(2)(1)(-4) = -8$ .
$x\left(x^2 - 6x + 9\right) = x(x - 3)^2$ . $a > 0$ ; crosses at the origin, touches at $(3, 0)$ .
$-8 = a(1)(4)(-4) = -16a$ , so $a = \tfrac{1}{2}$ .
$y = a(x + 2)(x - 1)(x - 3)$ . At $x = 0$ : $a(2)(-1)(-3) = 6a = -12$ , so $a = -2$ . $y = -2(x + 2)(x - 1)(x - 3)$ .
Stretch parallel to the $y$ -axis with scale factor $2$ , then translation by $\begin{pmatrix} -1 \\ -3 \end{pmatrix}$ (the stretch must come before the vertical translation). $y$ -intercept: $2(1) - 3 = -1$ , so $(0, -1)$ . $x$ -intercept: $(x + 1)^3 = 1.5$ , $x = \sqrt[3]{1.5} - 1 = 0.145$ , so $(0.145, 0)$ .
$y = x^2$ is a $\cup$ through the origin; $y = \dfrac{1}{x}$ has branches in quadrants 1 and 3. For $x < 0$ , $x^2 > 0$ but $\dfrac{1}{x} < 0$ , so no intersection there. For $x > 0$ , $x^2$ increases from $0$ and $\dfrac{1}{x}$ decreases, so they cross once. $x^2 = \dfrac{1}{x}$ gives $x^3 = 1$ , $x = 1$ : the point $(1, 1)$ .
$x(x - 4)^2 - 9x = 0$ , so $x\left[(x - 4)^2 - 9\right] = 0$ . $x = 0$ or $x - 4 = \pm 3$ , so $x = 0, 1, 7$ . Points $(0, 0)$ , $(1, 9)$ , $(7, 63)$ .
$y = 0$ when $x = k$ or $x^2 - 2x + 5 = 0$ . The quadratic has discriminant $4 - 20 = -16 < 0$ , so it has no real roots (indeed $x^2 - 2x + 5 = (x - 1)^2 + 4 > 0$ ). So the only root is $x = k$ , and the curve meets the $x$ -axis exactly once, at $(k, 0)$ .

Graphs of powers of x

The shape of a cubic

What the factors tell you

Finding the equation of a cubic from its graph

Transformed cubics

Using sketches to count solutions

Practice questions

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