Sketching cubics and higher polynomials — Mathematics

Before this

This leads to

The factor theorem · coming soon
Stationary points — finding and classifying · coming soon
Curve sketching using calculus · coming soon

A curve with no mirror line

The last lecture ended with a promise: the same features would sketch cubics, which can turn twice. Here is where that promise gets difficult.

Take $y = x^{3} - 3 x^{2} - 6 x + 8$ . The quadratic method starts well. Put $x = 0$ , and you get the $y$ -intercept $(0, 8)$ . Next, the quadratic method looks for the mirror point: where else does the curve reach height 8? Set $y = 8$ :

x^{3} - 3 x^{2} - 6 x x (x^{2} - 3 x - 6) = 0 = 0

This gives $x = 0$ , $x \approx - 1.37$ and $x \approx 4.37$ . The curve reaches height 8 three times, not twice. There is no single partner point.

In fact, no cubic has a mirror line. Far out on the left this curve goes down, and far out on the right it goes up. A mirror cannot swap a falling arm with a rising one. So the best tool of the last lecture, $x = - \frac{b}{2 a}$ , has nothing to say here.

This is the disruption. The expanded form $x^{3} - 3 x^{2} - 6 x + 8$ hides the shape. But the same curve can be written as

y = (x + 2) (x - 1) (x - 4)

In this factorised form, the whole shape is visible, once you know how to read it.

Polynomials, degree and the term that matters most

A polynomial is a sum of terms like $5 x^{3}$ , $- 2 x$ and $7$ , where every power of $x$ is a whole number: 0, 1, 2, 3, and so on. The highest power is the degree. The term with the highest power is the leading term, and its number is the leading coefficient. A degree 2 polynomial is a quadratic, degree 3 is a cubic, and degree 4 is a quartic.

The degree limits how complicated the graph can be. Every root $r$ gives a factor $(x - r)$ , and each factor carries one power of $x$ . So a polynomial of degree $n$ has room for at most $n$ such factors, and at most $n$ roots. The factor theorem, in a later lecture, proves that every root gives a factor.

The last lecture found that the gradient of a quadratic is $2 a x + b$ . That has degree 1, so it is zero once, and the parabola turns once. In the same way, the gradient of a degree $n$ polynomial has degree $n - 1$ , so the graph has at most $n - 1$ turning points. The lecture on stationary points proves this.

Degree	Name	Roots, at most	Turning points, at most
2	quadratic	2	1
3	cubic	3	2
4	quartic	4	3
$n$	degree $n$	$n$	$n - 1$

"At most" matters: $y = x^{3}$ has one root and no turning points.

Reading the roots, and the signs between them

A product is zero exactly when one of its factors is zero. So $y = (x + 2) (x - 1) (x - 4)$ is zero at $x = - 2$ , $x = 1$ and $x = 4$ . Those are the roots: the three places where the curve meets the $x$ -axis.

To know which side of the axis the curve is on in between, look at the sign of each factor. The factor $(x - 4)$ is negative when $x$ is less than 4, and positive when $x$ is more than 4. Each factor works the same way: negative to the left of its root, positive to the right. The sign of $y$ is then the sign of the product.

Factor	$x < - 2$	$- 2 < x < 1$	$1 < x < 4$	$x > 4$
$x + 2$	−	+	+	+
$x - 1$	−	−	+	+
$x - 4$	−	−	−	+
$y$	−	+	−	+

Look at what happens at each root. As $x$ moves past $- 2$ , exactly one factor, $(x + 2)$ , changes sign. So $y$ changes sign too. The curve goes from below the axis to above it, which means it must cross. The same happens at 1 and at 4.

y = (x + 2) (x - 1) (x - 4)

. The signs from the table are marked in each region. The curve crosses at every root, because one factor changes sign there. It must turn once between

- 2

and 1, and once between 1 and 4.

The signs also show where the turns must be. Between $- 2$ and 1 the curve is above the axis, but it is at height 0 at both ends. So it must rise and come back down: there is a maximum somewhere in between. In the same way, there is a minimum between 1 and 4. That makes two turning points, the most a cubic can have.

Touch or cross: what a squared factor does

Now take $y = (x - 3)^{2} (x + 1)$ . The root $x = 3$ comes from a squared factor. We call it a repeated root, or a root of multiplicity 2. The root $x = - 1$ has multiplicity 1.

A square is never negative. So as $x$ moves past 3, the factor $(x - 3)^{2}$ is positive on both sides. It does not change sign, so $y$ does not change sign either. The curve comes down to the axis at $x = 3$ and goes back the way it came. It touches the axis there, and does not cross.

There is a second way to see this, and it explains every case at once. Zoom in very close to $x = 3$ . There, the other factor, $(x + 1)$ , is almost exactly $3 + 1 = 4$ . So, close to the root,

y \approx 4 (x - 3)^{2}

That is a narrow U-shape with its vertex at $(3, 0)$ . Now zoom in near $x = - 1$ instead. There, $(x - 3)^{2}$ is almost $(- 4)^{2} = 16$ , so $y \approx 16 (x + 1)$ . That is a steep straight line through $(- 1, 0)$ , and a straight line crosses.

That is it. That is the whole idea. Close to any root, every other factor is almost a fixed number. So the curve looks like that number times $(x - r)$ raised to the multiplicity. A multiplicity of 1 looks like a straight line, so the curve crosses. A multiplicity of 2 looks like a parabola, so it touches. A multiplicity of 3 looks like $y = x^{3}$ : it crosses, but it flattens as it crosses.

What a curve looks like very close to a root of multiplicity 1, 2 and 3. Odd multiplicity: the curve crosses. Even multiplicity: it touches and turns back.

The fixed number also tells you the direction. In $y \approx 4 (x - 3)^{2}$ , the number 4 is positive, so the curve touches from above, like a U. If the number were negative, the curve would touch from below, like an ∩.

Stop and think

The curve $y = x^{2} (x - 3)$ touches the axis at $x = 0$ . Does it come down from above, touch, and bounce back up?

Show the answer

No. Zoom in near $x = 0$ . There, the factor $(x - 3)$ is almost $- 3$ , so $y \approx - 3 x^{2}$ . That is an ∩-shape, so the curve comes up from below, touches the axis at the origin, and goes back down. A test value agrees: at $x = 1$ , $y = 1 \times (- 2) = - 2$ , which is below the axis. "Touch" never tells you the direction by itself. The sign of the nearby number does.

Build some curves of your own and watch each root.

Graph builder — a polynomial from its factors

The curve is built from up to three factors, (x − p), (x − q) and (x − r), each raised to a power from 0 to 3. A power of 0 removes that factor. Start from Worked Example 1. First raise one power from 1 to 2, then to 3, and watch that root. Then slide one root onto another. Last, multiply by −1. The vertical scale adjusts itself, because a sketch shows shape, not scale.

Presets:

Root p =

Power of (x − p) =

Root q =

Power of (x − q) =

Root r =

Power of (x − r) =

What each factor does at its root, starting from y = (x + 2)(x − 1)(x − 4):

Curve	At the changed root
y = (x + 2)(x − 1)(x − 4)	Three single roots: the curve crosses at −2, 1 and 4.
y = (x + 2)(x − 1)²(x − 4)	Degree 4. At x = 1 the curve touches. Near 1, y ≈ (3)(−3)(x − 1)² = −9(x − 1)², so it touches from below.
y = (x + 2)(x − 1)³(x − 4)	Degree 5. At x = 1 the curve crosses again, but flat, like y = x³.
y = (x + 2)(x − 1)(x − 1)	Slide the root 4 onto 1: two crossings join into one touch, and the curve becomes (x + 2)(x − 1)².
y = −(x + 2)(x − 1)(x − 4)	Multiplying by −1 flips the curve upside down. The roots stay, and the ends swap: it now starts high and ends low.

What should you have seen? A power of 1 gives a crossing, a power of 2 a touch, and a power of 3 a flat crossing. When one root slides onto another, two crossings join into one touch, because two factors become one squared factor. Multiplying by −1 turns the curve upside down: the roots stay, but every "above" becomes "below".

The ends of the curve: the leading term wins

The roots describe the middle of the graph. The ends are decided by one term only. Put $x = 100$ into $x^{3} - 3 x^{2} - 6 x + 8$ . The four terms are $1000000$ , $- 30000$ , $- 600$ and $8$ . The total is $969408$ , which is about 97% of the first term alone. The further out you go, the more completely the leading term takes over, just as it did for the arms of a parabola.

So, far from the origin, the curve behaves like its leading term. Now compare odd and even powers. At $x = - 100$ , $x^{3} = - 1000000$ , which is negative, but $x^{4} = 100000000$ , which is positive. An odd power keeps the sign of $x$ . An even power is positive on both sides. That gives four cases.

Leading term	Example	Left end	Right end
odd degree, positive	$x^{3}$	down ↙	up ↗
odd degree, negative	$- x^{3}$	up ↖	down ↘
even degree, positive	$x^{4}$	up ↖	up ↗
even degree, negative	$- x^{4}$	down ↙	down ↘

In factorised form, you find the leading term without expanding. Multiply the $x$ from every factor, with its power and its sign. In $y = (x + 1) (3 - x)$ , it is $x \times (- x) = - x^{2}$ . A factor written as $(3 - x)$ hides a minus sign. Sort the curves below by their ends.

Sorter — where do the ends go?

Put each polynomial in the box that describes its two ends. Find the leading term first, without expanding anything.

Each polynomial, its leading term, and its ends:

Polynomial	Leading term	Ends
−2x³ + x²	−2x³	starts high, ends low
x⁴ − 5x² + 4	x⁴	both up
(x + 1)(x − 2)(3 − x)	−x³	starts high, ends low
−(x − 1)²(x + 2)²	−x⁴	both down
x²(x − 3)	x³	starts low, ends high
(2 − x)²(x + 1)	x³	starts low, ends high
x(1 − x)³	−x⁴	both down
x⁵ − 100x⁴	x⁵	starts low, ends high
3 + 2x − x²	−x²	both down
(x + 1)(x − 1)(x + 2)(x − 2)	x⁴	both up

Stop and think

A cubic can have two turning points or none. Can it have exactly one?

Show the answer

No. A positive cubic is rising at its far left end and rising at its far right end. Each turning point changes the direction, from rising to falling or back. To start rising and finish rising, the curve must change direction an even number of times: 0 or 2. By the same argument, a quartic has 1 or 3.

Sketching a polynomial from its factors

First: factorise, if the polynomial is not already factorised. For example, $x^{3} - 4 x = x (x - 2) (x + 2)$ .

Ends: find the leading term by multiplying the $x$ -terms of all the factors. The degree and the sign give one of the four end shapes.

Roots: each factor $(x - r)^{m}$ gives a root at $x = r$ with multiplicity $m$ . Odd $m$ : the curve crosses (flat when $m = 3$ ). Even $m$ : the curve touches.

Signs: the sign of $y$ changes only at a root of odd multiplicity. One test value, or the nearby fixed number, settles any doubt.

$y$ -intercept: put $x = 0$ into every factor and multiply.

A degree $n$ polynomial has at most $n$ roots and at most $n - 1$ turning points.

Worked examples

Worked Example 1 — The curve from the start

Sketch $y = (x + 2) (x - 1) (x - 4)$ .

Ends. $x \times x \times x = x^{3}$ : odd degree, positive. The curve starts low on the left and ends high on the right.

Roots. $x = - 2$ , $1$ and $4$ , each of multiplicity 1, so the curve crosses at each one.

$y$ -intercept. $(0 + 2) (0 - 1) (0 - 4) = 2 \times (- 1) \times (- 4) = 8$ , so $(0, 8)$ .

Shape. This is the sketch shown earlier: one turn above the axis between $- 2$ and 1, one below it between 1 and 4.

Check. At $x = 2$ , $y = 4 \times 1 \times (- 2) = - 8$ . That is negative, as the sign table says for $1 < x < 4$ ✓. The intercept 8 is positive, as it should be for $- 2 < x < 1$ ✓.

Worked Example 2 — A touch from below

Sketch $y = x^{2} (x - 3)$ .

Ends. $x^{2} \times x = x^{3}$ : starts low, ends high.

Roots. $x = 0$ with multiplicity 2, so the curve touches there. $x = 3$ with multiplicity 1, so it crosses there.

$y$ -intercept. $0^{2} \times (0 - 3) = 0$ . The curve passes through the origin, which is also the touching point.

Direction of the touch. Near $x = 0$ , $y \approx - 3 x^{2}$ , an ∩-shape. So the curve comes up from below, touches at the origin, and turns back down.

Check with test values. At $x = - 1$ , $y = 1 \times (- 4) = - 4$ . At $x = 1$ , $y = 1 \times (- 2) = - 2$ . Both are below the axis ✓.

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

. The double root at 0 is a touch from below, because the curve is negative on both sides of it. The hollow dots are the two test values.

Worked Example 3 — A quartic that never goes above the axis

Sketch $y = - (x + 1)^{2} (x - 2)^{2}$ .

Ends. $- x^{2} \times x^{2} = - x^{4}$ : even degree, negative. Both ends go down.

Roots. $x = - 1$ and $x = 2$ , both of multiplicity 2. The curve touches at both.

$y$ -intercept. $- (1)^{2} \times (- 2)^{2} = - 1 \times 4 = - 4$ , so $(0, - 4)$ .

Shape. Every value is minus a square, so $y$ is never positive. The curve rises from the bottom left, touches at $- 1$ , dips to a lowest point, rises to touch at 2, and falls to the bottom right. That is three turning points, the most a quartic can have.

Where is the dip? The curve is $y = - [(x + 1) (x - 2)]^{2}$ . The quadratic inside the bracket is symmetric about $x = \frac{1}{2}$ , halfway between its roots, so the whole quartic is too. At $x = \frac{1}{2}$ , $(x + 1) (x - 2) = 1.5 \times (- 1.5) = - 2.25$ , and $y = - (2.25)^{2} \approx - 5.06$ . The dip is at $(0.5, - 5.0625)$ .

Three traps

Trap 1 — Missing a hidden minus sign

In $y = (x + 1) (x - 2) (3 - x)$ , the leading term is not $x^{3}$ . The factor $(3 - x)$ contributes $- x$ , so the leading term is $- x^{3}$ , and the curve starts high and ends low.

Trap 2 — Assuming a touch always bounces upward

A double root means the curve touches. It does not say from which side. In $y = x^{2} (x - 3)$ , the touch at 0 is from below. Decide the side from the nearby fixed number, or from one test value on each side.

Trap 3 — Placing the turning points by guessing

The maximum of $y = (x + 2) (x - 1) (x - 4)$ is between $- 2$ and 1, but not halfway. The lecture on stationary points finds it at $x = 1 - 3 \approx - 0.73$ , with $y \approx 10.4$ . Draw turning points in the right interval, and label only points you have found exactly.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Sketch $y = - (x + 1) (x - 2) (x - 5)$ . Label the roots and the $y$ -intercept.

Hint 1
Start with the leading term. The minus sign in front multiplies $x \times x \times x$ .

Hint 2
For the intercept, put $x = 0$ : $- (1) (- 2) (- 5)$ . Count the minus signs carefully.

Full answer
The leading term is $- x^{3}$ , so the curve starts high on the left and ends low on the right. The roots are $- 1$ , 2 and 5, all crossings. The $y$ -intercept is $- (1) (- 2) (- 5) = - 10$ , so $(0, - 10)$ . Coming down from the top left, the curve crosses at $- 1$ and turns below the axis, passing through $(0, - 10)$ . It crosses at 2, turns above the axis, crosses at 5, and falls to the bottom right. Check: at $x = 3$ , $y = - (4) (1) (- 2) = 8$ , positive between 2 and 5 ✓. At $x = 6$ , $y = - (7) (4) (1) = - 28$ , negative ✓.
Sketch $y = (x + 2) (x - 1)^{3}$ . How many turning points does it have?

Hint 1
The degree is $1 + 3 = 4$ . The root at 1 has multiplicity 3.

Hint 2
Near $x = 1$ , the factor $(x + 2)$ is almost 3, so $y \approx 3 (x - 1)^{3}$ .

Full answer
Leading term $x \times x^{3} = x^{4}$ : both ends go up. Roots: $- 2$ (crosses) and 1 (multiplicity 3, crosses flat). The $y$ -intercept is $2 \times (- 1)^{3} = - 2$ . Signs: positive for $x < - 2$ , negative between $- 2$ and 1, positive for $x > 1$ . So the curve comes down from the top left, crosses at $- 2$ , and turns at a minimum between $- 2$ and 1. It then rises, flattening as it crosses at 1, like $y = 3 (x - 1)^{3}$ , and goes on up to the top right. It has one turning point. The flat point at 1 is not a turning point, because the curve keeps rising through it. One is allowed: a quartic with both ends up has 1 or 3. Check: at $x = 0$ , $y = - 2$ , negative ✓; at $x = 2$ , $y = 4 \times 1 = 4$ , positive ✓.
A cubic touches the $x$ -axis at $x = - 2$ , crosses it at $x = 3$ , and passes through $(0, - 6)$ . Find its equation, and say from which side it touches at $- 2$ .

Hint 1
A touch needs a squared factor. Write $y = a (x + 2)^{2} (x - 3)$ .

Hint 2
Put $x = 0$ and $y = - 6$ to find $a$ .

Full answer
$y = a (x + 2)^{2} (x - 3)$ . At $x = 0$ : $- 6 = a \times 4 \times (- 3) = - 12 a$ , so $a = \frac{1}{2}$ . The cubic is $y = \frac{1}{2} (x + 2)^{2} (x - 3)$ , which expands to $y = \frac{1}{2} x^{3} + \frac{1}{2} x^{2} - 4 x - 6$ . Near $x = - 2$ , $\frac{1}{2} (x - 3) \approx \frac{1}{2} \times (- 5) = - 2.5$ , so $y \approx - 2.5 (x + 2)^{2}$ : the curve touches from below. Check: at $x = - 3$ , $y = \frac{1}{2} \times 1 \times (- 6) = - 3$ , and at $x = - 1$ , $y = \frac{1}{2} \times 1 \times (- 4) = - 2$ . Both are below the axis ✓.
An open box is made from a square sheet of card 12 cm wide. A square of side $x$ cm is cut from each corner, and the sides are folded up. Show that the volume is $V = x (12 - 2 x)^{2}$ cm³. Sketch $V$ against $x$ . Which part of the sketch makes sense for the box? Use values to estimate the best $x$ .

Hint 1
The base is a square of side $12 - 2 x$ , and the height is $x$ .

Hint 2
$(12 - 2 x)^{2} = 4 (x - 6)^{2}$ . So $V = 4 x (x - 6)^{2}$ : find the roots and their multiplicities.

Full answer
Volume = base × height = $(12 - 2 x)^{2} \times x$ . Since $12 - 2 x = - 2 (x - 6)$ , $V = 4 x (x - 6)^{2}$ . The leading term is $4 x^{3}$ : starts low, ends high. Roots: 0 (crosses) and 6 (touches). Near 6, $4 x \approx 24$ , which is positive, so the curve touches from above. Only $0 < x < 6$ makes sense, because the base side $12 - 2 x$ must be positive. In that interval the curve rises from 0, reaches a maximum, and comes back down to touch at 6. Values: $x = 1$ gives 100, $x = 2$ gives 128, $x = 3$ gives 108, $x = 4$ gives 64, $x = 5$ gives 20. The best is near $x = 2$ , about 128 cm³. The lecture on stationary points shows it is exactly $x = 2$ . For $x > 6$ the formula gives positive values again, but no real box exists there.

The bigger idea

A cubic has no mirror line, but it has something just as useful: a centre. Look again at $y = (x + 2) (x - 1) (x - 4)$ . The average of its roots is $\frac{- 2 + 1 + 4}{3} = 1$ . Put $x = 1 + t$ and expand. Everything simplifies to

y = (t + 3) t (t - 3) = t^{3} - 9 t

Replacing $t$ by $- t$ reverses every height. So the curve looks the same after a half-turn about $(1, 0)$ , and its two turning points are mirror images through that centre: $(1 \pm 3, \mp 10.4)$ , approximately. Every cubic $a x^{3} + b x^{2} + c x + d$ has such a centre, at $x = - \frac{b}{3 a}$ . When the cubic has three roots, this is their average. Compare $x = - \frac{b}{2 a}$ for the parabola, the average of its two roots. The pattern is the same idea, one degree higher.

The deeper lesson is the method: read the shape from the structure of the equation before calculating anything. This lecture closes the topic of straight-line and quadratic graphs. Its tools, roots, signs, ends and turns, return when calculus finds turning points exactly.

Key concepts in this lesson

A polynomial of degree $n$ has at most $n$ roots and at most $n - 1$ turning points.
The leading term, found by multiplying the $x$ -terms of all the factors, decides the ends: odd degree sends them opposite ways, even degree the same way.
Close to a root of multiplicity $m$ , the curve looks like a fixed number times $(x - r)^{m}$ , so odd multiplicity crosses and even multiplicity touches.
The sign of $y$ changes only at roots of odd multiplicity, so a sign table or one test value shows which side of the axis the curve is on.
The $y$ -intercept comes from putting $x = 0$ into every factor and multiplying.