Properties of Quartics

Abstract: The aim of this paper is to investigate on a property of quartic polynomials. Considering a “W” shaped function with inflection points Q and R, a line was drawn through the points to meet the function again at P and S. The ratio of PQ:QR:RS is found to correspond to the Golden Ratio hence a conjecture is produced and formally proven in order to demonstrate this.

The function

€

f (x) = x 4 − 8x 3 +18x 2 −12x + 24

is a “W” shape quartic with two points of

inflection. Its first derivative

€

′ f (x) = 4x 3 − 24x 2 + 36x −12 is a cubic

function and its second derivative

€

′ ′ f (x) =12x 2 − 48x + 36 a quadratic.

Fig 1

Fig 1 shows the function

€

f (x) and its first and second derivatives. Note that the points where

€

′ f (x) = 0 are the function’s inflection points.

The quartic’s points of inflection are found by

equating the second derivative to zero.

€

′ ′ f (x) =12x 2 − 48x + 36

0 =12(x −1)(x − 3)

€

x =1 and x = 3 ,

Substituting for x,

€

f (1) = 23 and f (3) =15

hence points of inflection are Q(1,23) and

R(3,15) The equation of the straight line

joining Q and R is:

€

y − y1 = m(x − x1)

y − 23 = −4(x −1)

y = −4x + 27

Let points P and S be the intersection of the

line QR and the quartic function. They are

found by equating, factorizing using long

division and solving for x.

€

−4 x + 27 = x 4 − 8x 3 +18x 2 −12x + 24

0 = x 4 − 8x 3 +18x 2 − 8x − 3

0 = (x 2 − 4 x + 3)(x 2 − 4x +1)

x =4 ± 20

2

x = 2 + 5 and x = 2 − 5

Note that the other quadratic equation gives

the x-values of Q and R. The coordinates of

points P and S are therefore

€

P((2 − 5), f (2 − 5)) and

€

S((2 + 5), f (2 + 5)) .

The investigation must then consider the ratio

of PQ:QR:RS. This can be done using

Pythagoras Theorem, however, because the

coordinates of P and S are irrational, the

process is long and tedious. The features of

similar triangles enable us to concentrate on x

coordinates only. Since all four points lie on

the same line, the triangles they form are

similar and the ratio of their distances is the

same as the ratio of their components.

Fig 2

x

In Fig 2,points P, Q, R, and S lie on the line

€

y = −4x + 27 . The dotted lines illustrate the similar triangles they form.

The ratio PQ:QR:RS is found using the

previously calculated x-values. Let

€

xP = 2 − 5 ,

€

xS = 2 + 5 ,

€

xq =1

and

€

xR = 3

€

xP − xQ = (2 − 5) −1 = 5 −1

€

xQ − xR = 1− 3 = 2

€

xR − xS = 3 − (2 + 5) = 5 −1

PQ:QR:RS is

€

5 −1: 2 : 5 −1and dividing

through by

€

5 −1 simplifies to

€

1:1+ 5

2:1

Considering the simpler quartic function

€

g(x) = 2x 4 − 4x 3 , the points of inflection also

occur when the second derivative equals zero:

€

′ ′ g (x) = 24x 2 − 24x

€

0 = 24x(x −1)

x = 0 x =1

g(0) = 0 g(1) = −2

Hence coordinates are Q(0,0) and R(1,-2).

Since the line QR goes through the origin, its

equation is

€

y = −2x which intersects with the

quartic when:

€

2x 4 − 4x 3 = −2x

2x 4 − 4x 3 + 2x = 0

Long division using the factor

€

(x −1)

simplifies the equation into two quadratics,

which can then be solved using the formula.

€

2x 4 − 4x 3 + 2x = 0

(2x 2 − 2x)(x 2 − x −1) = 0

x =4 ± 4

4 and x =

1± 5

2

The x-coordinates of points P and S are

therefore

€

1− 5

2 and

€

1+ 5

2 respectively.

Calculating the ratio is simpler in this case

since the point of inflection Q lies on the

origin. The ratio PQ:QR:RS corresponds to

€

xP − xQ : xQ − xR : xR − xS

€

5 −1

2 :

€

1 :

€

5 −1

2

and dividing through by

€

5 −1

2 simplifies to

€

1:1+ 5

2:1

Fig 3

In both cases, we assist to the emergence of

the golden ratio defined as the positive

solution of

€

x 2 − x −1 = 0 or

€

1+ 5

2.

We therefore produce a conjecture based of

the previous investigations.

Conjecture: If

€

f (x) is a quartic polynomial

with two distinct points of inflection, Q and R,

then the straight line joining Q and R meets

the graph of

€

y = f (x) in two other points P

and S. In its simplest form, the ratio of

PQ:QR:RS is equal to

€

1:1+ 5

2:1 so that

€

QR

PQ=

1+ 5

2 or the Golden Ratio.

Let us consider the general case of a quartic

polynomial function

€

f (x) . It will have two

distinct points of inflection, Q and R, if its

second derivative has two real roots. This

means that the quadratic function

€

′ ′ f (x) must

be in the form of

€

y = bx(x − a) where a and b

are real numbers.

Fig 4

Fig 4 shows the relationship between the function and its second derivative.

The quartic function is obtained by

integrating

€

′ ′ f (x) twice.

€

bx(x − a)∫∫ dx =1

12b(x 4 − 2ax 3 + cx + d)

For simplicity, we assume that the point Q

lies on the origin (0,0). The constant d must

therefore equal zero so that

€

f (0) = 0 meaning

that the quartic is expressed as

€

f (x) =1

12b(x 4 − 2ax 3 + cx) .

Fig 3 shows the graph of

€

g(x) and

€

y = −2x . Points Q and R are inflection points and P and S intersection points.

€

y =1

12b(x 4 − ax 3 + cx)

Now if Q is the origin, the other root of the

second derivative is a hence the second point

of inflection R has coordinates

€

R(a, f (a)) .

Substituting for

€

f (a) , the coordinates are

€

R(a,1

12b(−a4 + ac)) .

The gradient of the straight line joining points

Q and R is

€

1

12bx(−a4 + ac)

a=

1

12bx(−a3 + c)

so its equation is

€

y =1

12bx(−a3 + c) because

the y-intercept is zero. The line intersects the

quartic at points P and S and the x-coordinates

of these points are the solutions of the

equation

€

1

12bx(−a3 + c) =

1

12bx(x 3 − 2ax 2 + c)

€

x(−a3) = x(x 3 − 2ax 2)

x 4 − 2ax 3 + a3x = 0

The easiest way to obtain the solutions of the

equation in terms of a is to use the quadratic

formula. This means simplifying the quartic

into a quadratic by long division using the

already known real factors

€

(x − a) and

€

(x − 0) .

Using factor

€

(x − a) first:

€

x − a x 4 − 2ax 3 + a3 x

−(x 4 − ax 3 )

− ax 3 + a3 x

−(−ax 3 + a2 x 2 )

− a2 x 2 + a3 x

−(−a2 x 2 + a3 x)

0

x 3 − ax 2 − a2 x

)

and then factor

€

(x − 0)

€

x − 0 x 3 − ax2 − a2x

−(x 3)

− ax2 − a2x

−(−ax2)

− a2x

−(−a2x)

0

x 2 − ax − a2

)

Simplifies the equation into the quadratic

€

y = x 2 − ax − a2 whose roots are found using the formula

€

x =a ± a2 − 4(−a2 )

2 hence

€

xS =a + 5a2

2,

€

xP =a − 5a2

2 ,

€

xQ = 0 and

€

xR = a.

Once again, properties of similar triangles

enable us to use x-coordinates only to find the

ratio PQ:QR:RS.

€

xP − xQ : xQ − xR : xR − xS

€

5a2 − a

2: a :

5a2 − a

2

The ratio can be greatly simplified:

€

5a2 − a

2: a :

5a2 − a

2

(dividing through by

€

5a2 − a

2)

€

1:2a

5a2 − a:1

(simplifying the fraction)

€

2a

5a2 − a =

a

a×

2

5 −1 =

1+ 5

2

Hence the ratio equals

€

1:1+ 5

2:1 or the

Golden Ratio

It is important to mention the limitations of

this conjecture that cannot be extended to

quartic functions that are not strictly of a “W”

shape. The ratio of lengths can only be found

if the function has two distinct points of

inflection meaning its second derivative must

have two real roots. There are therefore two

types of quartics that will not illustrate the

golden ratio.

The first type has no points of inflection since

its second derivative has no real roots. The

quadratic

€

′ ′ f (x) does not cross the x-axis

because its determinant

€

b2 − 4ac is negative. A

quartic with no points of inflection will have a

“U” shape as shown on fig.5

Fig 5

Fig 5 shows the graph of

€

y =x 4

12+

x 2

2 and its

second derivative

€

y = x 2 +1. Note that the quadratic has no real roots hence the quartic is “U” shaped.

The second type of quartic does not have any

points of inflection because its second

derivative has only one real solution. Indeed,

since its determinant equals zero, it touches the

x-axis at one point and does not change sign.

Quartics of such type have a flattened U shape.

Fig6

As a result, the conjecture only holds if the

determinant of the second derivative is greater

than zero. Integrating the general function

twice:

€

(ax 2 + bx + c)∫∫ dx =ax 4

12+

bx 3

6+

cx 2

2+ dx + e

Hence the conjecture only affects quartic

function with

€

b2 − 4ac > 0

Using the proof of the conjecture and the

analysis of its limitations, it is possible to

Fig6 shows the graph of

€

y = x 4 and its second derivative

€

y =12x 2 .

produce a formal theorem on this specific

property of quartic polynomials.

Theorem: Let

€

f (x) be a quartic polynomial

with two distinct points of inflection Q and R,

the straight line joining these points will meet

the function again at P and S. If points P,Q,R,S

are ordered by increasing x-

coordinates, then

€

PQ = RS and

€

QR

RS=

1+ 5

2, the Golden Ratio