PHANTOM GRAPHS
PART 2
Philip Lloyd
Epsom Girls Grammar School
Web site: www.phantomgraphs.weebly.com
Now we will consider: y = x (x – 3)2
(1, 4 )
3
As before, any horizontal line (or plane)
should cross this graph at 3 places because
any equation of the form :
x3 – 6x2 + 9x = “a constant”, has 3 solutions.
The maximum point is at (1, 4)
(1,4)
and the minimum point is at (3, 0).
We now realise that the phantom curves
will start to appear at the turning points
(1, 4) and (3, 0)
1
3
If x3 – 6x2 + 9x = 5 then x = 4.1 and 0.95 ± 0.6i
If x3 – 6x2 + 9x = 6 then x = 4.2 and 0.90 ± 0.8i
If x3 – 6x2 + 9x = 7 then x = 4.3 and 0.86 ± 0.9i
So the left hand phantom is leaning to the left.
The right hand phantom will be appearing
at the minimum point when x = 3, y = 0.
3
If x3 – 6x2 + 9x = – 1 then x = –0.1 and 3.05 ± 0.6i
If x3 – 6x2 + 9x = – 2 then x = –0.2 and 3.1 ± 0.8i
If x3 – 6x2 + 9x = – 3 then x = –0.3 and 3.14 ± 0.9i
So the right hand phantom is leaning to the right.
AUTOGRAPH VERSION.
y = x(x - 3)²
AUTOGRAPH VERSION.
Solutions of cubic crossing 3 planes

Notice how this typical cubic will cross any
horizontal plane exactly 3 times.

This shows how any equation of the form
y= x3 + ax2 + bx + c always has 3 solutions.
A very interesting case is the hyperbola :
y2 – x2 = 1
25
or
y2 = x2 + 25
25
If x = 0 then y = ±5
And, if x is large then y2 ≈ x2
producing asymptotes y = ± x
The graph of y2 = x2 + 25 for real x values is :
What x values produce
y values between
-5 and +5 ?
5
-5
y2 = x2 + 25
If y = 4 then 16 = x2 + 25
and
so
– 9 = x2
x = ± 3i
Similarly if y = 3 then 9 = x2 + 25
so x = ± 4i
And if y = 0
then 0 = x2 + 25
so x = ± 5i
These points should seem very
familiar as points on a CIRCLE of
radius 5 units.
(0, 5) (±3i, 4) (±4i, 3) (±5i, 0)
The circle has imaginary x values
but REAL y values.
This phantom circle is in the plane
at right angles to the hyperbola
and joining its two halves.
The graph of y2 = x2 + 25 for real y values is :
5
-5
AUTOGRAPH VERSION.
y² = x² + 25
SOME AFTERTHOUGHTS………………(not for Year 13 students)
Consider the graph
y = 2x2 = 2 +
2
x2 – 1
x2 – 1
This has a horizontal asymptote y = 2 and two vertical asymptotes x = ± 1
y
2
What x values can produce y values
from y = 0 to y = 2 ?
-1
1
x
If y = 1 then 2x2 = 1
x2 – 1
So
2x2 = x2 – 1
2
x2 = – 1
and
x= ±i
Producing
2x2 = 1.999
x2 – 1
2x2 = 1.999x2 – 1.999
0.001x2 = – 1.999
If y = 1.999 then
So
and
Producing
x2 = – 1999
x  ± 45 i
This is a “phantom graph” which approaches the horizontal
asymptotic plane y = 2 and is at right angles to the x, y plane,
resembling an upside down normal distribution curve…
Side view at right angles
to the x, y plane.
2
AUTOGRAPH VERSION.
Asymptotic plane
What about TRIGONOMETRICAL GRAPHS?
If
y = cos(x)
-π/2
what about
1
y values > 1 and
π/2
<-1 ?
π
Solving cos( x) = 2 seems impossible !
but with some intuition, we could try to find out what
cos( i) is equal to.
3π/2
Using cos(x) = 1 –
x2 + x4 – x6 + x8 - . . ….
2!
4!
6!
8!
Let’s find cos(± i ) = 1 + 1 + 1 + 1 + 1 . .  1.54 (ie > 1 )
2!
4!
6!
8!
Similarly cos(± 2i) = 1 + 4 + 16 + 64 + …  3.8
2! 4!
6!
Also find cos(π + i) = cos(π) cos(i) – sin(π) sin(i)
= – 1 × cos(i) – 0
 – 1.54
These results imply that the cosine graph also has its own “phantoms” in
vertical planes at right angles to the usual x, y graph, emanating from
each max/min point.
NB cos(i) = cosh(1)
-π/2
π/2
π
3π/2
2π
AUTOGRAPH VERSION.
y = cos(x)
Consider the exponential function y = e x
y(real)
1
x(real)
We need to be able to find x values which produce negative y values.
How can we find x if e x = – 1 ?
Using the expansion for e x = 1 + x + x2 + x3 + x4 + x5 +. . ….
2!
3! 4!
5!
We can find e x i = 1 + xi + (xi)2 + (xi)3 + (xi)4 + (xi)5 +. . ….
2!
3!
4!
5!
= (1 –
=
x2 + x4 – x6 + x8 +… ) +
2!
4! 6!
8!
cos x
+
i
( x – x3 + x5 – x7 + ...)
3! 5!
7!
i sin x
If we are to get REAL y values then using e xi = cos x + i sin x ,
we see that the imaginary part, sin x, has to be zero.
This only occurs when x = 0, π, 2π, 3π,… (or generally nπ)
e πi = cosπ + isinπ = –1 + 0i , e 2πi = cos2π + isin2π = +1 + 0i
e 3πi = cos3π + isin3π = –1 + 0i , e4πi = cos4π + isin4π = +1 + 0i
Now consider y = eX where X = x + 2nπ i
ie y = e x + 2nπ i = e x × e 2nπ i = e x × 1
(ie even numbers of π)
=ex
Also consider y = eX where X = x +(2n+1)π i (ie odd numbers of π)
ie y = e x + (2n+1)π i = e x × e (2n+1)π i = e x × –1 = – e x
This means that the graph of y = eX
consists of parallel identical curves
occurring at X = x + 2nπi
= x + even Nos of πi
and, upside down parallel identical curves
occurring at X = x + (2n + 1)πi
= x + odd Nos of πi
y(real)
1
-2πi
-πi
-1
1
x(real)
πi
-1
Graph of y = e X where X = x + nπi
1
2πi
x(unreal)
Graph of y = e X where X = x + nπi
x(unreal)
y(real)
2π
π
-π
x(real)
- 2π
AUTOGRAPH VERSION.
y = exp(x)
I call this THE ALPHA GRAPH y 2= x(x – 3) 2
y
3
2
1
1
2
3
4
x
y2 = x(x – 3)2
Using a technique from previous graphs:
4
1
I choose an x value such as x = 4,
calculate the y value, ie y = 4
then solve the equation x(x – 3)2 = 4
already knowing one factor is (x – 4)
ie If x(x – 3)2 = 4
then x3 – 6x2 + 9x – 4 = 0
so (x – 4)(x2 – 2x + 1) = 0
factorising (x – 4)(x – 1)2 = 0
x = 4 or 1
4
SIMILARLY :
If x = 5, y2= 20 and y = ±4.5 or ±2√5
So x(x – 3)2 = 20
x3 – 6x2 + 9x – 20 = 0
(x – 5)(x2 – x + 4) = 0
x = 5 or ½ ± 1.9i
If x = 6, y2= 54 and y = ±7.3
So x(x – 3)2 = 54
x3 – 6x2 + 9x – 54 = 0
(x – 6)(x2 + 9) = 0
x = 6 or
± 3i
If x = 7, y2= 112 and y = ±10.58
So x(x – 3)2 = 112
x3 – 6x2 + 9x – 112 = 0
(x – 7)(x2 + x + 16) = 0
x = 7 or – ½ ± 4i
6
4
2
1
2
3
4
AUTOGRAPH VERSION.
y² = x(x - 3)²
This is a more familiar graph: y =
x2
x–1
y
4
3
2
1
-1
1
2
x
y = x2
x–1
y
If y = any real number c
4
then
x2
=
c
3
x–1
2
1
which produces a
quadratic equation:
-1
1
2
x2 = cx – c
So for values of c from 0 to 4 we will get complex
conjugate solutions.
x
y = x2
x–1
If y = 0
If y = 1
y
x=0
4
x = 1 ± √3i
2
2
3
2
1
If y = 2
x=1±i
If y = 3
x = 3 ± √3i
2 2
x=2
If y = 4
-1
1
2
These points produce the phantom “oval”
shape as shown in the next picture.
x
5
4
3
2
1
1
2
3
AUTOGRAPH VERSION.
y = x²/(x - 1)
Consider an apparently “similar” equation
but with a completely different “Phantom”.
y = x2
=
x2
(x – 1)(x – 4)
x2 – 5x + 4
AUTOGRAPH VERSION.
Curve with 2 vertical asymptotes
y = x3/(x2 – 1)
AUTOGRAPH VERSION.
y = x³/(x² - 1)
y = x4/(x2 – 1)
AUTOGRAPH VERSION.
y = x^4/(x² - 1)
THE END
Please check out the Web site:
www.phantomgraphs.weebly.com
Not all graphs have phantoms!
Consider the equation : y =
x
x2 – 1
Briefly, if we consider any real y value,
for example y = 1 then if we solve :
x = 1 we just get a simple quadratic
x2 – 1
equation to solve which has 2 solutions.
Notice that ANY horizontal line will
always cross this curve twice already.
There are no “missing” values which is
the usual sign that there are “phantoms”.
The algebra for this case is as follows…
Let z =
x + iy
=
x + iy
(x + iy)2 – 1
(x2 – y2 – 1) + 2xyi
= x(x2 – y2 – 1) + 2xy2 + i ( y(x2 – y2 – 1 ) – 2x2 y
(x2 – y2 – 1)2 + 4x2 y2
= x3 + xy2 – x
+ i( -y(x2 + y2 + 1) )
(x2 – y2 – 1)2 + 4x2 y2
So if Im(z) = 0 then either y = 0 producing the original curve
OR x2 + y2 + 1 = 0 and since this has no real solutions
there are no phantoms.