MEI Maths Item of the Month
November 2014
Three's a Crowd
There is just one form of the graph of a quadratic needed to produce the graph of any
quadratic using vertical stretches and horizontal/vertical translations: i.e. the graph
of y = x2 can be transformed into the graph of any quadratic y = ax2 + bx + c.
The graph of y = x3 cannot be transformed into the graph of all cubics y = ax3 + bx2 + cx + d
using horizontal/vertical stretches and horizontal/vertical translations. How many basic forms
of a cubic are needed so that all cubics can be obtained and can you prove that these are
sufficient to do this?
Solution
The number of stationary points of a cubic
dy
y  ax 3  bx 2  cx  d has derivative
 3ax 2  2bx  c .
dx
2
3ax  2bx  c  0 has discriminant 4b2  12ac
y  ax 3  bx 2  cx  d has either 0, 1 or 2 stationary points corresponding to the cases
b 2  3ac , b 2  3ac or b 2  3ac .
Transforming a function with stretches and translations
 x p
sf 
  q is a transformation of y  f( x) by translating p units horizontally and q units
 r 
vertically and stretching by a factor of r units horizontally and s units vertically.
Transforming a basic cubic
Using f( x)  x 3 and expanding gives:
s 3 3 ps 2 3 p 2 s
p3s
 x p
sf 

q

x

x

x

q


r3
r3
r3
r3
 r 
Equating coefficients with y = ax3 + bx2 + cx + d gives
a
s
3 ps
3 p2s
,
b


,
c

 b 2  3ac
3
3
3
r
r
r
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MEI Maths Item of the Month
Therefore using this it is only possible to create values of a, b, c and d by selecting
appropriate values of p, q, r and s provided that b 2  3ac .
Transforming a cubic with two stationary points
Using an additional function g( x)  x 3  x generates:
s 3 3 ps 2 3 p 2 s  r 2 s
ps p 3 s
 x p
sg

q

x

x

x

q

 3

r3
r3
r3
r
r
 r 
2
s
3 ps
3p s s
a 3,b 3 , c 3 
r
r
r
r
2 2
2 2
2
9p s
9p s s
b 2  6 , 3ac  6  4  b 2  3ac
r
r
r
Using this it is possible to create all possible values of a, b, c and d by selecting appropriate
values of p, q, r and s, provided that b 2  3ac .
Transforming a cubic with no stationary points
Using an additional function h( x)  x3  x generates:
s 3 3 ps 2 3 p 2 s  r 2 s
ps p 3 s
 x p
sh
xq
 3
q  3 x  3 x 
r
r
r3
r
r
 r 
2
s
3 ps
3p s s
a 3,b 3 , c 3 
r
r
r
r
2 2
2 2
2
9p s
9p s s
b 2  6 , 3ac  6  4  b 2  3ac
r
r
r
Using this it is possible to create all possible values of a, b, c and d by selecting appropriate
values of p, q, r and s, provided that b 2  3ac .
Producing any cubic by transformation
Hence three forms of the cubic are needed to be transformed into the graph of all cubics
y = ax3 + bx2 + cx + d using horizontal/vertical stretches and horizontal/vertical translations.
These correspond to the cases of 0, 1 or 2 stationary points: e.g. h( x)  x3  x , f( x )  x 3
and g( x)  x 3  x .
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