# Homographic Functions

```Homographic Functions
A
( H1 ) f1 : x  y 
x
(H 2 )
A
f2 :x a y   H
x
(H 3 )
f3 : x a y 
(H 4 )
A
xL
A
f4 : x a y 
H
xL
ax  b
(H 5 ) f5 : x  y 
cx  d
Review sept.2010

1
Basic type
(Review 1)
A
( H1 ) f1 : x  y 
x
A&gt;0
y
1
x

• when x  +∞ then y  0 (+)
• when x  -∞ then y  0 (-)
• x-axis y = 0 is an asymptote for (H)
• when x  0 (+) then y  +∞
• when x  0 (-) then y  -∞
• y-axis x = 0 is an asymptote for (H)
• The vertex of the Hyperbola is the
point (√A,√A) on the blue Axis (y=x).
• The function is an odd function
• O is the center of symetry of (H).
Review sept.2010

2
Basic type
(Review 2)
A
( H1 ) f1 : x  y 
x
• when x  +∞ then y  0 (-)
• when x  -∞ then y  0 (+)
• x-axis y = 0 is an asymptote for (H)
A&lt;0
y
• when x  0 (+) then y  - ∞
• when x  0 (-) then y  + ∞
• y-axis x = 0 is an asymptote for (H)
• The vertex of the Hyperbola is the
point (-√(-A),√(-A) on the Axis (y=-x).
• The function is an odd function
• O is the center of symetry of (H).
Review sept.2010
4
x


3
First transformation (p.1)
(H1 )
A
f1 : x a y 
x
A=1
Review sept.2010
Vertical
Translation
 

(H 2 )
A
f2 : x a y   H
x
A=1

4
First transformation (p.1b)
(H1 )
A
f1 : x a y 
x
A=1
Review sept.2010
Vertical
Translation
 

(H 2 )
A
f2 : x a y   H
x
A=1
H = +2

5
First transformation
(H1 )
A
f1 :x a y 
x
Vertical
Translation
 

(H 2 )
A
f2 :x a y   H
x
A = -1
Review sept.2010
(p.2)
A = -1

6
First transformation
(H1 )
A
f1 :x a y 
x
Vertical
Translation
 

(H 2 )
A
f2 :x a y   H
x
A = -1
h = +2
A = -1
Review sept.2010
(p.3)

7
nd
2
transformation
A
f1 : x a y 
x
(H1 )
Horizontal
Translation
 

A&gt;0
y
A
f3 : x a y 
xL
(H 3 )
A&gt;0
1
x

(p.1)
y
1
x

Review sept.2010

Functions
8
nd
2
transformation
A
f1 : x a y 
x
(H1 )
Horizontal
Translation
 

A&gt;0
y
(H 3 )
y
1
x
(p.1b)
A
f3 : x a y 
xL
1
x 2


Review sept.2010

9
nd
2
(H1 )
transformation
A
f1 : x a y 
x
Horizontal
Translation
 

(H 3 )
A
f3 : x a y 
xL
A&lt;0
A&lt;0
4
y
x
4
y
x

Review sept.2010
(p.2)


Functions
10
nd
2
(H1 )
transformation
Horizontal
Translation
A
f1 : x a y 
x
 

(H 3 )
(p.2b)
A
f3 : x a y 
xL
A&lt;0
y
4
y
x


Review sept.2010
4
x 2

11
rd
3
transformation
ur
V (L;H )
Translation
A
A
(H1 ) f1: x a y 

 (H 4 ) f4 : x a y 
H
x
xL
A&gt;0
y
y
1
x
1
1
x 2


Review sept.2010

12
Change of center and variables
V ( l ;h )
A Translatio
A
n
( H1 ) f1 : x  y  ( H 4 ) f 4 : x  y 
h
x
x l
Let X = x – l and Y = y – h
then the equation becomes
Y
which
respect
0’(l,h),

function
original.
A
X
y
1
1
x 2
means that, with

to the new center
the graph of the
is the same as the
Review sept.2010

13
Limits &amp; Asymptotes
A
(H4 ) y 
h
xl
y
1
1

x 2
• when x  +∞ or x  - ∞
then y  h (&plusmn;)
the line y = h is an asymptote for (H)
• when x  l (&plusmn;) then y  &plusmn;∞
the line x = l is an asymptote for (H)
• The point (l,h) intersection of the
two asymptotes is the center of
symmetry of the hyperbola.
Review sept.2010

14
General case
ax  b
(H 5 ) f5 : x  y 
cx  d
•
It’s easy to check that all functions in the type of
can be changed into the form of f5(x).
Example :
y
1
x 1
1
x 2
x 2

•
Problem : prove that all functions defined by :
 into the previous one.
can be transformed
Example : y 
Review sept.2010
A
y
h
xl
ax  b
y
cx  d
4x  5 4(x 1)  9
9


4
x 1
x 1
x 1


15
General case
ax  b
(H 5 ) f5 : x  y 
cx  d
4x  5 4(x 1)  9
9
y


4
x 1
x 1
x 1
• In this example l = 1, h = 4, A = 9
•&laquo;Horizontal&raquo; Asymptote : y = 4
•&laquo;Vertical&raquo; Asymptote :
x=1
• Center : (1;4).
• A &gt; 0  function is decreasing.
• Only one point is necessary to be
able to place the whole graph !
• Interception with the Y-Axis : (0,-5)
or
• Interception with the X-Axis : ( 5 ; 0 )
4
Review sept.2010

16
General case
ax  b
(H 5 ) f5 : x  y 
cx  d
ax  b
A
y

H
cx  d x  L
d
L
c
a
H
c
• In fact one can find the asymptotes by
looking for the limits of the function in
the original form.
Then it’s not necessary to change the
equation to be able to plot the graph.
Review sept.2010

17
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