Section 1.3 New functions from old
Transformations of equations and functions
Shifting equations:
Consider a graph and an equation for the graph, such as the unit circle, x 2  y 2  1 .
For a>0 and b>0, if x is replaced by x-a in the equation, the graph shifts to the right a
units;
if y is replaced by y-b, the graph shifts up b units.
Ex. The graph of ( x  2 ) 2  ( y  3 ) 2  1 is the unit circle shifted to the right 2 units
and up 3 units so the new center is the point (2,3).
To shift the graph left, replace x by x+a. To shift down, replace y by y+b.
Ex. The graph of ( x  5 ) 2  ( y  1 ) 2  1 has center at ( - 5, - 1 ).
Distortions:
Stretching and shrinking equations. c > 0
x
If x is replaced by
in the equation, then all x values in the graph are multiplied by c.
c
Ex. The equation
 x 
 
 3 
2
 y
2
 1
is an ellipse in which x varies from -3 to 3. The unit
circle is stretched horizontally by a factor of 3.
Similarly, if y is replaced by
y
, then all y values in the graph are multiplied by c.
c
Ex. The equation
 x 
 
 3 
2
 y 
  
 2 
2
 1
is an ellipse in which x varies from -3 to 3 and y varies
from -2 to 2.
For c>1, the distortion stretches the graph. For c<1, the distortion shrinks the graph.
Shifts and distortions in functions are the same but since we solve y=f(x), the vertical
shifts and distortions look different in the function.
Ex. The top half of the unit circle is the equation x 2  y 2  1 for y>0.
Solving for
y  f ( x ),
y 
1 x
2
.
If we shift this right 2 units and up 3 units we have
y 
1  (x  2)
2
 3
( x  2)
 ( y  3)
2
1
for y>3, or
.
For the distortions: Consider the top half of the ellipse
Solving for
2
y  f ( x ),
y  2
 x 
1  
 3 
vertical stretch by a factor of 2.
 x 
 
 3 
2
 y 
  
 2 
2
 1
, y>0.
2
is the horizontal stretch by a factor of 3 and
Reflections:
To reflect the graph across the x-axis, replace y by -y in the equation; y=f(x) reflected
becomes y=-f(x).
To reflect the graph across the y-axis, replace x by -x in the equation; y=f(x) reflected
becomes y=f(-x).
Ex. y = 1-x reflected across the x-axis has equation y = -(1-x)= -1 + x.
y= 1-x reflected across the y-axis has equation y = 1+x.
y=| f(x) | , the absolute value of f(x), reflects all the negative y-values across the x-axis.
Compositions:
Example:
is f composed with g.
( f  g )( x )  f ( g ( x ))
f (u ) 
u
g(x)  1  x
2
( f  g )( x ) 
Ex. What two functions are composed to form
Answer:
f (u )  e
u
g(x)  x
2
1 x
2
( f  g )( x )  h ( x )  e
x
2
2x
?
 2x
Ex. Given the values in the tables for the functions f and g, find the values of
( f  g )( x )  f ( g ( x )) which can be found from this information.
x
g(x)
-1
3
0
5
2
7
3
-1
4
2
5
6
u
f(u)
-1
8
0
5
2
4
3
13
4
7
5
3
6
5
7
10
Answer:
x
-1
0
2
3
4
5
13
3
10
8
4
5
( f  g )( x )
We do not have enough info to find the values at 6 or 7.
Supplementary Example
Any quadratic is a transformation of y  x 2 .
Ex: List the transformations of y  x 2 that result in
Recall
2
2
( x  a )  x  2 ax  a
2
f ( x )  x  6 x  19
2
.
So
2
2
2
f ( x )  x  6 x  19
( x  3)  x  6 x  9
.
.
2
= ( x  3 )  10
The transformations are shift right 3 units and shift up 10 units.
Ex. List the transformations of
y  x
2
that result in
2
g ( x)  2 x  4 x  7
g (x)  2(x
Factoring out -2 from the first two terms we have
2
.
 2 x)  7
2
  2 ( x  1)  2  7
So g ( x )   2 ( x  1) 2  9 is the result of a left shift by 1 unit, a reflection about the
x-axis, a vertical stretching by a factor of 2 and last a shift up by 9 units. The first three
transformations may be done in any order but the vertical shift must be done last, since it
is not reflected or stretched.