Learning objectives

The student will perform parameter changes on a parent function. The student will observe the effects of the parameter changes on the transformed graph.

Real world application

Transforming a parent function can produce mathematical models that are useful in many different applications. Transformations provide a way to create relations that model relationships as diverse as population growth, projectile motion, constant rate of change, and inverse variation.

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Choose a type of transformation from the tabs below. Then select a parent function from the drop-down menu. Use the arrows to change the parameter(s) in the equation. Changing a parameter will have a predictable effect on the graph of the function.
 
Function:
af(x) = ax =
1
1
x
af(x) = ax2 =
1
1
x2
af(x) = a x  =
1
1
 x 
af(x) = a
1
x
=
1
1
1
x
af(x) = ax3 =  
1
1
x3
af(x) = a
=
1
1
af(x) = a2x =
1
1
2x
af(x) = a10x =
1
1
10x
af(x) = aex =
1
1
ex
af(x) = a|x| =
1
1
|x|
af(x) = alog2x =
1
1
log2x
af(x) = alogx =
1
1
logx
af(x) = alnx =
1
1
lnx
f(x) = x  
af(x) =
x
f(x) = √ x   
af(x) =
 x 
f(x) =
1
x
 
af(x) =
1
x
f(x) = |x|  
af(x) =
|x|
f(x) = logx  
af(x) =
logx
f(x) = lnx  
af(x) =
lnx
 
Function:
f(bx) = bx =
1
1
x
f(bx) = (bx)2 = (
1
1
x)2
f(bx) = √ bx  =  
1
1
x
f(bx) =
1
bx
=
1
(
1
1
x )
f(bx) = (bx)3 = (
1
1
x)3
f(bx) =
=  
1
1
x
f(bx) = 2bx = 2
1
1
x
f(bx) = 10bx = 10
1
1
x
f(bx) = ebx = e
1
1
x
f(bx) = |bx| = |
1
1
x|
f(bx) = log2 (bx) = log2(
1
1
x)
f(bx) = log(bx) = log(
1
1
x)
f(bx) = ln(bx) = ln(
1
1
x)
f(x) = x  
f(bx) =
x
f(x) = x2  
f(bx) = (
x)2
f(x) = √ x   
f(bx) =
x
f(x) =
1
x
 
f(bx) =
1
(
x)
f(x) = x3  
f(bx) = (
x)3
f(x) =
 
f(bx) =
x
f(x) = 2 x  
f(bx) = 2
x
f(x) = 10 x  
f(bx) = 10
x
f(x) = e x  
f(bx) = e
x
f(x) = | x |  
f(bx) = |
x |
f(x) = log2 x  
f(bx) = log2 (
x)
f(x) = log x  
f(bx) = log (
x)
f(x) = ln x  
f(bx) = ln (
x)
 
Function:
f(x c ) = xc =
  x
0
f(xc) = (xc)2 =
  (x
0
)2
f(xc) = √ xc  =
   x
0
f(xc) =
1
xc
=
 
1
x
0
f(xc) = (xc)3 =
  (x
0
)3
f(xc) =
=
  x
0
f(xc) = 2xc =
  2 x
0
f(xc) = 10xc =
  10 x
0
f(xc) = exc =
  e x
0
f(xc) = |xc| =
  |x
0
|
f(xc) = log2(xc) =
  log2(x
0
)
f(xc) = log(xc) =
  log(x
0
)
f(xc) = ln(xc) =
  ln(x
0
)
f(x) = x  
f(x – c) = x
f(x) = x2  
f(x – c) = x2
f(x) =
(
x
)
2
f(x) = √ x   
f(x – c) =  x 
f(x) =
1
x
 
f(x – c) =
1
x
f(x) = x3  
f(x) = x3
f(x – c) = (x
)3
f(x) =
 
f(x – c) =
x
f(x) = 2x  
f(x – c) = 2x
f(x) = 10x  
f(x – c) = 10x
f(x) = ex  
f(x – c) = ex
f(x) = |x|  
f(x – c) = |x|
f(x) = log2 x  
f(x – c) = log2
f(x) = log x  
f(x – c) = log 0
f(x) = ln x  
f(x – c) = ln 0
 
Function:
f(x) + d = x + d =
  x +
0
f(x) + d = x2 + d =
  x2 +
0
f(x) + d = √ x  + d =
   x +
0
f(x) + d =
1
x
+ d =
 
1
x
+
0
f(x) + d = x3 + d =
  x3 +
0
f(x) + d =
+ d  =
 
+
0
f(x) + d = 2x + d =
  2x +
0
f(x) + d = 10x + d =
  10x +
0
f(x) + d = ex + d =
  ex +
0
f(x) + d = |x| + d =
  |x| +
0
f(x) + d = log2(x) + d =
  log2(x) +
0
f(x) + d = log(x) + d =
  log(x) +
0
f(x) + d = ln(x) + d =
  ln(x) +
0
f(x) = x  
f(x) + d = x
f(x) = x2  
f(x) + d = x2
f(x) = √ x   
f(x) + d =  x 
f(x) =
1
x
 
f(x) + d =
1
x
f(x) = x3  
f(x) + d = x3
f(x) =
 
f(x) + d =
f(x) = 2x  
f(x) + d = 2x
f(x) = 10x  
f(x) + d = 10x
f(x) = ex  
f(x) + d = ex
f(x) = |x|  
f(x) + d = |x|
f(x) = log2(x)  
f(x) + d = log2(x)
f(x) = log(x)  
f(x) + d = log(x)
f(x) = ln(x)  
f(x) + d = ln(x)
 
Function:
a
f
b (x c )
+ d =
a
b (x c )
+ d =
1
1
1
1
(x
0
)
+
0
a
f
b (x c )
+ d =
a
b (x c )
2
+ d =
1
1
1
1
(x
0
)
2
+
0
a
f
b ( x c )
+ d =
a
b ( x c )
+ d =
 
1
1
 
1
1
( x
0
)
+
0
a
f
b (x c )
+ d =
a
1
b (x c )
+ d =
1
1
1
1
1
(x
0
)
+
0
a
f
b (x c )
+ d =
a
b (x c )
3
+ d =
1
1
1
1
(x
0
)
3
+
0
a
f
b (x c )
+ d =
a
 
b ( x c )
+ d =
1
1
 
1
1
( x
0
)
+
0
a
f
b (x c )
+ d =
a
2
b (x c )
+ d =
1
1
2
1
1
(x
0
)
+
0
a
f
b (x c )
+ d =
a
10
b (xc)
+ d =
1
1
10
1
1
(x
0
)
+
0
a
f
b (x c )
+ d =
a
e
b (xc)
+ d =
1
1
e
1
1
(x
0
)
+
0
a
f
b (x c )
+ d =
a
b (x c )
+ d =
1
1
1
1
(x
0
)
+
0
a
f
b (x c )
+ d =
a
log2
b (x c )
+ d =
1
1
log2
1
1
(x
0
)
+
0
a
f
b (x c )
+ d =
a
log
b (x c )
+ d =
1
1
log
1
1
(x
0
)
+
0
a
f
b (x c )
+ d =
a
ln
b (x c )
+ d =
1
1
ln
1
1
(x
0
)
+
0
f(x) = x  
(a(f(b(x – c)))) + d = ( ( ( x ) ) )
f(x) = x 2  
(a(f(b(x – c)))) + d = (
(
(
( x ) ) 2
)
)
f(x) = √ x   
(a(f(b(x – c)))) + d = ( (
 
(
( x ) )
) )
f(x) =
1
x
 
(a(f(b(x – c)))) + d = ( (
1
( x )
)
)
f(x) = x 3  
(a(f(b(x – c)))) + d = (
(
(
( x ) ) 3
)
)
f(x) =
 
(a(f(b(x – c)))) + d = ( (
 
(
( x ) )
) )
f(x) = 2 x  
(a(f(b(x – c)))) + d = ( ( 2
(
( x ) ) ) )
f(x) = 10 x  
(a(f(b(x – c)))) + d = ( ( 10
(
( x ) ) ) )
f(x) = e x  
(a(f(b(x – c)))) + d = ( ( e
(
( x ) ) ) )
f(x) = | x |  
(a(f(b(x – c)))) + d = ( ( |
(
( x ) ) | ) )
f(x) = log2 x  
(a(f(b(x – c)))) + d = ( ( log2
(
( x ) ) ) )
f(x) = log x  
(a(f(b(x – c)))) + d = ( ( log
(
( x ) ) ) )
f(x) = lnx  
(a(f(b(x – c)))) + d = ( ( ln
(
( x ) ) ) )