Transformation of Functions

You should know the graph of the following basic functions:

f(x) = x2

f(x) = x3

f(x) =√x

f(x) =1

x

f(x) = |x|

If we know the graph of a basic function f(x), we can draw the graph of morecomplicated functions, like−2f(−x+1), using the method of transformation offunctions. In order to do that, we do algebra to change f(x), one step at a time,to the new function. Each algebraic change causes a corresponding geometricalchange to the graph of the previous function. We discuss the different types ofgeometrical changes that would occur to a function when we perform certain typeof algebraic operation to a function:

Vertical Shift:

Given a function f(x), if c > 0 is a constant, then:

The graph of f(x) + c is the graph of f moved up c units.

The graph of f(x)− c is the graph of f moved down c units.

Example: Starting with f1(x) = x2, then f2(x) = f1(x) + 3 = x2 + 3. The graphof f2(x) = x2 + 3 is the graph of x2 moved up by 3 units.

x2 + 3

x2

Example: Starting with f1(x) = x2, we can obtain f2(x) = f1(x) − 1 = x2 − 1.The graph of x2 − 1 is the graph of x2 moved down by 1 unit.

x2 − 1

x2

Horizontal Shift

Again assuming c > 0.

The graph of f(x+ c) is the graph of f moved left c units. Notice that the graphof f actually moves to the left if we add a positive constant to the argument(input) of f .

The graph of f(x − c) is the graph of f moved right c units. Notice that thegraph of f actually moves to the right if we subtract a positive constant from theargument (input) of f .

Example: Starting with f1(x) =√x, we can obtain f2(x) = f1(x− 3) =

√x− 3.

The graph of√x− 3 is the graph of

√x moved to the right by 3 units. Notice

that the new graph is actually moved to the right when a negative constant isadded to the argument of a function.

√x

√x− 3

Example: Starting with f1(x) =√x, we can obtain f2(x) = f1(x + 1) =

√x + 1.

The graph of√x + 1 is the graph of

√x moved to the left by 3 units. Notice that

the new graph is actually moved to the left when a positive constant is added tothe argument of a function.

√x + 1

√x

Vertical stretch/compression

The graph of cf(x) is the graph of f stretched vertically (from the x−axis)by a factor of c if c > 1. To visualize a vertical stretch, imagine that you arepulling the graph of f away from the x−axis in both up and down directions.In a vertical stretch, the x−intercepts are unchanged.

Example: Starting with f1(x) =√x, we can obtain f2(x) = 2f1(x) = 2

√x. The

graph of 2√x is the graph of

√x stretched by a factor of 2 vertically.

2√x

√x

The graph of cf(x) is the graph of f compressed vertically by a factor of1

cif c < 1 To visualize a vertical compression, imagine you push the graph of ftoward the x−axis from both up and down directions. In a vertical compression,the x−intercepts are unchanged.

Example: Starting with f1(x) =√x, we can obtain f2(x) =

1

3f1(x) =

1

3

√x. The

graph of1

3

√x is the graph of

√3 compressed by a factor of 3 vertically.

1

3

√x

√x

Horizontal stretch/compression

The graph of f(cx) is the graph of f compressed horizontally by a factor of cif c > 1. To visualize a horizontal compression, imagine that you push the graphof the function toward the y−axis from both the left and the right hand side.In a horizontal compression, the y−intercept is unchanged.

Example: Starting with f1(x) =1

x, we can obtain f2(x) = f1(2x) =

1

2x. The

graph of1

2xis the graph of

1

xcompressed by a factor of 2 horizontally.

1

x

1

2x

The graph of f(cx) is the graph of f strectched horizontally by a factor of1

cif c < 1. To visualize a horizontal stretch, imagine that you pull the graph of thefunction away from the y−axis from both the left and the right hand side. Ina horizontal stretch, the y−intercept is unchanged.

Example: Starting with f1(x) = x3, we can obtain f2(x) = f1

(1

2x

)=

(1

2x

)3

.

The graph of

(1

2x

)3

is the graph of x3 stretched by a factor of 2 horizontally.

(1

2x

)3

x3

Again, pay attention to the fact that, in a horizontal stretch/compression that,if the constant c multiplied to x is greater than 1, the resulting graph is actuallycompressed by a factor of 1/c. On the other hand, if c is less than 1, theresulting graph is stretched by a factor of c.

Reflection

The graph of −f(x) is the graph of f reflected with respect to the x-axis. To drawa reflection with respect to the x−axis, draw the resulting graph by reflectingeverything in the original graph using the x−axis as the mirror.

Example: Starting with f1(x) = x2, we can obtain f2(x) = −f1(x) = −x2. Thegraph of −x2 is the graph of x2 reflected with respect to the x−axis:

x2

−x2

Example: Starting with f1(x) =√x, we can obtain f2(x) = −f1(x) = −

√x. The

graph of −√x is the graph of

√x reflected with respect to the x−axis:

√x

−√x

The graph of f(−x) is the graph of f reflected with respect to the y-axis. To drawa reflection with respect to the y−axis, draw the resulting graph by reflectingeverything in the original graph using the y−axis as the mirror.

Example: Starting with f1(x) =√x, we can obtain f2(x) = f1(−x) =

√−x. The

graph of√−x is the graph of

√x reflected with respect to the y−axis:

√x√

−x

Example: Starting with f1(x) = x3, we can obtain f2(x) = f1(−x) = (−x)3. Thegraph of (−x)3 is the graph of x3 reflected with respect to the y−axis:

x3(−x)3

To draw the graph of a complicated function, start with a simple function whosegraph is known. Algebraically change the known function one step at the time,by adding or subtracting a constant, multiply by a constant ...etc, to the functionin question. The geometry must be dictated by the algebraic changes, notthe other way around. You must first finish the algebra as to how the simplefunction changes to the complicated function, then do the geometry.

Example:

Graph f(x) =√−2x + 1− 1

Ans:

We start with f1(x) =√x. We know the graph of f1(x)

Let f2(x) = f1(x + 1) =√x + 1. The graph of f2(x) can be obtained from the

graph of f1(x) by a horizontal shift one unit to the left.

f1(x) =√x

f2(x) =√x + 1

f3(x) = f2(2x) =√

2x + 1. The graph of f3(x) can be obtained by horizontallycompress f2(x) by a factor of 2.

f3(x) =√

2x + 1

f2(x) =√x + 1

f4(x) = f3(−x) =√

2(−x) + 1 =√−2x + 1. The graph of f4(x) can be obtained

by reflecting the graph of f3(x) with respect to the y−axis.

f3(x) =√

2x + 1f4(x) =√−2x + 1

f5(x) = f4(x)− 1 =√−2x + 1− 1. The graph of f5(x) is obtained by vertically

shift f4(x) one unit down.

f4(x) =√−2x + 1

f5(x) =√−2x + 1− 1

The graph of f is obtained from the graph of√x by moving left one unit, followed

by horizontal compression by a factored of 2, reflected with respect to the y axis,then moved down one unit.

In the previous example, it is possible to change the order of how you wouldchange the functions, but you must let the algebra dictate the geometry.

Second approach to the same problem:

Let g1(x) =√x

g2(x) = g1(2x) =√

2x. The graph of g2(x) can be obtained by horizontallycompress g1(x) by a factor of 2.

g1(x) =√x

g2(x) =√

2x

g3(x) = g2(−x) =√

2(−x) =√−2x. The graph of g3(x) can be obtained by

reflecting g2(x) with respect to the y−axis.

g3(x) =√−2x g2(x) =

√2x

g4(x) = g3

(x− 1

2

)=

√−2

(x− 1

2

)=√−2x + 1. The graph of g4(x) is

obtained by moving g3(x) half unit to the right.

g4(x) =√−2x + 1

g3(x) =√−2x

Notice in this step that we added −1

2to the argument of g3, not 1. The reason

is because if we added 1 to the argument of g3, the distributive property wouldgive us

√−2x− 2, which is not what we want. You must always actually work

out the algebra of how each subsequent function is obtained from the previousfunction.

g5(x) = g4(x) − 1 =√−2x + 1 − 1. The graph of g5(x) is obtained by moving

the graph of g4(x) 1 unit down.

g4(x) =√−2x + 1

g5(x) =√−2x + 1− 1

The graph of f can be obtained from the graph of√x by first horizontally

compress by a factor of 2, reflected with respect to y, moved to the right by1

2unit, then moved down 1 unit.

Example: Draw the graph of − 2

3x− 1

Ans: We start with f1(x) =1

x

f2(x) = f1(x − 1) =1

x− 1. The graph of g2(x) can be obtained by horizontally

shift f1(x) 1 unit to the right.

f1(x) =1

x

f2(x) =1

x− 1

f3(x) = f2(3x) =1

(3x)− 1=

1

3x− 1. The graph of f3(x) can be obtained by

horizontally compress f2(x) by a factor of 3.

f2(x) =1

x− 1

f3(x) =1

3x− 1

f4(x) = 2f3(x) = 2

(1

3x− 1

)=

2

3x− 1. The graph of f4(x) can be obtained by

vertically stretch f3(x) by a factor of 2.

f4(x) =2

3x− 1

f3(x) =1

3x− 1

f5(x) = −f4(x) = − 2

3x− 1. The graph of f4(x) can be obtained by reflecting

f4(x) with respect to the x−axis.

f4(x) =2

3x− 1f5(x) = − 2

3x− 1

The method of transformation of functions can be applied to any function ingeneral. As long as we have the graph of f , we can draw the graphs of otherfunctions that are transformed from f . We do not need to have an expressionexplicitely defined for f .

Example: The graph of a function f is drawn below. Draw the graph of

−2f(−2x + 1) + 1

f(x)

Ans: We change f(x) using algebra and in each stage keep track of the changewe made and the corresponding geometric changes:

f1(x) = f(x + 1). The graph of f1(x) is obtained by horizontally shift the graphof f(x) 1 unit to left.

f(x)f1(x)

f2(x) = f1(2x) = f(2x + 1). The graph of f2(x) is obtained by horizontallycompress f1(x) by factor of 2.

f1(x)f2(x)

f3(x) = f2(−x) = f(2(−x) + 1) = f(−2x + 1). The graph of f3(x) is obtainedby reflecting the graph of f2(x) with respect to the y−axis.

f3(x) f2(x)

f4(x) = 2f3(x) = 2f(−2x + 1). The graph of f4(x) is obtained by verticallystretch f3(x) by factor of 2

f3(x)

f4(x)

f5(x) = −f4(x) = −2f(−2x + 1). The graph of f5(x) is obtained by reflectingthe graph of f4(x) with respect to x−axis.

f5(x)

f4(x)

f6(x) = f5(x)+1 = −2f(−2x+1)+1. The graph of f6(x) is obtained by verticallyshift f5(x) 1 unit up.

f6(x)

f5(x)

Below is a graph of the original function f and the final result, −2f(−2x+ 1) + 1

f(x)

−2f(−2x + 1) + 1