Section 2.5 Transformations of Functions

Vertical Shifting

EXAMPLE: Use the graph of f(x) = x2 to sketch the graph of each function.

(a) g(x) = x2 + 3 (b) h(x) = x2 − 2

EXAMPLE: Use the graph of f(x) = x3 − 9x to sketch the graph of each function.

(a) g(x) = x3 − 9x+ 10 (b) h(x) = x3 − 9x− 20

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EXAMPLE: Use the graph of f(x) = x3 − 9x to sketch the graph of each function.

(a) g(x) = x3 − 9x+ 10 (b) h(x) = x3 − 9x− 20

Horizontal Shifting

EXAMPLE: Use the graph of f(x) = x2 to sketch the graph of each function.

(a) g(x) = (x+ 4)2 (b) h(x) = (x− 2)2

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EXAMPLE: Use the graph of f(x) = x2 to sketch the graph of each function.

(a) g(x) = (x+ 4)2 (b) h(x) = (x− 2)2

EXAMPLE: How is the graph of y = f(x− 3) + 2 obtained from the graph of f?

Answer: The graph shifts right 3 units, then shifts upward 2 units.

EXAMPLE: Sketch the graph of f(x) =√x− 3 + 4.

Reflecting Graphs

EXAMPLE: Sketch the graph of each function.

(a) f(x) = −x2 (b) g(x) =√−x

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EXAMPLE: Sketch the graph of each function.

(a) f(x) = −x2 (b) g(x) =√−x

EXAMPLE: Given the graph of f(x) =√x, use transformations to graph f(x) = 1−

√1 + x.

Step 1: f(x) =√x Step 2: f(x) =

√1 + x (horizontal shift)

Step 3: f(x) = −√1 + x (reflection) Step 4: f(x) = 1−

√1 + x (vertical shift)

EXAMPLE: Given the graph of f(x) =√x, use transformations to graph f(x) = 1−

√1− x.

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EXAMPLE: Given the graph of f(x) =√x, use transformations to graph f(x) = 1−

√1− x.

Step 1: f(x) =√x Step 2: f(x) =

√1 + x (horizontal shift)

Step 3: f(x) = −√1 + x (reflection) Step 4: f(x) = 1−

√1 + x (vertical shift)

Step 5: f(x) = 1−√1− x (reflection about the y-axis)

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Vertical Stretching and Shrinking

EXAMPLE: Use the graph of f(x) = x2 to sketch the graph of each function.

(a) g(x) = 3x2 (b) h(x) =1

3x2

EXAMPLE: Given the graph of f(x) below, sketch the graph of 1

2f(x) + 1.

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EXAMPLE: Given the graph of f(x) below, sketch the graph of 1

2f(x) + 1.

EXAMPLE: Sketch the graph of the function f(x) = 1− 2(x− 3)2.

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Horizontal Stretching and Shrinking

EXAMPLE: The graph of y = f(x) is shown below.

Sketch the graph of each function.

(a) y = f(3x) (b) y = f(

1

3x)

EXAMPLE: The graph of y = f(x) is shown below.

Sketch the graph of each function.

(a) y = f(2x) (b) y = f(

1

2x)

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EXAMPLE: The graph of y = f(x) is shown below.

Sketch the graph of each function.

(a) y = f(2x) (b) y = f(

1

2x)

Even and Odd Functions

REMARK: Any function is either even, or odd, or neither.

PROPERTY: Graphs of even functions are symmetric with respect to the y-axis. Graphs ofodd functions are symmetric with respect to the origin.

IMPORTANT: Do NOT confuse even/odd functions and even/odd integers!

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EXAMPLES:

1. Functions f(x) = x2, x4, x8, x4 − x2, x2 + 1, |x|, cos x, etc. are even. In fact,

• if f(x) = x2, then f(−x) = (−x)2 = x2 = f(x)

• if f(x) = x4, then f(−x) = (−x)4 = x4 = f(x)

• if f(x) = x8, then f(−x) = (−x)8 = x8 = f(x)

• if f(x) = x4 − x2, then f(−x) = (−x)4 − (−x)2 = x4 − x2 = f(x)

• if f(x) = x2 + 1, then f(−x) = (−x)2 + 1 = x2 + 1 = f(x)

• if f(x) = |x|, then f(−x) = | − x| = |x| = f(x)

• if f(x) = cos x, then f(−x) = cos(−x) = cos x = f(x)

One can see that graphs of all these functions are symmetric with respect to the y-axis.

2. Functions f(x) = x, x3, x5, x3 − x7, sin x, etc. are odd. In fact,

• if f(x) = x, then f(−x) = −x = −f(x)

• if f(x) = x3, then f(−x) = (−x)3 = −x3 = −f(x)

• if f(x) = x5, then f(−x) = (−x)5 = −x5 = −f(x)

• if f(x) = x3 − x7, then f(−x) = (−x)3 − (−x)7 = −x3 + x7 = −(x3 − x7) = −f(x)

• if f(x) = sin x, then f(−x) = sin(−x) = − sin x = −f(x)

One can see that graphs of all these functions are symmetric with respect to the origin.

3. Functions f(x) = x+ 1, x3 + x2, x5 − 2, |x− 2| etc. are neither even nor odd. In fact,

• if f(x) = x+ 1, then

f(−1) = −1 + 1 = 0, f(1) = 1 + 1 = 2

so f(−1) 6= ±f(1). Therefore f(x) = x+ 1 is neither even nor odd.

• if f(x) = x3 + x2, then

f(−1) = (−1)3 + (−1)2 = −1 + 1 = 0, f(1) = 13 + 12 = 2

so f(−1) 6= ±f(1). Therefore f(x) = x3 + x2 is neither even nor odd.

• if f(x) = x5 − 2, then

f(−1) = (−1)5 − 2 = −1− 2 = −3, f(1) = 15 − 2 = 1− 2 = −1

so f(−1) 6= ±f(1). Therefore f(x) = x5 − 2 is neither even nor odd.

• if f(x) = |x− 2|, then

f(−1) = | − 1− 2| = | − 3| = 3, f(1) = |1− 2| = | − 1| = 1

so f(−1) 6= ±f(1). Therefore f(x) = |x− 2| is neither even nor odd.

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