Section 2.5 Transformations of...
Transcript of Section 2.5 Transformations of...
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 12f(x) + 1.
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EXAMPLE: Given the graph of f(x) below, sketch the graph of 12f(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 is shown in the Figure below.
Sketch the graph of each function.
(a) y = f(2x) (b) y = f(12x)
EXAMPLE: The graph of y = f(x) is shown below.
Sketch the graph of each function:
(a) y = f(3x) (b) y = f(13x)8
EXAMPLE: The graph of y = f(x) is shown below.
Sketch the graph of each function:
(a) y = f(3x) (b) y = f(13x)
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) = cosx, then f(−x) = cos(−x) = cosx = 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) = sinx, then f(−x) = sin(−x) = − sinx = −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) ̸= ±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) ̸= ±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) ̸= ±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) ̸= ±f(1). Therefore f(x) = |x− 2| is neither even nor odd.
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