4. Using image manipulation software, we can drag and
drop or stretch images.
Transformations of Graphs
5. Using image manipulation software, we can drag and
drop or stretch images. For example, from the original
image below,
Transformations of Graphs
y = f(x)
6. Using image manipulation software, we can drag and
drop or stretch images. For example, from the original
image below, we can elongate it,
stretch
Transformations of Graphs
y = f(x)
7. Using image manipulation software, we can drag and
drop or stretch images. For example, from the original
image below, we can elongate it, drag and lower it,
stretch
lower
Transformations of Graphs
y = f(x)
8. Using image manipulation software, we can drag and
drop or stretch images. For example, from the original
image below, we can elongate it, drag and lower it,
stretch
lower
vertically
reflected
and reflect it vertically to
create another
pattern.
Transformations of Graphs
y = f(x)
9. Using image manipulation software, we can drag and
drop or stretch images. For example, from the original
image below, we can elongate it, drag and lower it,
stretch
lower
vertically
reflected
and reflect it vertically to
create another
pattern.
If the original image
is the graph of the
function y = f(x),
then these
transformations
can be tracked
easily with the
notation of functions.
Transformations of Graphs
y = f(x)
10. Given a function y = f(x) and P = (x, y = f(x))
a generic point on the graph as shown,
the output f(x) = y represents the height of the point.
x
P = (x, f(x))
f(x) = ht y= f(x)
x
Vertical Translations
11. Given a function y = f(x) and P = (x, y = f(x))
a generic point on the graph as shown,
the output f(x) = y represents the height of the point.
Hence expressions in terms of f(x)
may be translated precisely into
the corresponding manipulation
of the graph.
x
P = (x, f(x))
f(x) = ht y= f(x)
x
Vertical Translations
12. Given a function y = f(x) and P = (x, y = f(x))
a generic point on the graph as shown,
the output f(x) = y represents the height of the point.
Hence expressions in terms of f(x)
may be translated precisely into
the corresponding manipulation
of the graph.
Vertical Translations x
P = (x, f(x))
f(x) = ht y= f(x)
x
Vertical Translations
13. Given a function y = f(x) and P = (x, y = f(x))
a generic point on the graph as shown,
the output f(x) = y represents the height of the point.
x
P = (x, f(x))
f(x) = ht y= f(x)
Hence expressions in terms of f(x)
may be translated precisely into
the corresponding manipulation
of the graph.
Vertical Translations
Changing the y–coordinate to
f(x) + 3 moves P vertically up 3 units.
(x, f(x)+3)
x
Vertical Translations
14. Given a function y = f(x) and P = (x, y = f(x))
a generic point on the graph as shown,
the output f(x) = y represents the height of the point.
x
P = (x, f(x))
f(x) = ht y= f(x)
Hence expressions in terms of f(x)
may be translated precisely into
the corresponding manipulation
of the graph.
Vertical Translations
Changing the y–coordinate to
f(x) + 3 moves P vertically up 3 units.
(x, f(x)+3)
y= f(x) + 3
Hence setting y = f(x) + 3 to all the points on the
graph means to move the entire graph 3 units up as
shown.
x
Vertical Translations
15. Given a function y = f(x) and P = (x, y = f(x))
a generic point on the graph as shown,
the output f(x) = y represents the height of the point.
x
P = (x, f(x))
f(x) = ht y= f(x)
Hence expressions in terms of f(x)
may be translated precisely into
the corresponding manipulation
of the graph.
Vertical Translations
Changing the y–coordinate to
f(x) + 3 moves P vertically up 3 units.
Hence setting y = f(x) + 3 to all the points on the
graph means to move the entire graph 3 units up as
shown. Likewise changing the y–coordinate to f(x)
– 3 corresponds to moving y = f(x) down 3 units.
(x, f(x)+3)
(x, f(x)–3)
y= f(x) – 3
y= f(x) + 3
x
Vertical Translations
16. The graph of (x, y = f(x) + c) is the vertical translation
of the graph (x, f(x)).
Vertical Translations
Vertical Translations
17. The graph of (x, y = f(x) + c) is the vertical translation
of the graph (x, f(x)).
Vertical Translations
P = (x, f(x)) y= f(x)
(x, f(x)+c)
where c > 0
(x, f(x)+c)
where c < 0
Vertical Translations
18. The graph of (x, y = f(x) + c) is the vertical translation
of the graph (x, f(x)).
Vertical Translations
P = (x, f(x)) y= f(x)
(x, f(x)+c)
where c > 0
(x, f(x)+c)
where c < 0
Here are the graphs of:
y = f(x) = x2 vs.
y = f(x) + 5 = x2 + 5
y = x2
y = x2 + 5
(0, 0)
(0, 5)
Vertical Translations
x
19. Vertical Translations
P = (x, f(x)) y= f(x)
(x, f(x)+c)
where c > 0
(x, f(x)+c)
where c < 0
Here are the graphs of:
y = f(x) = x2 vs.
y = f(x) + 5 = x2 + 5
y = x2
y = x2 + 5
(0, 0)
(0, 5)
x
Vertical Translations
The graph of (x, y = f(x) + c) is the vertical translation
of the graph (x, f(x)). Assuming c > 0,
move the graph (x, f(x)) up to
obtain the graph (x, f(x) + c).
(0, 5)
20. Vertical Translations
P = (x, f(x)) y= f(x)
(x, f(x)+c)
where c > 0
(x, f(x)+c)
where c < 0
Here are the graphs of:
y = f(x) = x2 vs.
y = f(x) + 5 = x2 + 5
y = f(x) = x2 vs.
y = f(x) – 5 = x2 – 5
y = x2
y = x2 + 5
y = x2 – 5
y = x2
(0, 0)
(0, 5)
(0, 0)
(0, –5)
x
x
Vertical Translations
The graph of (x, y = f(x) + c) is the vertical translation
of the graph (x, f(x)). Assuming c > 0,
move the graph (x, f(x)) up to
obtain the graph (x, f(x) + c).
(0, 5)
21. The graph of (x, y = f(x) + c) is the vertical translation
of the graph (x, f(x)). Assuming c > 0,
move the graph (x, f(x)) up to
obtain the graph (x, f(x) + c).
Vertical Translations
move the graph (x, f(x)) down
to obtain the graph (x, f(x) – c).
P = (x, f(x)) y= f(x)
(x, f(x)+c)
where c > 0
(x, f(x)+c)
where c < 0
Here are the graphs of:
y = f(x) = x2 vs.
y = f(x) + 5 = x2 + 5
y = f(x) = x2 vs.
y = f(x) – 5 = x2 – 5
y = x2
y = x2 + 5
y = x2 – 5
y = x2
(0, 0)
(0, 5)
(0, 0)
(0, –5)
x
x
Vertical Translations
22. Given a function y = f(x) and P = (x, y = f(x))
a generic point on the graph as shown.
Vertical Stretches and Compressions
P = (x, f(x))
y= f(x)
Vertical Stretches and Compressions
23. Given a function y = f(x) and P = (x, y = f(x))
a generic point on the graph as shown.
Changing the y–coordinate to 3f(x)
would triple the height of the point P.
Vertical Stretches and Compressions
P = (x, f(x))
y= f(x)
Vertical Stretches and Compressions
24. Given a function y = f(x) and P = (x, y = f(x))
a generic point on the graph as shown.
Changing the y–coordinate to 3f(x)
would triple the height of the point P.
Vertical Stretches and Compressions
P = (x, f(x))
y= f(x)
Hence setting y = 3f(x) to all the
points on the graph means to
elongate the entire graph 3 times
while the x–intercepts (x, 0)’s
remain fixed because 3(0) = 0.
y= 3f(x)
Vertical Stretches and Compressions
25. Given a function y = f(x) and P = (x, y = f(x))
a generic point on the graph as shown.
Changing the y–coordinate to 3f(x)
would triple the height of the point P.
Vertical Stretches and Compressions
P = (x, f(x))
y= f(x)
Hence setting y = 3f(x) to all the
points on the graph means to
elongate the entire graph 3 times
while the x–intercepts (x, 0)’s
remain fixed because 3(0) = 0.
y= 3f(x)
Vertical Stretches and Compressions
26. Given a function y = f(x) and P = (x, y = f(x))
a generic point on the graph as shown.
Changing the y–coordinate to 3f(x)
would triple the height of the point P.
Vertical Stretches and Compressions
P = (x, f(x))
y= f(x)
Hence setting y = 3f(x) to all the
points on the graph means to
elongate the entire graph 3 times
while the x–intercepts (x, 0)’s
remain fixed because 3(0) = 0.
Likewise setting y = (1/3)f(x) would
compress the entire graph to a third of it’s
original size while the x–intercepts
or (x, 0)’s remain fixed.
y= 3f(x)
Vertical Stretches and Compressions
27. Given a function y = f(x) and P = (x, y = f(x))
a generic point on the graph as shown.
Changing the y–coordinate to 3f(x)
would triple the height of the point P.
Vertical Stretches and Compressions
P = (x, f(x))
y= f(x)
Hence setting y = 3f(x) to all the
points on the graph means to
elongate the entire graph 3 times
while the x–intercepts (x, 0)’s
remain fixed because 3(0) = 0.
Likewise setting y = (1/3)f(x) would
compress the entire graph to a third of it’s
original size while the x–intercepts
or (x, 0)’s remain fixed.
y= 3f(x)
y= f(x)/3
Vertical Stretches and Compressions
28. Vertical Stretches and Compressions
Assuming c > 0, the graph of y = cf(x) is the
vertical-stretch or compression of y = f(x).
Vertical Stretches and Compressions
29. Vertical Stretches and Compressions
Assuming c > 0, the graph of y = cf(x) is the
vertical-stretch or compression of y = f(x).
Here are the graphs of:
y = f(x) = 4 – x2 vs.
y = 3f(x) = 3(4 – x2)
y = 4 – x2
y = 3(4 – x2)
(0, 4)
(0, 12)
(–2, 0) (2, 0) x
Vertical Stretches and Compressions
30. Vertical Stretches and Compressions
Assuming c > 0, the graph of y = cf(x) is the
vertical-stretch or compression of y = f(x).
If c > 1, it is a vertical stretch by a factor of c.
Here are the graphs of:
y = f(x) = 4 – x2 vs.
y = 3f(x) = 3(4 – x2)
y = 4 – x2
y = 3(4 – x2)
(0, 4)
(0, 12)
(–2, 0) (2, 0)
c = 3
x
Vertical Stretches and Compressions
31. Vertical Stretches and Compressions
Assuming c > 0, the graph of y = cf(x) is the
vertical-stretch or compression of y = f(x).
If c > 1, it is a vertical stretch by a factor of c.
Here are the graphs of:
y = f(x) = 4 – x2 vs.
y = 3f(x) = 3(4 – x2)
y = f(x) = 4 – x2 vs.
y = f(x)/2 = (4 – x2)/2
y = 4 – x2
y = 3(4 – x2)
y = 4 – x2
y = (4 – x2)/2
(0, 4)
(0, 12)
(0, 4)
(0, 2)
(–2, 0) (2, 0)
(–2, 0) (2, 0)
c = 3 c = 1/2
x
x
Vertical Stretches and Compressions
32. Vertical Stretches and Compressions
Assuming c > 0, the graph of y = cf(x) is the
vertical-stretch or compression of y = f(x).
If c > 1, it is a vertical stretch by a factor of c.
If 0 < c < 1, it is a vertical compression by a factor of c.
Here are the graphs of:
y = f(x) = 4 – x2 vs.
y = 3f(x) = 3(4 – x2)
y = f(x) = 4 – x2 vs.
y = f(x)/2 = (4 – x2)/2
y = 4 – x2
y = 3(4 – x2)
y = 4 – x2
y = (4 – x2)/2
(0, 4)
(0, 12)
(0, 4)
(0, 2)
(–2, 0) (2, 0)
(–2, 0) (2, 0)
c = 3 c = 1/2
x
x
Vertical Stretches and Compressions
33. Changing the y-coordinate to –f(x) reflects the point P
vertically across the x–axis to Q(x, –f(x)) as shown.
P = (x, f(x))
y= f(x)
x
Vertical Stretches and Compressions
34. Changing the y-coordinate to –f(x) reflects the point P
vertically across the x–axis to Q(x, –f(x)) as shown.
P = (x, f(x))
y= f(x)
Q = (x, –f(x))
x
Vertical Stretches and Compressions
35. Changing the y-coordinate to –f(x) reflects the point P
vertically across the x–axis to Q(x, –f(x)) as shown.
P = (x, f(x))
y= f(x)
Hence setting y = –f(x) to all the
points on the graph means to
reflect the entire graph vertically
across the x–axis.
y= –f(x)Q = (x, –f(x))
x
Vertical Stretches and Compressions
36. Changing the y-coordinate to –f(x) reflects the point P
vertically across the x–axis to Q(x, –f(x)) as shown.
P = (x, f(x))
y= f(x)
Hence setting y = –f(x) to all the
points on the graph means to
reflect the entire graph vertically
across the x–axis. Hence setting
y = –cf(x) = c*(–f(x)) to all the
points means to reflect the
entire graph (x, f(x)) vertically,
then stretch the reflection
by a factor of c.
y= –f(x)Q = (x, –f(x))
x
Vertical Stretches and Compressions
37. Changing the y-coordinate to –f(x) reflects the point P
vertically across the x–axis to Q(x, –f(x)) as shown.
P = (x, f(x))
y= f(x)
Hence setting y = –f(x) to all the
points on the graph means to
reflect the entire graph vertically
across the x–axis. Hence setting
y = –cf(x) = c*(–f(x)) to all the
points means to reflect the
entire graph (x, f(x)) vertically,
then stretch the reflection
by a factor of c.
y= –f(x)
y= –2f(x)
Q = (x, –f(x))
(x, –2f(x))
x
Vertical Stretches and Compressions
38. Changing the y-coordinate to –f(x) reflects the point P
vertically across the x–axis to Q(x, –f(x)) as shown.
P = (x, f(x))
y= f(x)
Hence setting y = –f(x) to all the
points on the graph means to
reflect the entire graph vertically
across the x–axis. Hence setting
y = –cf(x) = c*(–f(x)) to all the
points means to reflect the
entire graph (x, f(x)) vertically,
then stretch the reflection
by a factor of c. The order of applying stretching vs.
reflecting does not matter, “reflect then stretch” or
“stretch then reflect” yields the same result.
This is not the case for “stretch” vs. “vertical shift”.
y= –f(x)
y= –2f(x)
Q = (x, –f(x))
(x, –2f(x))
x
Vertical Stretches and Compressions
39. Example A.
a. Given the graph of y = f(x),
graph y = g(x) = –2f(x) + 3
(–3, 1)
(–1, –1)
(1, 1) (2, 1)
x
Vertical Stretches and Compressions
40. Example A.
a. Given the graph of y = f(x),
graph y = g(x) = –2f(x) + 3
(–3, 1)
(–1, –1)
(1, 1) (2, 1)
“–2f(x)" corresponds stretching the graph by a factor
of 2, then reflecting the entire graph across the x axis.
Afterwards move the graph vertically up by 3.
x
Vertical Stretches and Compressions
41. Example A.
a. Given the graph of y = f(x),
graph y = g(x) = –2f(x) + 3
(–3, 1)
(–1, –1)
(1, 1) (2, 1)
“–2f(x)" corresponds stretching the graph by a factor
of 2, then reflecting the entire graph across the x axis.
Afterwards move the graph vertically up by 3.
To draw the graph, track the “important points” on the
graph.
x
Vertical Stretches and Compressions
42. Example A.
a. Given the graph of y = f(x),
graph y = g(x) = –2f(x) + 3
(–3, 1)
(–1, –1)
(1, 1) (2, 1)
“–2f(x)" corresponds stretching the graph by a factor
of 2, then reflecting the entire graph across the x axis.
Afterwards move the graph vertically up by 3.
To draw the graph, track the “important points” on the
graph. Plug in their x–values to find the corresponding
y–values, then plot their destinations.
x
Vertical Stretches and Compressions
43. Example A.
a. Given the graph of y = f(x),
graph y = g(x) = –2f(x) + 3
(–3, 1)
(–1, –1)
(1, 1) (2, 1)
“–2f(x)" corresponds stretching the graph by a factor
of 2, then reflecting the entire graph across the x axis.
Afterwards move the graph vertically up by 3.
To draw the graph, track the “important points” on the
graph. Plug in their x–values to find the corresponding
y–values, then plot their destinations.
(–3, 1)
x
Vertical Stretches and Compressions
y = g(x) = –2f(x) + 3y = f(x)
44. Example A.
a. Given the graph of y = f(x),
graph y = g(x) = –2f(x) + 3
(–3, 1)
(–1, –1)
(1, 1) (2, 1)
“–2f(x)" corresponds stretching the graph by a factor
of 2, then reflecting the entire graph across the x axis.
Afterwards move the graph vertically up by 3.
To draw the graph, track the “important points” on the
graph. Plug in their x–values to find the corresponding
y–values, then plot their destinations.
(–3, 1) (–3, –2f(–3) + 3 = 1)
x
Vertical Stretches and Compressions
y = g(x) = –2f(x) + 3y = f(x)
45. Example A.
a. Given the graph of y = f(x),
graph y = g(x) = –2f(x) + 3
(–3, 1)
(–1, –1)
(1, 1) (2, 1)
“–2f(x)" corresponds stretching the graph by a factor
of 2, then reflecting the entire graph across the x axis.
Afterwards move the graph vertically up by 3.
To draw the graph, track the “important points” on the
graph. Plug in their x–values to find the corresponding
y–values, then plot their destinations.
(–3, 1) (–3, –2f(–3) + 3 = 1)
(–1, –1) (–1, –2f(–1) + 3 = 5)
x
Vertical Stretches and Compressions
y = g(x) = –2f(x) + 3y = f(x)
46. Example A.
a. Given the graph of y = f(x),
graph y = g(x) = –2f(x) + 3
(–3, 1)
(–1, –1)
(1, 1) (2, 1)
“–2f(x)" corresponds stretching the graph by a factor
of 2, then reflecting the entire graph across the x axis.
Afterwards move the graph vertically up by 3.
To draw the graph, track the “important points” on the
graph. Plug in their x–values to find the corresponding
y–values, then plot their destinations.
(–3, 1) (–3, –2f(–3) + 3 = 1)
(–1, –1) (–1, –2f(–1) + 3 = 5)
(1, 1) (1, –2f(1) + 3 = 1)
(2, 1) (2, –2f(2) + 3 = 1)
x
Vertical Stretches and Compressions
y = g(x) = –2f(x) + 3y = f(x)
47. Example A.
a. Given the graph of y = f(x),
graph y = g(x) = –2f(x) + 3
(–3, 1)
(–1, –1)
(1, 1) (2, 1)
“–2f(x)" corresponds stretching the graph by a factor
of 2, then reflecting the entire graph across the x axis.
Afterwards move the graph vertically up by 3.
To draw the graph, track the “important points” on the
graph. Plug in their x–values to find the corresponding
y–values, then plot their destinations.
(–3, 1) (–3, –2f(–3) + 3 = 1)
(–1, –1) (–1, –2f(–1) + 3 = 5)
(1, 1) (1, –2f(1) + 3 = 1)
(2, 1) (2, –2f(2) + 3 = 1)
(–3, 1)
(–1, 5)
(1, 1)
(2, 1)
x
x
Vertical Stretches and Compressions
y = g(x) = –2f(x) + 3y = f(x) Graph of
y = g(x)
50. Horizontal Translations
x
y= f(x)
x
Let y = f(x) be as shown.
Let’s define g(x) = f(x + 1),
and the goal is to graph g(x).
For the point x, the output is
y = g(x) = f(x + 1),
51. Horizontal Translations
x
(x+1, f(x+1))
ht =f(x+1)
y= f(x)
x
Let y = f(x) be as shown.
Let’s define g(x) = f(x + 1),
and the goal is to graph g(x).
For the point x, the output is
y = g(x) = f(x + 1),
x+1
52. Horizontal Translations
x
(x+1, f(x+1))
ht =f(x+1)
y= f(x)
x
Let y = f(x) be as shown.
Let’s define g(x) = f(x + 1),
and the goal is to graph g(x).
For the point x, the output is
y = g(x) = f(x + 1),
so the point (x, g(x) = f(x +1))
is on the graph of g(x). x+1
53. Horizontal Translations
x
(x+1, f(x+1))
ht =f(x+1)
y= f(x)
x
Let y = f(x) be as shown.
Let’s define g(x) = f(x + 1),
and the goal is to graph g(x).
For the point x, the output is
y = g(x) = f(x + 1),
so the point (x, g(x) = f(x +1))
is on the graph of g(x). x+1
(x, f(x +1))
54. Horizontal Translations
x
(x+1, f(x+1))
ht =f(x+1)
y= f(x)
x
Let y = f(x) be as shown.
Let’s define g(x) = f(x + 1),
and the goal is to graph g(x).
For the point x, the output is
y = g(x) = f(x + 1),
so the point (x, g(x) = f(x +1))
is on the graph of g(x). ux+1
(x, f(x +1))
Likewise if the input is u,
55. Horizontal Translations
x
(x+1, f(x+1))
ht =f(x+1)
y= f(x)
x
Let y = f(x) be as shown.
Let’s define g(x) = f(x + 1),
and the goal is to graph g(x).
For the point x, the output is
y = g(x) = f(x + 1),
so the point (x, g(x) = f(x +1))
is on the graph of g(x). u
(u+1, f(u+1))
ht = f(u+1)
u+1x+1
(x, f(x +1))
Likewise if the input is u,
then y = g(u) = f(u + 1)
56. Horizontal Translations
x
(x+1, f(x+1))
ht =f(x+1)
y= f(x)
x
Let y = f(x) be as shown.
Let’s define g(x) = f(x + 1),
and the goal is to graph g(x).
For the point x, the output is
y = g(x) = f(x + 1),
so the point (x, g(x) = f(x +1))
is on the graph of g(x). u
(u+1, f(u+1))
ht = f(u+1)
u+1x+1
(x, f(x +1))
(u, f(u +1))
Likewise if the input is u,
then y = g(u) = f(u + 1) so the point (u, f(u +1)) is on
the graph of g(x).
57. Hence to obtain the graph of y = f(x + 1),
shift the entire graph of y = f(x) left by 1 unit.
Horizontal Translations
x
(x+1, f(x+1))
ht =f(x+1)
y= f(x)
x
Let y = f(x) be as shown.
Let’s define g(x) = f(x + 1),
and the goal is to graph g(x).
For the point x, the output is
y = g(x) = f(x + 1),
so the point (x, g(x) = f(x +1))
is on the graph of g(x). u
(u+1, f(u+1))
ht = f(u+1)
u+1x+1
y = g(x) or
y = f(x + 1)
(x, f(x +1))
(u, f(u +1))
Likewise if the input is u,
then y = g(u) = f(u + 1) so the point (u, f(u +1)) is on
the graph of g(x).
Shifting left by 1
58. Hence to obtain the graph of y = f(x + 1),
shift the entire graph of y = f(x) left by 1 unit.
Horizontal Translations
x
(x+1, f(x+1))
ht =f(x+1)
y= f(x)
x
Let y = f(x) be as shown.
Let’s define g(x) = f(x + 1),
and the goal is to graph g(x).
For the point x, the output is
y = g(x) = f(x + 1),
so the point (x, g(x) = f(x +1))
is on the graph of g(x). u
(u+1, f(u+1))
ht = f(u+1)
u+1x+1
y = g(x) or
y = f(x + 1)
(x, f(x +1))
(u, f(u +1))
Likewise if the input is u,
then y = g(u) = f(u + 1) so the point (u, f(u +1)) is on
the graph of g(x).
Shifting left by 1
Similarly demonstrations show that to obtain the graph
of y = f(x – 1), shift the graph of y = f(x) right by 1 unit.
59. Horizontal Translations
The graphs of y = f(x + c) are the horizontal shifts
(or translations) of y = f(x) by c units,
Horizontal Translations
60. Horizontal Translations
The graphs of y = f(x + c) are the horizontal shifts
(or translations) of y = f(x) by c units,
Horizontal Translations
Example B. Graph the following functions by shifting
the graph of y = f(x) = x2. Label their vertices.
a. g(x) = (x + 2)2 = f(x + 2)
x
y=x 2
b. h(x) = (x – 2)2 = f(x – 2)
Horizontal Shifts
61. Horizontal Translations
The graphs of y = f(x + c) are the horizontal shifts
(or translations) of y = f(x) by c units, assuming c > 0:
Horizontal Translations
moves y = f(x) to the left for y = f(x + c).
Example B. Graph the following functions by shifting
the graph of y = f(x) = x2. Label their vertices.
a. g(x) = (x + 2)2 = f(x + 2)
b. h(x) = (x – 2)2 = f(x – 2) x
y=x 2
Horizontal Shifts
62. Horizontal Translations
The graphs of y = f(x + c) are the horizontal shifts
(or translations) of y = f(x) by c units, assuming c > 0:
Horizontal Translations
moves y = f(x) to the left for y = f(x + c).
b. h(x) = (x – 2)2 = f(x – 2)
Example B. Graph the following functions by shifting
the graph of y = f(x) = x2. Label their vertices.
a. g(x) = (x + 2)2 = f(x + 2)
Shift of the graph of y = x2
left 2 units.
x
y=x 2
Horizontal Shifts
63. Horizontal Translations
The graphs of y = f(x + c) are the horizontal shifts
(or translations) of y = f(x) by c units, assuming c > 0:
y=(x + 2)2
Horizontal Translations
moves y = f(x) to the left for y = f(x + c).
Example B. Graph the following functions by shifting
the graph of y = f(x) = x2. Label their vertices.
a. g(x) = (x + 2)2 = f(x + 2)
Shift of the graph of y = x2
left 2 units.
x
y=x 2
b. h(x) = (x – 2)2 = f(x – 2)
Horizontal Shifts
64. Horizontal Translations
The graphs of y = f(x + c) are the horizontal shifts
(or translations) of y = f(x) by c units, assuming c > 0:
y=(x + 2)2
Horizontal Translations
moves y = f(x) to the left for y = f(x + c).
Example B. Graph the following functions by shifting
the graph of y = f(x) = x2. Label their vertices.
a. g(x) = (x + 2)2 = f(x + 2)
Shift of the graph of y = x2
left 2 units. The vertex of
g(x) = (x + 2)2 is (–2, 0).
x
y=x 2
b. h(x) = (x – 2)2 = f(x – 2)
Horizontal Shifts
(–2, 0)
65. Horizontal Translations
The graphs of y = f(x + c) are the horizontal shifts
(or translations) of y = f(x) by c units, assuming c > 0:
y=(x + 2)2
Horizontal Translations
moves y = f(x) to the left for y = f(x + c).
moves y = f(x) to the right for y = f(x – c).
Example B. Graph the following functions by shifting
the graph of y = f(x) = x2. Label their vertices.
a. g(x) = (x + 2)2 = f(x + 2)
Shift of the graph of y = x2
left 2 units. The vertex of
g(x) = (x + 2)2 is (–2, 0).
x
y=x 2
b. h(x) = (x – 2)2 = f(x – 2)
Horizontal Shifts
(–2, 0)
66. Horizontal Translations
The graphs of y = f(x + c) are the horizontal shifts
(or translations) of y = f(x) by c units, assuming c > 0:
y=(x + 2)2
Horizontal Translations
moves y = f(x) to the left for y = f(x + c).
moves y = f(x) to the right for y = f(x – c).
Example B. Graph the following functions by shifting
the graph of y = f(x) = x2. Label their vertices.
a. g(x) = (x + 2)2 = f(x + 2)
Shift of the graph of y = x2
left 2 units. The vertex of
g(x) = (x + 2)2 is (–2, 0).
x
y=x 2
b. h(x) = (x – 2)2 = f(x – 2)
Shift the graph of y = x2 right
2 units. The vertex of h(x) is (2, 0). Horizontal Shifts
(–2, 0)
67. Horizontal Translations
The graphs of y = f(x + c) are the horizontal shifts
(or translations) of y = f(x) by c units, assuming c > 0:
y=(x + 2)2 y=(x – 2)2
Horizontal Translations
moves y = f(x) to the left for y = f(x + c).
moves y = f(x) to the right for y = f(x – c).
Example B. Graph the following functions by shifting
the graph of y = f(x) = x2. Label their vertices.
a. g(x) = (x + 2)2 = f(x + 2)
Shift of the graph of y = x2
left 2 units. The vertex of
g(x) = (x + 2)2 is (–2, 0).
x
y=x 2
b. h(x) = (x – 2)2 = f(x – 2)
Shift the graph of y = x2 right
2 units. The vertex of h(x) is (2, 0). Horizontal Shifts
(–2, 0) (2, 0)
68. Horizontal Stretches and Compressions
x
Let y = f(x) with its graph
shown here and let
g(x) = f(2x).
0
y= f(x)
Horizontal Stretches and Compressions
69. Horizontal Stretches and Compressions
x
Let y = f(x) with its graph
shown here and let
g(x) = f(2x).
For the point x,
the output is g(x) = f(2x)
x0
y= f(x)
Horizontal Stretches and Compressions
x
70. Horizontal Stretches and Compressions
x
Let y = f(x) with its graph
shown here and let
g(x) = f(2x).
For the point x,
the output is g(x) = f(2x)
so the point (x, y = f(2x)) is on the graph of g(x).
x0
y= f(x)
Horizontal Stretches and Compressions
x
71. Horizontal Stretches and Compressions
x
Let y = f(x) with its graph
shown here and let
g(x) = f(2x).
For the point x,
the output is g(x) = f(2x)
so the point (x, y = f(2x)) is on the graph of g(x).
0
y= f(x)
Horizontal Stretches and Compressions
2x
ht =f(2x)
(2x, f(2x))
x
72. Horizontal Stretches and Compressions
2x
ht =f(2x) x
Let y = f(x) with its graph
shown here and let
g(x) = f(2x).
For the point x,
the output is g(x) = f(2x)
so the point (x, y = f(2x)) is on the graph of g(x).
x0
y= f(x)
(2x, f(2x))
Horizontal Stretches and Compressions
(x, g(x)=f(2x))
73. Horizontal Stretches and Compressions
2x
ht =f(2x) x
Let y = f(x) with its graph
shown here and let
g(x) = f(2x).
For the point x,
the output is g(x) = f(2x)
so the point (x, y = f(2x)) is on the graph of g(x).
x0
y= f(x)
(2x, f(2x))
Horizontal Stretches and Compressions
x
Likewise for the point u, the output is g(u) = f(2u)
so the point (u, y = f(2u)) is on the graph of g(x).
(x, g(x)=f(2x))
u
74. Horizontal Stretches and Compressions
2x
ht =f(2x) x
Let y = f(x) with its graph
shown here and let
g(x) = f(2x).
For the point x,
the output is g(x) = f(2x)
so the point (x, y = f(2x)) is on the graph of g(x).
x0
y= f(x)
(2x, f(2x))
Horizontal Stretches and Compressions
x
2u
(2u, f(2u))
ht = f(2u)
Likewise for the point u, the output is g(u) = f(2u)
so the point (u, y = f(2u)) is on the graph of g(x).
(x, g(x)=f(2x))
u
75. Horizontal Stretches and Compressions
2x
ht =f(2x) x
Let y = f(x) with its graph
shown here and let
g(x) = f(2x).
For the point x,
the output is g(x) = f(2x)
so the point (x, y = f(2x)) is on the graph of g(x).
x0
y= f(x)
(2x, f(2x))
Horizontal Stretches and Compressions
x
2u
(2u, f(2u))
ht = f(2u)
Likewise for the point u, the output is g(u) = f(2u)
so the point (u, y = f(2u)) is on the graph of g(x).
(x, g(x)=f(2x))
u
(u,g(u))=f(2u)
76. Horizontal Stretches and Compressions
2x
(u,g(u))=f(2u)
ht =f(2x)
y=g(x)=f(2x)
x
2u
(2u, f(2u))
ht = f(2u)
u
Let y = f(x) with its graph
shown here and let
g(x) = f(2x).
For the point x,
the output is g(x) = f(2x)
so the point (x, y = f(2x)) is on the graph of g(x).
Likewise for the point u, the output is g(u) = f(2u)
so the point (u, y = f(2u)) is on the graph of g(x).
x0
y= f(x)
Horizontal stretch by a factor of 2
Hence we see that the graph of y = f(2x)
is the horizontal compression of the graph y = f(x)
by a factor of ½ .
(x, g(x)=f(2x))
(2x, f(2x))
Horizontal Stretches and Compressions
77. Horizontal Stretches and Compressions
2x
(u,g(u))=f(2u)
ht =f(2x)
y=g(x)=f(2x)
x
2u
(2u, f(2u))
ht = f(2u)
u
Let y = f(x) with its graph
shown here and let
g(x) = f(2x).
For the point x,
the output is g(x) = f(2x)
so the point (x, y = f(2x)) is on the graph of g(x).
Likewise for the point u, the output is g(u) = f(2u)
so the point (u, y = f(2u)) is on the graph of g(x).
x0
y= f(x)
Horizontal stretch by a factor of 2
Hence we see that the graph of y = f(2x)
is the horizontal compression of the graph y = f(x)
by a factor of ½ . Similarly, the graph of y = f(½ * x)
is the horizontal stretch the graph of y = f(x)
by a factor of 2. (Convince yourself of this fact.)
(x, g(x)=f(2x))
(2x, f(2x))
Horizontal Stretches and Compressions
79. Horizontal Reflections
Horizontal Stretches and Compressions
Let y = f(x) with its graph
shown here and let
g(x) = f(–x).
For the point x,
the output is g(x) = f(–x)
x
y= f(x)
x
y
0
Horizontal reflection
80. Horizontal Reflections
Horizontal Stretches and Compressions
Let y = f(x) with its graph
shown here and let
g(x) = f(–x).
For the point x,
the output is g(x) = f(–x)
so the point (x, y = f(–x)) is on the graph of g(x).
x
y= f(x)
x
y
–x
0
Horizontal reflection
81. Horizontal Reflections
Horizontal Stretches and Compressions
Let y = f(x) with its graph
shown here and let
g(x) = f(–x).
For the point x,
the output is g(x) = f(–x)
so the point (x, y = f(–x)) is on the graph of g(x).
x
y= f(x)
x
y
–x
(x, g(x)=f(–x))
0
Horizontal reflection
82. Horizontal Reflections
x
Let y = f(x) with its graph
shown here and let
g(x) = f(–x).
For the point x,
the output is g(x) = f(–x)
so the point (x, y = f(–x)) is on the graph of g(x).
0
y= f(x)
Horizontal Stretches and Compressions
x–x
(x, g(x)=f(–x))
u
Likewise for the point u, the output is g(u) = f(–u)
so the point (u, y = f(–u)) is on the graph of g(x).
y
Horizontal reflection
83. Horizontal Reflections
x
Let y = f(x) with its graph
shown here and let
g(x) = f(–x).
For the point x,
the output is g(x) = f(–x)
so the point (x, y = f(–x)) is on the graph of g(x).
0
y= f(x)
Horizontal Stretches and Compressions
x–x
(x, g(x)=f(–x))
–uu
Likewise for the point u, the output is g(u) = f(–u)
so the point (u, y = f(–u)) is on the graph of g(x).
y
Horizontal reflection
84. Horizontal Reflections
x
Let y = f(x) with its graph
shown here and let
g(x) = f(–x).
For the point x,
the output is g(x) = f(–x)
so the point (x, y = f(–x)) is on the graph of g(x).
0
y= f(x)
Horizontal Stretches and Compressions
x–x
(x, g(x)=f(–x))
–uu
Likewise for the point u, the output is g(u) = f(–u)
so the point (u, y = f(–u)) is on the graph of g(x).
(u, g(u)=f(–u))
y
Horizontal reflection
85. Horizontal Reflections
x
Let y = f(x) with its graph
shown here and let
g(x) = f(–x).
For the point x,
the output is g(x) = f(–x)
so the point (x, y = f(–x)) is on the graph of g(x).
0
y= f(x)
Horizontal Stretches and Compressions
x–x
(x, g(x)=f(–x))
–uu
Likewise for the point u, the output is g(u) = f(–u)
so the point (u, y = f(–u)) is on the graph of g(x).
(u, g(u)=f(–u))
y
Horizontal reflection
Hence we reflect the graph of y = f(x) horizontally
across the y–axis to obtain the graph of y = f(–x).
86. Horizontal Reflections
x
Let y = f(x) with its graph
shown here and let
g(x) = f(–x).
For the point x,
the output is g(x) = f(–x)
so the point (x, y = f(–x)) is on the graph of g(x).
0
y= f(x)
Horizontal Stretches and Compressions
x–x
(x, g(x)=f(–x))
–uu
(u, g(x)=f(–x))
Likewise for the point u, the output is g(u) = f(–u)
so the point (u, y = f(–u)) is on the graph of g(x).
(u, g(u)=f(–u))
y
Horizontal reflection
Hence we reflect the graph of y = f(x) horizontally
across the y–axis to obtain the graph of y = f(–x).
87. Horizontal Reflections
x
Let y = f(x) with its graph
shown here and let
g(x) = f(–x).
For the point x,
the output is g(x) = f(–x)
so the point (x, y = f(–x)) is on the graph of g(x).
0
y= f(x)
Horizontal Stretches and Compressions
x–x
(x, g(x)=f(–x))
–uu
(u, g(x)=f(–x))
Likewise for the point u, the output is g(u) = f(–u)
so the point (u, y = f(–u)) is on the graph of g(x).
(u, g(u)=f(–u))
y
Horizontal reflection
Hence we reflect the graph of y = f(x) horizontally
across the y–axis to obtain the graph of y = f(–x).
To graph y = f(–2x), we compress y = f(x)
by a factor of ½ to obtain the graph of y = f(2x),
then reflect the result to obtain the graph of y = f(–2x).
88. Summary of vertical transformations of graphs (c > 0).
Transformations of Graphs
Vertical Shifts Vertical Stretches
and Compressions
x
y= f(x)
y= f(x)
x
89. Summary of vertical transformations of graphs (c > 0).
Transformations of Graphs
Vertical Shifts Vertical Stretches
and Compressions
x
y= f(x)
y= f(x)
y = f(x) + c moves f up by c
y = f(x) – c move f down by c
x
90. Summary of vertical transformations of graphs (c > 0).
Transformations of Graphs
Vertical Shifts Vertical Stretches
and Compressions
y= f(x) + 1
x
y= f(x) + 2
y= f(x)
y= f(x)
y = f(x) + c moves f up by c
y = f(x) – c move f down by c
x
91. Summary of vertical transformations of graphs (c > 0).
Transformations of Graphs
Vertical Shifts Vertical Stretches
and Compressions
y= f(x)–1
y= f(x) + 1
x
y= f(x) + 2
y= f(x)–2
y= f(x)–3
y= f(x)
y= f(x)
y = f(x) + c moves f up by c
y = f(x) – c move f down by c
x
92. Summary of vertical transformations of graphs (c > 0).
Transformations of Graphs
Vertical Shifts Vertical Stretches
and Compressions
y= f(x)–1
y= f(x) + 1
x
y= f(x) + 2
y= f(x)–2
y= f(x)–3
y= f(x)
y= f(x)
y = f(x) + c moves f up by c
y = f(x) – c move f down by c
x
c > 1, y = cf(x) stretches f vertically
93. Summary of vertical transformations of graphs (c > 0).
Transformations of Graphs
Vertical Shifts Vertical Stretches
and Compressions
y= f(x)–1
y= f(x) + 1
x
y= f(x) + 2
y= f(x)–2
y= f(x)–3
y= f(x)
y= 2f(x)
y= f(x)
y = f(x) + c moves f up by c
y = f(x) – c move f down by c c > 1, y = cf(x) stretches f vertically
x
94. Summary of vertical transformations of graphs (c > 0).
Transformations of Graphs
Vertical Shifts Vertical Stretches
and Compressions
y= f(x)–1
y= f(x) + 1
x
y= f(x) + 2
y= f(x)–2
y= f(x)–3
y= f(x)
y= 3f(x)
y= 2f(x)
y= f(x)
y = f(x) + c moves f up by c
y = f(x) – c move f down by c c > 1, y = cf(x) stretches f vertically
x
95. Summary of vertical transformations of graphs (c > 0).
Transformations of Graphs
Vertical Shifts Vertical Stretches
and Compressions
y= f(x)–1
y= f(x) + 1
x
y= f(x) + 2
y= f(x)–2
y= f(x)–3
y= f(x)
y= 3f(x)
y= f(x)/3
y= 2f(x)
y= f(x)
y = f(x) + c moves f up by c
y = f(x) – c move f down by c c > 1, y = cf(x) stretches f vertically
0 < c < 1, y = cf(x) compresses f
x
96. Summary of vertical transformations of graphs (c > 0).
Transformations of Graphs
Vertical Shifts Vertical Stretches
and Compressions
y= f(x)–1
y= f(x) + 1
x
y= f(x) + 2
y= f(x)–2
y= f(x)–3
y= f(x)
y= 3f(x)
y= f(x)/3
y= 2f(x)
y= –f(x)
y= f(x)
y = f(x) + c moves f up by c
y = f(x) – c move f down by c c > 1, y = cf(x) stretches f vertically
y = –f(x) reflects
f vertically
0 < c < 1, y = cf(x) compresses f
x
97. Summary of vertical transformations of graphs (c > 0).
Transformations of Graphs
Vertical Shifts Vertical Stretches
and Compressions
y= f(x)–1
y= f(x) + 1
x
y= f(x) + 2
y= f(x)–2
y= f(x)–3
y= f(x)
y= 3f(x)
y= f(x)/3
y= 2f(x)
y= –f(x)
y= –2f(x)
y= –3f(x)
y= f(x)
y = f(x) + c moves f up by c
y = f(x) – c move f down by c c > 1, y = cf(x) stretches f vertically
y = –f(x) reflects
f vertically
0 < c < 1, y = cf(x) compresses f
x
98. x
–1
Transformations of Graphs
Horizontal Shifts
y=f(x)
y=f(x)
–2–3
y
Horizontal Stretches
and Compressions
Summary of horizontal transformations of graph (c > 0).
x
99. x
–1
Transformations of Graphs
Horizontal Shifts
y=f(x)
y=f(x)
–2–3
y
Horizontal Stretches
and Compressions
y = f(x + c) moves f left by c
y = f(x – c) moves f right by c
Summary of horizontal transformations of graph (c > 0).
x
100. Transformations of Graphs
Horizontal Shifts
y=f(x)
y=f(x+2)
y=f(x+1)
Horizontal Stretches
and Compressions
y = f(x + c) moves f left by c
y = f(x – c) moves f right by c
Summary of horizontal transformations of graph (c > 0).
x
x
–1
y=f(x)
–2–3
y
101. Transformations of Graphs
Horizontal Shifts
y=f(x)
y=f(x+2)
y=f(x+1)
y=f(x–2)
y=f(x–1)
Horizontal Stretches
and Compressions
y = f(x + c) moves f left by c
y = f(x – c) moves f right by c
Summary of horizontal transformations of graph (c > 0).
x
x
–1
y=f(x)
–2–3
y
102. Transformations of Graphs
Horizontal Shifts
y=f(x)
y=f(x+2)
y=f(x+1)
y=f(x–2)
y=f(x–1)
Horizontal Stretches
and Compressions
y = f(x + c) moves f left by c
y = f(x – c) moves f right by c
Summary of horizontal transformations of graph (c > 0).
x
x
–1
y=f(x)
–2–3
y
c > 1, y = f(cx) compresses f horizontally
0 < c < 1, y = f(cx) stretches f horizontally.
103. –1
Transformations of Graphs
Horizontal Shifts
y=f(x)
y=f(x+2)
y=f(x+1)
y=f(x–2)
y=f(x–1)
y=f(x) y=f(2x)
–2–3
y
Horizontal Stretches
and Compressions
(0,f(0))
y = f(x + c) moves f left by c
y = f(x – c) moves f right by c
c > 1, y = f(cx) compresses f horizontally
0 < c < 1, y = f(cx) stretches f horizontally.
Summary of horizontal transformations of graph (c > 0).
x
104. x
–1
Transformations of Graphs
Horizontal Shifts
y=f(x)
y=f(x+2)
y=f(x+1)
y=f(x–2)
y=f(x–1)
y=f(x) y=f(2x)
–2–3
y=f(3x)
y
Horizontal Stretches
and Compressions
(0,f(0))
y = f(x + c) moves f left by c
y = f(x – c) moves f right by c
c > 1, y = f(cx) compresses f horizontally
0 < c < 1, y = f(cx) stretches f horizontally.
Summary of horizontal transformations of graph (c > 0).
x
105. x
–1
Transformations of Graphs
Horizontal Shifts
y=f(x)
y=f(x+2)
y=f(x+1)
y=f(x–2)
y=f(x–1)
y=f(x)y=f(x/2)y=f(x/3) y=f(2x)
–2–3
y=f(3x)
y
Horizontal Stretches
and Compressions
(0,f(0))
y = f(x + c) moves f left by c
y = f(x – c) moves f right by c
c > 1, y = f(cx) compresses f horizontally
0 < c < 1, y = f(cx) stretches f horizontally.
Summary of horizontal transformations of graph (c > 0).
x
106. x
–1
Transformations of Graphs
Horizontal Shifts
y=f(x)
y=f(x+2)
y=f(x+1)
y=f(x–2)
y=f(x–1)
y=f(x)y=f(x/2)y=f(x/3) y=f(2x)
–2–3
y=f(3x)
y
Horizontal Stretches
and Compressions
y = f(–x) reflect f horizontally
y=f(–x/3)
(0,f(0))
y = f(x + c) moves f left by c
y = f(x – c) moves f right by c
c > 1, y = f(cx) compresses f horizontally
0 < c < 1, y = f(cx) stretches f horizontally.
Summary of horizontal transformations of graph (c > 0).
x
107. Horizontal Translations
Let y = f(x) be a function with the interval [0, 1]
as its domain as shown.
y
1
x
2 3½
y = f(x)
108. Horizontal Translations
Let y = f(x) be a function with the interval [0, 1]
as its domain as shown.
The graph of y = g(x) = f(½ * x) is the horizontal stretch,
by a factor of 2, of the graph of y = f(x).
y
1
x
2 3½
y = f(x) y=g(x)=f(½ * x)
109. Horizontal Translations
Let y = f(x) be a function with the interval [0, 1]
as its domain as shown.
The graph of y = g(x) = f(½ * x) is the horizontal stretch,
by a factor of 2, of the graph of y = f(x).
The domain of y = g(x) = f(½ * x) correspondingly is the
stretch of the domain of y = f(x), from [0, 1] to [0, 2].
y
1
x
2 3½
y = f(x) y=g(x)=f(½ * x)
110. Horizontal Translations
Let y = f(x) be a function with the interval [0, 1]
as its domain as shown.
y
y = f(x)
1
x
2 3
The graph of y = g(x) = f(½ * x) is the horizontal stretch,
by a factor of 2, of the graph of y = f(x).
The domain of y = g(x) = f(½ * x) correspondingly is the
stretch of the domain of y = f(x), from [0, 1] to [0, 2].
y=g(x)=f(½ * x)
½
y = f(x/3)
111. Horizontal Translations
Let y = f(x) be a function with the interval [0, 1]
as its domain as shown.
The graph of y = g(x) = f(½ * x) is the horizontal stretch,
by a factor of 2, of the graph of y = f(x).
The domain of y = g(x) = f(½ * x) correspondingly is the
stretch of the domain of y = f(x), from [0, 1] to [0, 2].
Similarly the domain of y = h(x) = f(2x) is the
compression from [0, 1] to [0, ½ ].
y
1
x
2 3½
y = f(x) y=g(x)=f(½ * x)y = f(2x) y = f(x/3)
112. Horizontal Translations
Let y = f(x) be a function with the interval [0, 1]
as its domain as shown.
y
y = f(x)
1
x
2 3
The graph of y = g(x) = f(½ * x) is the horizontal stretch,
by a factor of 2, of the graph of y = f(x).
The domain of y = g(x) = f(½ * x) correspondingly is the
stretch of the domain of y = f(x), from [0, 1] to [0, 2].
Similarly the domain of y = h(x) = f(2x) is the
compression from [0, 1] to [0, ½ ].
y = f(x/3)y=g(x)=f(½ * x)y = f(2x)
y = f(3x)
½
113. Horizontal Translations
Let y = f(x) be a function with the interval [0, 1]
as its domain as shown.
y
y = f(x)
1
x
2 3
The graph of y = g(x) = f(½ * x) is the horizontal stretch,
by a factor of 2, of the graph of y = f(x).
The domain of y = g(x) = f(½ * x) correspondingly is the
stretch of the domain of y = f(x), from [0, 1] to [0, 2].
Similarly the domain of y = h(x) = f(2x) is the
compression from [0, 1] to [0, ½ ].
y=g(x)=f(½ * x)
The Domain of y = f(cx), c > 0
If the domain of
y = f(x) is [0, a],
then the domain of
y = f(cx) is [0, a/c].
y = f(2x)
½
y = f(x/3)
y = f(3x)
114. Horizontal Translations
Example C.
a. Given the graph of the function
y = f(x) with the domain [–2, 2],
graph y = (x – 3)2 – 1 by applying
the transformation rules. Give the new domain and
label the end points of the graph.
f(x)=x2
x
(2,4)(–2,4)
2–2 (0,0)
115. Horizontal Translations
Example C.
a. Given the graph of the function
y = f(x) with the domain [–2, 2],
graph y = (x – 3)2 – 1 by applying
the transformation rules. Give the new domain and
label the end points of the graph.
f(x)=x2
x
(2,4)(–2,4)
2–2
i. Shift right 3 units the
graph of f(x) = x2 to obtain
the graph of y = (x – 3)2.
(0,0)
2–2
Shift right 3 units
116. Horizontal Translations
Example C.
a. Given the graph of the function
y = f(x) with the domain [–2, 2],
graph y = (x – 3)2 – 1 by applying
the transformation rules. Give the new domain and
label the end points of the graph.
f(x)=x2
x
(2,4)(–2,4)
2–2
i. Shift right 3 units the
graph of f(x) = x2 to obtain
the graph of y = (x – 3)2.
(0,0)
ii. Lower the graph of
y = (x – 3)2 by 1 unit for
the graph of y = (x – 3)2 – 1.
2–2
(5,3)(1,3)
(3,–1)
Shift right 3 units
Lower by
1 unit
117. Horizontal Translations
Example C.
a. Given the graph of the function
y = f(x) with the domain [–2, 2],
graph y = (x – 3)2 – 1 by applying
the transformation rules. Give the new domain and
label the end points of the graph.
f(x)=x2
x
(2,4)(–2,4)
2–2
i. Shift right 3 units the
graph of f(x) = x2 to obtain
the graph of y = (x – 3)2.
(0,0)
ii. Lower the graph of
y = (x – 3)2 by 1 unit for
the graph of y = (x – 3)2 – 1.
2–2
(5,3)(1,3)
(3,–1)
Shift right 3 units
Lower by
1 unit
The new domain is [–2 + 3, 2 + 3] = [1, 5]. The new
vertex is (3, –1) and end points (1, 3) and (5, 3).
119. Horizontal Translations
The graph of y = G(x) is
obtained by horizontally
compressing the graph of
x
(4,2)
(0,0)
y=g(x)=√x
4
b. Given the graphs of the function y = g(x) = √x
and y = G(x) a transformation of y = g(x) as shown,
express G(x) using g(x).
cy=G(x)
(6,2)
62
y = g(x) by a factor of ½,
which gives the graph of
h(x) = g(2x) = √2x as shown, x
(4,2)
(0,0)
y=g(x)=√x
4 62
(2,2)
c
y=h(x)=√2x horizontal
compression
120. Horizontal Translations
The graph of y = G(x) is
obtained by horizontally
compressing the graph of
x
(4,2)
(0,0)
y=g(x)=√x
4
b. Given the graphs of the function y = g(x) = √x
and y = G(x) a transformation of y = g(x) as shown,
express G(x) using g(x).
cy=G(x)
(6,2)
62
y = g(x) by a factor of ½,
which gives the graph of
h(x) = g(2x) = √2x as shown, x
(4,2)
(0,0)
y=g(x)=√x
4 62
(2,2)
c
then moving h(x) to the right
by 4 units.
y=h(x)=√2x
x
(0,0) 4
cy=G(x)
(6,2)
62
(2,2)
c
y=h(x)=√2x
horizontal
compression
horizontal
shift
121. Horizontal Translations
The graph of y = G(x) is
obtained by horizontally
compressing the graph of
x
(4,2)
(0,0)
y=g(x)=√x
4
b. Given the graphs of the function y = g(x) = √x
and y = G(x) a transformation of y = g(x) as shown,
express G(x) using g(x).
cy=G(x)
(6,2)
62
y = g(x) by a factor of ½,
which gives the graph of
h(x) = g(2x) = √2x as shown, x
(4,2)
(0,0)
y=g(x)=√x
4 62
(2,2)
c
then moving h(x) to the right
by 4 units. Hence
G(x) = h(x – 4)
= √2(x – 4) or that
G(x) = √2x – 8.
y=h(x)=√2x
x
(0,0) 4
cy=G(x)
(6,2)
62
(2,2)
c
y=h(x)=√2x
horizontal
compression
horizontal
shift
125. Absolute-Value Flip
y = f(x) = x y = |f(x)| = |x|
x -2 -1 0 1
y -2 -1 0 1
x -2 -1 0 1
y 2 1 0 1
126. Absolute-Value Flip
y = f(x) = x y = |f(x)| = |x|
x -2 -1 0 1
y -2 -1 0 1
x -2 -1 0 1
y 2 1 0 1
127. Absolute-Value Flip
y = f(x) = x
The graph of y = |f(x)| is obtained by reflecting the
portion of the graph below the x-axis to above the
x-axis.
y = |f(x)| = |x|
x -2 -1 0 1
y -2 -1 0 1
x -2 -1 0 1
y 2 1 0 1
135. Horizontal Flip
The graph of y = f(–x) is the horizontal reflection of
the graph of y = f(x) across the y axis.
136. Horizontal Flip
The graph of y = f(–x) is the horizontal reflection of
the graph of y = f(x) across the y axis.
y = f(x) = x3 – x2
137. Horizontal Flip
The graph of y = f(–x) is the horizontal reflection of
the graph of y = f(x) across the y axis.
y = f(x) = x3 – x2
y = f(-x) = (-x)3 – (-x)2
138. Horizontal Flip
The graph of y = f(–x) is the horizontal reflection of
the graph of y = f(x) across the y axis.
y = f(x) = x3 – x2
y = f(-x) = (-x)3 – (-x)2
y = f(-x) = – x3 – x2
x
139. Horizontal Flip
The graph of y = f(–x) is the horizontal reflection of
the graph of y = f(x) across the y axis.
y = f(x) = x3 – x2
y = f(-x) = (-x)3 – (-x)2
y = f(-x) = – x3 – x2
140. Horizontal Flip
The graph of y = f(–x) is the horizontal reflection of
the graph of y = f(x) across the y axis.
y = f(x) = x3 – x2
y = f(-x) = (-x)3 – (-x)2
y = f(-x) = – x3 – x2
A function is said to be
even if f(x) = f(– x).
Graphs of even functions
are symmetric to the
y-axis.
141. Horizontal Flip
The graph of y = f(–x) is the horizontal reflection of
the graph of y = f(x) across the y axis.
y = f(x) = x3 – x2
y = f(-x) = (-x)3 – (-x)2
y = f(-x) = – x3 – x2
A function is said to be
even if f(x) = f(– x).
Graphs of even functions
are symmetric to the
y-axis. Graph of an even function
x
(x, f(x))
–x
(–x, f(–x))
142. Horizontal Flip
The graph of y = f(–x) is the horizontal reflection of
the graph of y = f(x) across the y axis.
y = f(x) = x3 – x2
y = f(-x) = (-x)3 – (-x)2
y = f(-x) = – x3 – x2
A function is said to be
even if f(x) = f(– x).
Graphs of even functions
are symmetric to the
y-axis. Graph of an even function
x
(x, f(x))
–x
(–x, f(–x))
144. Horizontal Flip
A function is said to be odd
iff f(–x) = – f(x).
Polynomial-functions whose
terms are all even powers
are even. The graph on the
right is the even polynomial–
function y = x4 – 4x2. y = x4 – 4x2
145. Horizontal Flip
y = x4 – 4x2
A function is said to be odd
iff f(–x) = – f(x).
Graphs of odd functions
are symmetric to the origin,
Polynomial-functions whose
terms are all even powers
are even. The graph on the
right is the even polynomial–
function y = x4 – 4x2.
146. Horizontal Flip
A function is said to be odd
iff f(–x) = – f(x).
Graphs of odd functions
are symmetric to the origin,
that is, they're the same as
reflecting across the y-axis
followed by reflecting
across the x-axis.
Polynomial-functions whose
terms are all even powers
are even. The graph on the
right is the even polynomial–
function y = x4 – 4x2. y = x4 – 4x2
147. Horizontal Flip
Graph of an odd function
A function is said to be odd
iff f(–x) = – f(x).
Graphs of odd functions
are symmetric to the origin,
that is, they're the same as
reflecting across the y-axis
followed by reflecting
across the x-axis.
x–x
0
(x, f(x))
(–x, –f(x))
Polynomial-functions whose
terms are all even powers
are even. The graph on the
right is the even polynomial–
function y = x4 – 4x2. y = x4 – 4x2
148. Horizontal Flip
Graph of an odd function
A function is said to be odd
iff f(–x) = – f(x).
Graphs of odd functions
are symmetric to the origin,
that is, they're the same as
reflecting across the y-axis
followed by reflecting
across the x-axis.
x–x
0
(x, f(x))
(–x, –f(x))
u
(u, f(u))
(–u, –f(u))
–u
Polynomial-functions whose
terms are all even powers
are even. The graph on the
right is the even polynomial–
function y = x4 – 4x2. y = x4 – 4x2
149. Horizontal Flip
Polynomial-functions whose
terms are all even powers
are even. The graph on the
right is the even polynomial–
function y = x4 – 4x2. y = x4 – 4x2
Graph of an odd function
A function is said to be odd
iff f(–x) = – f(x).
Graphs of odd functions
are symmetric to the origin,
that is, they're the same as
reflecting across the y-axis
followed by reflecting
across the x-axis.
x–x
0
(x, f(x))
(–x, –f(x))
u
(u, f(u))
(–u, –f(u))
–u
151. Horizontal Flip
y = x3 – 4x
Polynomial-functions whose
terms are all odd powers are
odd. The graph on the right is
the odd polynomial–function
y = x3 – 4x.
153. Horizontal Flip
y = x3 – 4x
Theorem (even and odd):
I. The sum of even functions is even.
The sum of odd functions is odd.
Polynomial-functions whose
terms are all odd powers are
odd. The graph on the right is
the odd polynomial–function
y = x3 – 4x.
154. Horizontal Flip
y = x3 – 4x
Theorem (even and odd):
I. The sum of even functions is even.
The sum of odd functions is odd.
II. The product of even functions is even.
Polynomial-functions whose
terms are all odd powers are
odd. The graph on the right is
the odd polynomial–function
y = x3 – 4x.
155. Horizontal Flip
y = x3 – 4x
Theorem (even and odd):
I. The sum of even functions is even.
The sum of odd functions is odd.
II. The product of even functions is even.
The product of odd functions is even.
Polynomial-functions whose
terms are all odd powers are
odd. The graph on the right is
the odd polynomial–function
y = x3 – 4x.
156. Horizontal Flip
y = x3 – 4x
Theorem (even and odd):
I. The sum of even functions is even.
The sum of odd functions is odd.
II. The product of even functions is even.
The product of odd functions is even.
The product of an even function with an odd
function is odd.
Polynomial-functions whose
terms are all odd powers are
odd. The graph on the right is
the odd polynomial–function
y = x3 – 4x.
157. Horizontal Flip
y = x3 – 4x
Theorem (even and odd):
I. The sum of even functions is even.
The sum of odd functions is odd.
II. The product of even functions is even.
The product of odd functions is even.
The product of an even function with an odd
function is odd. (The same hold for quotient.)
Polynomial-functions whose
terms are all odd powers are
odd. The graph on the right is
the odd polynomial–function
y = x3 – 4x.
158. Horizontal Flip
y = x3 – 4x
Theorem (even and odd):
I. The sum of even functions is even.
The sum of odd functions is odd.
II. The product of even functions is even.
The product of odd functions is even.
The product of an even function with an odd
function is odd. (The same hold for quotient.)
is odd,x
x4 + 1
Polynomial-functions whose
terms are all odd powers are
odd. The graph on the right is
the odd polynomial–function
y = x3 – 4x.
159. Horizontal Flip
y = x3 – 4x
Theorem (even and odd):
I. The sum of even functions is even.
The sum of odd functions is odd.
II. The product of even functions is even.
The product of odd functions is even.
The product of an even function with an odd
function is odd. (The same hold for quotient.)
is odd, is even,x
x4 + 1
x2
x4 + 1
Polynomial-functions whose
terms are all odd powers are
odd. The graph on the right is
the odd polynomial–function
y = x3 – 4x.
160. Horizontal Flip
y = x3 – 4x
Theorem (even and odd):
I. The sum of even functions is even.
The sum of odd functions is odd.
II. The product of even functions is even.
The product of odd functions is even.
The product of an even function with an odd
function is odd. (The same hold for quotient.)
is odd, is even, x + 1 is neither.x
x4 + 1
x2
x4 + 1
Polynomial-functions whose
terms are all odd powers are
odd. The graph on the right is
the odd polynomial–function
y = x3 – 4x.
161. Transformations of Graphs
Function Calisthenics
(Unknown Artist )
Exercise A.
Use the graphs shown on the list
for sketching the following graphs.
1. y = 3x2 2. y = –2x2
3. y = –0.5x2 4. y = x2 – 1
5. y = 2x2 – 1
8. y = –x3 – 2
6. y = (x+1)2
7. y = 2(x – 3)2
10. y = –(x – 2)3 – 29. y = –(x – 2)3
11. y = l x – 2 l + 1
12. y = –2l x + 2 l + 3
13. y = 14. y =
x
–1
+ 1 x + 1
1 – 1
15. y = 16. y =x
–1
+ 1 x + 1
1
– 1l l l l
162. Transformations of Graphs
B. The following problems assumes the knowledge of
graphs of trig-functions. Graph at least two periods of each
function. Label the high and low points.
1. y = sin(x – π/2) 2. y = cos(x + π/4)
3. y = cos(x – 3π/4) 4. y = –3sin(x – π/2)
8. y = cos(2x)
9. y = 3sin(4x) 10. y = cos(x/3)
7. y = –sin(x/2)
11. y = –2cos(3x)
5. y = tan(2x) 6. y = –cot(x/2)
12. y = 3cos(x + π/4) – 2
13. y = –3sin(x – 3π/4) + 1 14. y = 4cos(x/2) – 2
15. y = –2sin(2x) + 1
163. C. Given the graphs of the following functions,
draw the graphs of the following functions.
Transformations of Graphs
(1,0)
(0,1)
(2,0)
(3,2)y = f(x): y = g(x):
(0,0)
(–1,1) (1,1)
(2, –1)
1. y = 2f(x – 4) 2. y = –f(x – 2)
3. y = –3g(x + 4) 4. y = –1/2 g(x – 2)
5. y = 2g(x + 2) – 1 6. y = –3f(x – 1) + 1
7. y = –4f(x + 4) + 3 8. y = –1/2 g(x – 3/2) – 4
9. What’s the domain of f(x)? What is the domain of f(–x)?
a. Draw f(–x). b. Draw –f(–x).
10. What’s the domain of g(x)? What is the domain of g(–x)?
a. Draw g(–x). b. Draw –g(–x).
164. Transformations of Graphs
(Answers to odd problems) Exercise A.
1. y = 3x2 3. y = –0.5x2 5. y = 2x2 – 1
7. y = 2(x – 3)2 9. y = –(x – 2)3 11. y = l x – 2 l + 1
170. Exercise C.
Transformations of Graphs
(5,0)
(4,2)
(6,0)
(7,4)
(-4,0)
(–5,-3) (-3,-3)
(-2, 3)
1. y = 2f(x – 4) 3. y = –3g(x + 4)
5. y = 2g(x + 2) – 1 7. y = –4f(x + 4) + 3
(-2,-1)
(–3,1) (-1,1)
(0, –3)
(-3,0)
(-4,-4)
(-2,0)
(-1,-8)
171. Transformations of Graphs
9. domain of f(x): [0, 3]
domain of f(–x): [-3, 0]
10. domain of g(x): [-1, 2]
domain of g(–x): [-2, 1]
a. y = f(–x): b. y = –f(–x):
a. y = g(–x): b. y = –g(–x):
(-3,2)
(-3,-2)
(-2,0)
(-2,0)
(-1,0)
(-1,0)
(0,1)
(0,-1)
(-2,-1)
(-1,1)
(0,0)
(0,0)
(1,1)
(1,-1)(-1,-1)
(-2,1)
