Pre – Calculus Math 40S: Explained!
www.math40s.com
110
Transformations Lesson 3
Part I: Algebraic Transformations
Algebraic Transformations of Functions: If you know the
equation of a particular function, you can “insert” a transformation to
derive the new function.
Example 1: Given the function 2x - 1, find the equation of:
a) y = 2f(x)
The 2 in front of f(x) tells you to multiply the entire function by 2.
y = 2f(x)
y = 2(2x-1)
y = 4x - 2
b) y = f(3x)
The 3x inside the function brackets tells you that wherever there
is an x in the original function, you must replace it with 3x.
y = f(3x)
y = 2(3x)-1
y = 6x -1
c) y = –f(x)
The - in front of f(x) tells you to multiply the entire function by -1.
y = -f(x)
y = -1(2x-1)
y = -2x + 1
d) f(-x)
The -x inside the function brackets tells you that wherever there
is an x in the original function, you must replace it with -x.
y = f(-x)
y = 2(-x) -1
y = -2x - 1
e) y = f(x) +3
The +3 tells you to add 3 units to the original function.
y = f(x) + 3
y = 2x — 1 + 3
y = 2x + 2
f) y = f(x – 4)
The x-4 inside the brackets tells you that wherever there is
an x in the original function, you must replace it with x - 4.
y = f(x — 4)
y = 2(x — 4) — 1
y = 2x — 8 — 1
y = 2x — 9
Pre – Calculus Math 40S: Explained!
www.math40s.com
111
Transformations Lesson 3
Part I: Algebraic Transformations
Example 2: Given the function
f(x) = 3x2 – 3x + 4,
find the equation of:
Example 3: Given the function
f(x) = 3x + 6 , find the equation of:
a) 2f(x)
y = 4 3(x - 2)+ 6
2
y = 2(3x — 3x + 4)
y = 6x2 — 6x + 8
b) f(3x)
a) y = 4f(x - 2)
y = 4 3x - 6 + 6
y = 4 3x
b) y = -f(-3x)
y = 3(3x)2 — 3(3x) + 4
y = 3(9x2) — 3(3x) + 4
y = 27x2 — 9x + 4
y = - -9x + 6
c) –f(x)
c) y = –f(x + 2) + 4
2
y = -(3x — 3x + 4)
y = -3x2 + 3x - 4
d) f(-x)
y = - 3(-3x)+ 6
y = - 3(x + 2)+ 6 + 4
y = - 3x + 6 + 6 + 4
y = - 3x + 12 + 4
y = 3(-x)2 — 3(-x) + 4
y = 3x2 + 3x + 4
d) y =
e) y = f(x) +3
y=
y = 3x2 — 3x + 4 + 3
y = 3x2 — 3x + 7
f) y = f(x-4)
y = 3(x — 4)2 — 3(x — 4) + 4
y = 3(x2 — 8x + 16) — 3x + 12 + 4
y = 3x2 - 24x + 48 — 3x + 12 + 4
y = 3x2 -27x + 64
1
f(-x - 2)
2
1
3(-x - 2)+ 6
2
1
y=
-3x - 6 + 6
2
1
y=
-3x
2
e) y = -2f(x) +3
y = -2 3x + 6 + 3
f) y = -4f(-2x) - 5
y = -4 3(-2x)+ 6 - 5
y = -4 -6x + 6 - 5
Pre – Calculus Math 40S: Explained!
www.math40s.com
112
Transformations Lesson 3
Part I: Algebraic Transformations
Questions:
1) f(x) = 3x – 7
a) y = 3f(x)
2) f(x) = (2x – 3)2
a) y =
a) y = 2f(x - 3)
1
f(x)
2
4) f(x) = 3x - 4
a) y = 2f(x) - 8
b) y = -f(7x - 1)
b) y = f(3x + 4)
1
b) y = f(
x)
2
b) y = f(6x)
3) f(x) = - -2x - 3
c) y = –
c) y = –f(-x)
c) y = –3f(x)
c) y = –f(x)
d) y =
d) y = 2f(-x) + 4
1
f(-x)
2
d) y = f(-2x)
d) y = f(-x)
e) y = f(x) +5
f) y = f(x – 6)
Answers:
1. a) y = 9x - 21
b) y = 18x - 7
c) y = - 3x + 7
1
f(x) - 3
3
e) y = f(x) – 4
1
e) y = f(x) +3
2
f) y = f(x + 5)
f) y = f(-x – 4) + 9
2. a) y =
1
2
( 2x - 3)
2
b) y = ( x - 3 )
2
c) y = - 3 ( 2x - 3 )
d) y = - 3x - 7
e) y = 3x - 2
d) y = ( -4x - 3 )
f) y = 3x - 25
f) y = ( 2x + 7 )
2
2
e) y = ( 2x - 3) - 4
2
2
e) y = 2f(x - 5) +3
f) y = -f(-x – 4)
3. a) y = - 2 -2x + 3
4. a) y = 2 3x - 4 - 8
b) y = - -6x - 11
b) y = - 21x - 7
c) y = 2x - 3
c) y = -
d) y = - 2 2x - 3 + 4
1
e) y =
-2x - 3 + 3
2
f) y = - 2x + 5 + 9
1
3x - 4 - 3
3
1
d) y = -3x - 4
2
e) y = 2 3x - 19 + 3
f) y = - -3x - 16
Pre – Calculus Math 40S: Explained!
www.math40s.com
113
Transformations Lesson 3
Part II: Describing Transformations
Describing Transformations: This section deals with verbal
descriptions of transformations.
Example 1: How does the graph of y = -3f(x – 4) + 2 compare to y = f(x)?
Vertical Stretch by a factor of 3
Reflection in the x-axis
Translation of 4 Right and 2 Up.
Example 2: How does the graph of y = 2(x + 4)2 – 1 compare to the graph of y = x2
Vertical Stretch by a factor of 2
Translation of 4 Left and 1 Down.
Example 3: How does the graph of y = (3x + 12)2 compare to the graph of y = x2
2
First factor out the 3 from the x: y = [ 3(x + 4)]
Horizontal stretch by a factor of 1/3.
Translation of 4 units left
Example 4: Given the graph of y = x , write the new equation after a vertical stretch by
a factor of 1/2, a horizontal stretch by a factor of 1/3, and a vertical translation of 3
units up.
1
First apply the stretches: y =
3x
2
1
Now apply the translations: y =
3x + 3
2
Example 5: The graph of y = (x + 2)2 + 1 is shifted 6 units left and 4 units
down. Determine the equation of the transformed function.
The best way to do this type of question is to find a point on the graph, then apply the
transformation to that point.
We know the point (-2, 1) is on the graph. (It’s the vertex of the parabola)
After moving 6 left & 4 down, it will become (-8, -3)
Rewrite the equation using these values: y = (x + 8)2 - 3
Pre – Calculus Math 40S: Explained!
www.math40s.com
114
Transformations Lesson 3
Part II: Describing Transformations
Questions:
1) Describe how each transformed
function compares to the original:
g) Original is y = x
Transformed is y = -3 x
a) Original is y = f(x)
Transformed is y = -3f(
1
x)
4
h) Original is y = x2
Transformed is y = 2(-3x – 12)2 + 3
b) Original is y = f(x)
Transformed is y = -
1
f(-x) – 4
2
i) Original is y =
x
Transformed is y = - 2x + 4
c) Original is y = f(x)
Transformed is y = f(-x + 3)
j) Original is y = (x – 3)2
Transformed is y = (x – 4)2
d) Original is y = f(x - 2)
Transformed is y = f(x +5)
k) Original is y = (x + 8)2 - 4
Transformed is y = (x + 6)2 - 3
e) Original is y = f(x + 7) + 2
Transformed is y = f(x + 4) - 6
l) Original is y = x2
Transformed is y + 3 = x2
f) Original is y = f(x)
Transformed is 4y = f(x)
Pre – Calculus Math 40S: Explained!
www.math40s.com
115
Transformations Lesson 3
Part II: Describing Transformations
Questions: Continued
f) The graph of y = (x – 4)2 is shifted 5 units to the right.
2) Write out the transformation given
the following information:
a) The graph of y = x2 is transformed by a vertical stretch
of a factor of
2
, a horizontal stretch by a factor of 3, and a
3
g) The graph of y = (x + 4)2 – 6 is shifted 2 units to the
left and 5 units up.
horizontal translation of 3 units right.
b) The graph of y = x is reflected in the x-axis, and
shifted down by 4 units.
c) The graph of y = f(x) is vertically stretched by a factor of
4, reflected in the y-axis, and moved 7 units left.
h) The graph of y = f(x – 1) - 3 is shifted 7 units to the
right and 3 units down.
i) The graph of y = f(x) has y replaced with
1
y
2
d) The graph of y = x3 is horizontally stretched by a factor
of 3, reflected in the x-axis, and then moved 5 units down.
j) The graph of y = f(x) has y replaced with y - 2
e) The graph of y = x is vertically stretched by a factor of
4/3, horizontally stretched by a factor of 6, reflected in
both the x & y axis, then shifted 2 units right and 2 units
down.
Pre – Calculus Math 40S: Explained!
www.math40s.com
116
Transformations Lesson 3
Part II: Describing Transformations
Answers:
1.
2.
1
a) Vertical Stretch by a factor of 3 f) Rewrite as y = f ( x )
4
Horizontal Stretch by a factor of 4
Vertical
Stretch
by
a factor of 1/4
Reflection in the x-axis
b) Vertical Stretch by a factor of ½ g) Vertical Stretch by a factor of 3
Reflected in the x-axis
Reflection in both the x & y axis
Translated 4 units down
h) Rewrite as y = 2 [ -3(x + 4) ]2 + 3
c) Rewrite as y = f [ −( x − 3)]
Vertical Stretch by a factor of 2
Horizontal stretch by a factor of
Reflection in the y-axis
1/3
Translated 3 units right
Reflected in the y-axis
Translated 4 units left and 3 units
d) Original point = (2, 0)
up
Transformed point = (-5, 0)
Translated 7 units left
i) Rewrite as y = − 2( x + 2)
e) Original Point = (-7, 2)
Horizontal stretch by a factor of
Transformed Point = (-4, -6)
1/2
Translated 3 units right and
Reflection in the x-axis
8 units down.
Translated 2 units left
j) Original Point = (3, 0)
Transformed Point = (4, 0)
2⎛ 1
⎞
a) y = ⎜ (x - 3) ⎟
3⎝3
⎠
b) y = - x - 4
c) y = 4f ⎡⎣ - ( x + 7 ) ⎤⎦
3
⎛1 ⎞
d) y = - ⎜ x ⎟ - 5
⎝3 ⎠
4 1
e) y = - - (x - 2) - 2
3 6
f) Original Point = (4, 0)
Transformed Point = (9, 0)
y = (x - 9)2
g) Original Point = (-4, -6)
Transformed Point = (-6, -1)
y = (x + 6)2 - 1
h) Original Point = (1, -3)
Transformed Point = (8, -6)
y = f(x - 8) - 6
Translated 1 unit right
i) Replace y with
k) Original Point = (-8, -4)
Transformed Point = (-6, -3)
Translated 2 units right and 1 unit
up.
l) Rewrite as y = x2 — 3
Translated 3 units down
2
1
y:
2
1
y = f(x)
2
y = 2f(x)
j) Replace y with y — 2:
y - 2 = f(x)
y = f(x)+ 2
Pre – Calculus Math 40S: Explained!
www.math40s.com
117
Transformations Lesson 3
Part III: Transforming a point
Transforming a point: Always transform a point by doing
stretches / reflections first, followed by translations.
Example 1: What will the point (-3, 4) become after a transformation of
y = -2f(-x - 4)?
First rewrite the transformation as y = - 2f [ -(x + 4)]
Multiply the x-values by -1, and the y-values by -2 to get (+3, -8)
Move four units left to get (-1, -8)
Example 2: What will a y-intercept of -2 become after a transformation of
y = -f(4x - 28) + 5?
First rewrite the transformation as y = - f [4(x - 7)] + 5
The original point is (0, -2)
Multiply the x-values by ¼ and the y-values by -1 to get (0, 2)
Move 7 right and 5 up to get (7, 7)
Example 3: If the function f(x) = 2x 2 + 3x - 5 is multiplied by a constant
value m, the graph of g(x) = mf(x) passes through the point (2, -27).
Determine the value of m.
First rewrite the equation as y = m(2x 2 - 3x + 5)
Then plug in the given point:
-27 = m ( 2(2)2 + 3(2) - 5)
-27 = 9m
m = -3
Questions:
Given the point (-5, 12), find the new point
after each of the following transformations:
1
1
2) y = - f(-x) – 4
1) y = -3f( x)
2
4
4) y = -2f(-5x – 15) – 6
7) If the function
5) y =
1
2
f(-x - 2)
3) f(-x – 4) + 9
6) y = f(
1
2
x)
2
f(x) = x + 3x - 7 is multiplied by a constant value m,
the graph of g(x) = mf(x) passes through the point (-1, -18). Determine the value of m.
Answers:
1) (-20, -36)
2) (5, -10)
3) (1, 21)
4) (-2, -30)
5) (3, 6)
6) (-10, 12)
7) m = 2
Pre – Calculus Math 40S: Explained!
www.math40s.com
118
Transformations Lesson 3
Part IV: Further Properties of Functions
Even/Odd Functions: These terms are used to describe the
symmetry of a function.
If f(-x) = f(x), the function is said to be even. Even functions are symmetric with respect to
the y-axis, and remain unchanged in a reflection about the y-axis
If f(-x) = -f(x), the function is said to be odd. Odd functions are symmetric with respect to
the origin, and the graph remains unchanged upon a rotation of 180 degrees.
(Instead of a rotation, you can think of it as a reflection in y-axis, then a second reflection in the x-axis)
Example 1: Determine if the function y = 2x 3 + 3x is even, odd, or neither.
f ( -x ) = 2 ( -x ) + 3 ( -x ) → Replace the variable with - x
3
f ( -x ) = -2x 3 - 3x
f ( -x ) = - ( 2x 3 + 3x )
→ Factor out the negative
f ( -x ) = -f ( x ) → This is an odd function
Example 2: Given the partial graph on the right,
draw in the rest of the graph if it‘s:
a) An even function
b) An odd function
Questions: Determine if the following graphs are even, odd, or neither:
1.
y = x3 - x
2.
y = x2
Answers:
y = x3 - x
f ( -x ) = ( -x ) - ( -x )
y = x2
f ( -x ) = -x 3 + x
f ( -x ) = ( -x )
3
3.
y = x3 - x2
f ( -x ) = - ( x 3 - x )
f ( -x ) = -f ( x ) → ODD
y = x3 - x2
2
f ( -x ) = x 2
f ( -x ) = f ( x ) → EVEN
f ( -x ) = ( -x ) - ( -x )
3
f ( -x ) = -x 3 - x
f ( -x ) = ? → NEITHER
Pre – Calculus Math 40S: Explained!
www.math40s.com
2
119
Transformations Lesson 3
Part IV: Further Properties of Functions
x & y intercepts: These are the points on a graph that cross the
x — axis and the y — axis.
To find the x — intercepts, substitute y = 0, then solve for x.
To find the y — intercepts, substitute x = 0, then solve for y.
Example 1: Determine the x & y intercepts of the function f(x) = 2x - 8
*Note for these questions you should rewrite the function as y = 2x — 8 to keep things simple.
x-intercept
y-intercept
Replace y with zero:
Replace x with zero:
y = 2x — 8
0 = 2x — 8
8 = 2x
x=4
y = 2x — 8
y = 2(0) — 8
y = -8
The y — intercept is
the point (0, -8)
The x — intercept is
the point (4, 0)
Example 2: Determine the x & y intercepts of the function f(x) = x + 9 - 2
x-intercept
y-intercept
Replace y with zero:
Replace x with zero:
y = x +9 - 2
y = x+3-2
0 = x +9 - 2
y = 0+9 - 2
2 = x +9
y = 9-2
→ Now square both sides
y = 3- 2
4 = x +9
y =1
x = -5
The x — intercept is
the point (-5, 0)
The y — intercept is
the point (0, 1)
Questions: Determine the x & y intercepts of the following functions
1) f(x) = 4x - 12
2) f(x) = x 2 - 4
Answers:
1) x-int: (3, 0) ; y-int: (0, -12)
2) x-int: (±2, 0) ; y-int: (0, -4)
3) x-int: (-3, 0) ; y-int: (0, 1)
3) f(x) = x + 4 - 1
Pre – Calculus Math 40S: Explained!
www.math40s.com
120
Transformations Lesson 3
Part IV: Further Properties of Functions
Graphs Containing Multiple Functions: Some graphs are a
composite of two or more functions.
⎧⎪ x 2 , x < 0
Example 1: Draw the graph of: f(x) = ⎨
⎩⎪ x , x ≥ 0
⎫⎪
⎬
⎭⎪
This information says that the region to the left of the origin is going to be y = x2, while the region
to the right of the origin is y = x .
Note that the dashed line is only
used for illustrative purposes.
In your graphs, draw a solid line.
⎧⎪ x 2 , x < 0
Example 2: Determine the value of f(-3) in f(x) = ⎨
⎩⎪ x , x ≥ 0
⎫⎪
⎬
⎭⎪
The equation f(x) = x2 must be used, since that’s the function occurring when x = -3
f(x) = (-3)2 = 9
⎧ -x - 3, x < -1 ⎫
f(x)
=
⎨ 2
⎬
Example 3: Draw the graph of:
⎩ x , x ≥ -1 ⎭
When the graph “jumps”, it is necessary to
indicate if the endpoint is part of the graph
or not.
The left part of the graph does not include
the x-value -1, so use an open circle to
indicate this.
The right part of the graph does include the
x-value -1, so use a closed circle to indicate
this.
Pre – Calculus Math 40S: Explained!
www.math40s.com
121
Transformations Lesson 3
Part IV: Further Properties of Functions
Questions: Draw the following graphs:
1.
2.
⎧ -x, x < -1
⎫
f(x) = ⎨
⎬
2
⎩(x + 1) + 1, x ≥ -1 ⎭
Answers:
1.
⎧−
⎪ x + 4, x < 2 ⎫⎪
f(x) = ⎨
⎬
⎩⎪2x - 2, x ≥ 2 ⎭⎪
3.
⎧1/x, x < -1 ⎫
f(x) = ⎨
⎬
⎩ x, x ≥ -1 ⎭
4.
5.
⎧ -x 2 , x ≤ 1 ⎫
f(x) = ⎨
⎬
⎩ x, x > 1 ⎭
⎪⎧ x - 5, x < 1 ⎫⎪
f(x) = ⎨
⎬
⎩⎪− x , x ≥ 1 ⎭⎪
2.
3.
4.
5.
Pre – Calculus Math 40S: Explained!
www.math40s.com
122