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Pre – Calculus Math 40S: Explained! www.math40s.com 110
Pre – Calculus Math 40S: Explained! www.math40s.com 111
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 112
Transformations Lesson 3 Part I: Algebraic Transformations
Example 2: Given the function f(x) = 3x2 – 3x + 4, find the equation of: a) 2f(x) y = 2(3x2 — 3x + 4) y = 6x2 — 6x + 8 b) f(3x) y = 3(3x)2 — 3(3x) + 4 y = 3(9x2) — 3(3x) + 4 y = 27x2 — 9x + 4 c) –f(x) y = -(3x2 — 3x + 4) y = -3x2 + 3x - 4 d) f(-x) y = 3(-x)2 — 3(-x) + 4 y = 3x2 + 3x + 4 e) y = f(x) +3 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
Example 3: Given the function f(x) = 3x + 6 , find the equation of: a) y = 4f(x - 2) y = 4 3(x - 2)+6
y = 4 3x - 6+6 y = 4 3x b) y = -f(-3x) y = - 3(-3x)+6
y = - -9x +6 c) y = –f(x + 2) + 4 y = - 3(x + 2)+6 + 4
y = - 3x +6+6 + 4 y = - 3x +12 + 4
d) y = 12
f(-x - 2)
y = 1
3(-x - 2)+62
y = 1
-3x - 6+62
y = 1
-3x2
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 113
Transformations Lesson 3 Part I: Algebraic Transformations
Questions: 1) f(x) = 3x – 7 a) y = 3f(x) b) y = f(6x) c) y = –f(x) d) y = f(-x) e) y = f(x) +5 f) y = f(x – 6)
Answers:
4) f(x) =
3) f(x) = - -2x - 3 a) y = 2f(x - 3) b) y = f(3x + 4) c) y = –f(-x) d) y = 2f(-x) + 4
e) y = -12
f(x) +3
f) y = f(-x – 4) + 9
3x - 4 a) y = 2f(x) - 8 b) y = -f(7x - 1)
c) y = –
3. a) y = - 2 -2x + 3
b) y = - -6x - 11
c) y = 2x - 3
d) y = - 2 2x - 3 +4
1e) y = -2x - 3 + 3
2f) y = - 2x +5 +9
4. a) y = 2 3x - 4 -8
b) y = - 21x - 7
1c) y = - 3x - 4 - 3
31
d) y = -3x - 42
e) y = 2 3x - 19 + 3
f) y = - -3x - 16
a) y = 9x - 21
b) y = 18x - 7
c) y = - 3x +7
d) y = - 3x - 7
e) y = 3x - 2
f) y = 3x - 25
1.
( )
( )( )
( )( )( )
2. 2
2
2
2
2
2
1a) y = 2x - 3
2
b) y = x - 3
c) y = - 3 2x - 3
d) y = -4x - 3
e) y = 2x - 3 - 4
f) y = 2x +7
13
f(x) - 3
d) y = 12
f(-x)
e) y = 2f(x - 5) +3 f) y = -f(-x – 4)
2) f(x) = (2x – 3)2
a) y = 12
f(x)
b) y = f(12
x)
c) y = –3f(x) d) y = f(-2x) e) y = f(x) – 4 f) y = f(x + 5)
Pre – Calculus Math 40S: Explained! www.math40s.com 114
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
First factor out the 3 from the x: [ ]2y = 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.
First apply the stretches: y = 1
3x2
Now apply the translations: y = 1
3x + 32
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 115
Transformations Lesson 3 Part II: Describing Transformations
Questions: 1) Describe how each transformed function compares to the original:
xg) Original is y =
Transformed is y = -3 x
h) Original is y = x2
Transformed is y = 2(-3x – 12)2 + 3
i) Original is y =
a) Original is y = f(x)
Transformed is y = -3f(
1
4x)
b) Original is y = f(x)
Transformed is y = -1
2f(-x) – 4
c) Original is y = f(x) Transformed is y = f(-x + 3) d) Original is y = f(x - 2) Transformed is y = f(x +5) e) Original is y = f(x + 7) + 2 Transformed is y = f(x + 4) - 6 f) Original is y = f(x) Transformed is 4y = f(x)
x
Transformed is y = - 2x + 4 j) Original is y = (x – 3)2 Transformed is y = (x – 4)2 k) Original is y = (x + 8)2
- 4 Transformed is y = (x + 6)2
- 3 l) Original is y = x2
Transformed is y + 3 = x2
Pre – Calculus Math 40S: Explained! www.math40s.com 116
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
3, a horizontal stretch by a factor of 3, and a
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.
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.
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.
g) The graph of y = (x + 4)2 – 6 is shifted 2 units to the left and 5 units up. 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
2y
j) The graph of y = f(x) has y replaced with y - 2
Pre – Calculus Math 40S: Explained! www.math40s.com 117
Transformations Lesson 3 Part II: Describing Transformations
Answers: 2.
a) ⎛ ⎞⎜ ⎟⎝ ⎠
22 1
y = (x - 3)3 3
b) y = - x - 4
c) y = ( )⎡ ⎤⎣ ⎦
1. a) Vertical Stretch by a factor of 3 Horizontal Stretch by a factor of 4 Reflection in the x-axis b) Vertical Stretch by a factor of ½ Reflection in both the x & y axis Translated 4 units down c) Rewrite as [ ]( 3)y f x= − − Reflection in the y-axis Translated 3 units right d) Original point = (2, 0) Transformed point = (-5, 0) Translated 7 units left e) Original Point = (-7, 2) Transformed Point = (-4, -6) Translated 3 units right and 8 units down.
f) Rewrite as y = 1
( )4
f x
Vertical Stretch by a factor of 1/4 g) Vertical Stretch by a factor of 3 4f - x +7
d)⎛ ⎞⎜ ⎟⎝ ⎠
31
y = - x - 53
e) 4 1
y = - - (x - 2) - 23 6
f) Original Point = (4, 0) Transformed Point = (9, 0)
2y = (x - 9) g) Original Point = (-4, -6) Transformed Point = (-6, -1)
2y = (x +6) - 1 h) Original Point = (1, -3) Transformed Point = (8, -6) y = f(x -8) - 6
i) Replace y with 1 y2
:
Reflected in the x-axis
h) Rewrite as [ ]2y = 2 -3(x + 4) + 3
Vertical Stretch by a factor of 2 Horizontal stretch by a factor of 1/3 Reflected in the y-axis Translated 4 units left and 3 units up
i) Rewrite as 2( 2)y x= − + Horizontal stretch by a factor of 1/2 Reflection in the x-axis Translated 2 units left j) Original Point = (3, 0) Transformed Point = (4, 0) Translated 1 unit right
k) Original Point = (-8, -4) 1
y = f(x)2y = 2f(x)
Transformed Point = (-6, -3) Translated 2 units right and 1 unit
up.
l) Rewrite as y = x2 — 3 j) Replace y with y — 2: Translated 3 units down y - 2 = f(x)
y = f(x)+2
Pre – Calculus Math 40S: Explained! www.math40s.com 118
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)? [ ]y = - 2f -(x +4)
First rewrite the transformation as
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? [ ] y = - f 4(x - 7) + 5
First rewrite the transformation as
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 2f(x) = 2x + 3x - 5 is multiplied by a constant g(x) = f(x)mvalue m, the graph of passes through the point (2, -27).
Determine the value of m. First rewrite the equation as 2y = m(2x - 3x +5)Then plug in the given point:
( )2-27 = m 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
21
42) y = - f(-x) – 4 3) f(-x – 4) + 9 1) y = -3f( x)
Answers: 4) y = -2f(-5x – 15) – 6
7) If the function 2
f(x) = x + 3x - 7 is multiplied by a constant value m,
the graph of passes through the point (-1, -18). Determine the value of m. g(x) = f(x)m
1) (-20, -36) 1
21
2f(-x - 2) 5) y = x) 6) y = f( 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 119
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 3y = 2x + 3x is even, odd, or neither.
( ) ( ) ( )( )( ) ( )( ) ( )
→
→
→
3
3
3
f -x = 2 -x + 3 -x Replace the variable with - x
f -x = -2x - 3x
f -x = - 2x + 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:
y 3 -= x x1. Answers:
y = 2x2.
( ) ( ) ( )( )( ) ( )( ) ( ) ODD→
3
3
3
3
y = x - x
f -x = -x - -x
f -x = -x +x
f -x = - x - x
f -x = -f x
( ) ( ) ( )( )( )
2
NEITHER→
3 2
3
3
y = x - x
f -x = -x - -x
f -x = -x - x
f -x = ?
( ) ( )( )( ) ( ) EVEN→
2
2
2
y = x
f -x = -x
f -x = x
f -x = f x
y3. 3 2-= x x
Pre – Calculus Math 40S: Explained! www.math40s.com 120
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 Example 2: Determine the x & y intercepts of the function f(x) = x + 9 - 2 Questions: Determine the x & y intercepts of the following functions
1) f(x) = 4x -12
2) 2f(x) = x - 4
3) f(x) = x + 4 -1
y-interceptReplace y with zero: Replace x with zero:
y = 2x — 8 y = 2x — 8 0 = 2x — 8 y = 2(0) — 8
8 = 2x y = -8 x = 4
The y — intercept is the point (0, -8) The x — intercept is
the point (4, 0)
x-intercept y-interceptReplace y with zero: Replace x with zero:
→ Now square both sides
y = x +9 - 2
0 = x +9 - 2
2 = x +94 = x +9x = -5
y = x +3 - 2
y = 0+9 - 2
y = 9 - 2y = 3- 2y =1
The y — intercept is The x — intercept is
the point (0, 1) the point (-5, 0)
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)
Pre – Calculus Math 40S: Explained! www.math40s.com 121
Transformations Lesson 3 Part IV: Further Properties of Functions
Graphs Containing Multiple Functions:
Some graphs are a composite of two or more functions.
,
,
⎧ ⎫⎪ ⎪⎨ ⎬
≥⎪ ⎪⎩ ⎭
2x x < 0f(x) =
x x 0 Example 1:
Draw the graph of:
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.
,
,
⎧ ⎫⎪ ⎪⎨ ⎬
≥⎪ ⎪⎩ ⎭
2x x < 0f(x) =
x x 0Example 2: Determine the value of f(-3) in
The equation f(x) = x2 must be used, since that’s the function occurring when x = -3 f(x) = (-3)2
= 9
, ,
⎧ ⎫⎨ ⎬
≥⎩ ⎭2
-x - 3 x < -1f(x) =
x x -1Example 3: Draw the graph of:
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 122
Transformations Lesson 3 Part IV: Further Properties of Functions
Questions: Draw the following graphs:
1.
, ,
⎧ ⎫⎨ ⎬
≥⎩ ⎭2
-x x < -1f(x)
(x +1) +1 x -1=
, ,
⎧− + ⎫⎪ ⎪⎨ ⎬
≥⎪ ⎪⎩ ⎭
x 4 x < 2f(x) =
2x - 2 x 2
2.
, ,
⎧ ⎫⎨ ⎬≥⎩ ⎭
1/x x < -1f(x) =
x x -1
3.
4. 5. ,
,
⎧ ⎫⎪ ⎪⎨ ⎬− ≥⎪ ⎪⎩ ⎭
x - 5 x < 1f(x) =
x x 1
, ,
⎧ ⎫≤⎨ ⎬⎩ ⎭
2-x x 1f(x) =
x x > 1
Answers: 1. 2. 3. 4. 5.