Pre-Calculus Math 40s - Transformations Lesson 3

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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



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



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)


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



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



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



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


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


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



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)



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.



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.

Pre-Calculus Math 40s - Transformations Lesson 3

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Transcript of Pre-Calculus Math 40s - Transformations Lesson 3