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3.1: Changing Windows

I. Calculator Windows

A. Example 1: Consider the following graphs of

( ) 3f x x=10

8

6

4

2

10

8

6

4

2

10

8

6

4

2

-2

-4

-6

-8

-10

-10 -5 5 10

-2

-4

-6

-8

-10

-40 -20 20 40

-2

-4

-6

-8

-10

-4 -2 2 4

10 1010 10

xy

− ≤ ≤− ≤ ≤

50 5010 10

xy

− ≤ ≤− ≤ ≤

5 510 10

xy

− ≤ ≤− ≤ ≤

B. Example 1:Graph the following function:

Asymptote – a line

the graph approaches

i. As x increases or decreases

without bound.

ii. As x approaches some fixed value

1( )f xx

=

6

4

2

f x( ) = 1x

The function has two asymptotes:-2

-4

-5 5

Horizontal: 0Vertical: 0

yx

==

II. The Parent Functions

A. A parent function is the simplest version of a commonly used function (centered at the origin).

B. It gives the general shape of the function.

1. The Identity Function

( )f x x=Domain:

Range:a ge:

Asymptotes?

Points of Discontinuity?

none

none

2. The Squaring Function2( )f x x=

Domain:

Range: , 0y ≥

Asymptotes?

Points of Discontinuity?

none

none

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3. The Cubing Function3( )f x x=

Domain:

Range:

Asymptotes?

Points of Discontinuity?

none

none

4. The Square Root Function

( )f x x=Domain:

Range:

, 0x ≥, 0y ≥

Asymptotes?

Points of Discontinuity?

none

none

5. The Absolute Value Function

( )f x x=Domain:

Range: , 0y ≥

Asymptotes?

Points of Discontinuity?

none

none

6. The Exponential Function

( ) , 1xf x b b= >Domain:

Range: , 0y >a ge:

Asymptotes?

Points of Discontinuity?

, y

0y =

none

7. The Reciprocal Function

1( )f xx

=

Domain: , 0x ≠

Range:

Asymptotes?

Points of Discontinuity?

, 0y ≠

0, 0x y= =

yes, when 0x =

8. The Inverse Square Function

2

1( )f xx

=

Domain: , 0x ≠

Range:

Asymptotes?

Points of Discontinuity?

, 0y >

0, 0x y= =

yes, when 0x =

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9. The Greatest Integer Function

( )f x x= ⎢ ⎥⎣ ⎦Domain:

Range:Range:

Asymptotes?

Points of Discontinuity?

, all integersnone

yes, at every integer

Note: What you should include on a graph from a calculator:

Axes are labeled

Scales on the axes are shown

Characteristic shape can be seen

Intercepts are shown

Points of discontinuity are shown

HW: p. 164-166 # 4,5, 8, 10, 11, 13, 14, 15, 17, 18, 19

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3.2: The Graph Translation Theorem

I. Translating graphs.

A. Example1: Write a rule for translating the point (x, y) 8 units down and 5 right.

( )( , ) 8, 5f x y x y= − +

B. Example 2: Let and

Graph each

2( )f x x= ( )2( ) 3 4g x x= + +

6

4

8

6

4

g x( ) = x+3( )2+4

2

-2

-4

-5 5

f x( ) = x2

2

-2

-5 5

f x( ) = x2

C. Note: f(x) is the pre-image and g(x) is the image.

D. Definition: A translation in the plane is a transformation that maps each point (x,y) onto

(x+h, y+k).

Fi d th t l ti th t f t Find the translation that maps f onto g

( ): ( , ) 3, 4T x y x y→ − +

E. The Graph Translation Theorem:

In a relation described by a sentence in x and y, the following two processes yield the same graph.

1. replacing x by x-h and y by y-k in the same sentence

2. applying the translation below to the graph or h i i l l ithe original relation

( , ) ( , )x y x h y k→ + −

F. Example 3: Find an equation for each of the functions below using the following steps:

1. Determine the parent function

2. Determine how the function has moved from the original location (h units in the x-directionand k in the y-direction)and k in the y direction)

3. replace x with (x-h) and y by (y-k) in the equation

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1. Parent function:

Moved:

Equation:

4

2

-2

-5

( )f x x=

: ( , ) ( 4, 3)T x y x y→ − −

4, 3h k= − = −

Asymptotes:

( ) 3 4f x x+ = + or ( ) 4 3f x x= + −

none

2. Parent function:

Moved:

Equation:

3( )f x x=

: ( , ) ( 3, 3)T x y x y→ − +

3, 3h k= − =

6

4

2

-5 5

Asymptotes:

( )3( ) 3 3f x x− = + or ( )3( ) 3 3f x x= + +none

3. Parent function:

Moved:

Equation:

1( )f xx

=

: ( , ) ( 5, 2)T x y x y→ + +

5, 2h k= =1

4

2

-2

5

Asymptotes:

1( ) 25

f xx

− =−

or

5, 2x y= =

1( ) 25

f xx

= +−

4. Parent function:

Moved:

Equation:

( )f x x=

: ( , ) ( 4, 3)T x y x y→ − −

4, 3h k= − = −

4

2

-2

5

Asymptotes:

( ) 3 4f x x+ = + or

none( ) 4 3f x x= + −

5. Parent function:

Moved:

Equation:

2

1( )f xx

=

: ( , ) ( 4, 2)T x y x y→ + −

4, 2h k= = −

4

2

-2

5

1

Asymptotes: ( )2

1( ) 24

f xx

+ =−

or

4, 2x y= = −

( )21( ) 2

4f x

x= −

−

G. EX 4: Find equations for all asymptotes of the graph of

When is the denominator 0 (this is where there will be vertical asymptotes)?

1( )( 5)( 2)

f xx x

=+ −

when 5,2x = − 5, 2x x= − =What happens to the function as x gets large (this will give you the horizontal asymptotes)

, ,

it approaches 0 0y =

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Homework: p. 170-171 # 1-17

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3 3: Translations of Data3.3: Translations of Data

I. Data TranslationsA. Example 1: Suppose a small class yields the following set of test scores.87, 86, 85, 81, 78, 75, 75, 73, 70, 68, 67, 63.Calculate each of the following:Mean Min75.6 63Mode Q1

Variance Median

Standard Deviation

Q3

Range Max

IQR

75

62.79

7.92

24

14

69

75

83

87

B. Example 2: add 12 points to each score, then recalculate the information from before

Original +12 Original +12

Mean 75.67 Min 63

Mode 75 Q1 69

87.6

87

75

81Mode 75 Q1 69

Variance 62.79 Median 75

St. Dev. 7.92 Q3 83

Range 24 Max 87

IQR 14

87

62.79

7.92

24

14

81

87

95

99

C. Theorem: Adding (h) to each number in a data set adds (h) to each of the mean, median, and mode

D. Theorem: Adding (h) to each number in a data set does not change the range interquartiledata set does not change the range, interquartilerange, variance, or standard deviation. They are called invariant under a translation.

II. Reducing an entire list of data on the TI-83.A. Example 4: Find the line of best fit:

B. Find the line of best fit where x = # of years after 1900.

L1 L2

1960 45

1963 43

( ) 0.485 985.830f x x= − +

1963 43

1965 42

1969 40.5

1972 39

( ) 0.485 73.496f x x= − +

III. Summation Notation ReviewEX 2: Evaluate the following for the set

p1 = 4 p2 = 7 p3 = 6 p4 = 9

4

1

a. 6ii

p=

+∑ ( )4 7 6 9 6= + + + + 32=

4

1

b. ( 6)ii

p=

+∑ (4 6) (7 6) (6 6) (9 6)= + + + + + + +

50=

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Homework: p. 178 # 1, 4, 5, 7-9, 11, 12.

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I. SymmetriesA. A function is said to be even if:

1. for every point (x,y) on the graph, (-x,y) is also on the graph.

2 2.

3. The function is symmetric to the y-axis

( ) ( )f x f x− =

B. A functions is said to be odd if:

1. for every point (x,y) on the graph, (-x,-y) is also on the graph.

2. ( ) ( )f x f x− = −

3. The function is symmetric to the origin

C. Example 1: List another point on the function containing (-1,3) if it is an:

even function

odd function

(1,3)

(1, 3)−odd function (1, 3)

D. Example 2: Prove that is symmetric to the y-axis.

2y x=

( )2( )f x x− = −

( ) ( )2 21 x= −

( )( )21( )( )1 x=2x=( )f x=

( ) is an even functionf x∴

You can verify by graphing

4

2

f x( ) = x2

since the graph is symmetric to the y-axis,it is an even function

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E. Example 3: Prove that is odd. 3y x x= −

( ) ( )3( )f x x x− = − − −

( ) ( ) ( )( )3 31 1x x= − − −

( )( ) ( )( )31 1x x= − − −

( )31 x x= − −

( )f x= −( ) is an odd functionf x∴

( )( ) ( )( )1 1x x= − − −

You can verify by graphing

2

f x( ) = x3-x

since the graph is symmetric to the origin,it is an odd function

-2

F. Example 4: Consider the following functions. Decide if they are even, odd, or neither.

4( )f x x x= + 2( ) 3f x x= −4

2 2

g x( ) = 3-x2

g x( ) = x4+x

-2

neither even

5

3

2( ) x xf xx−

=

6

4

2

2f x( ) = x+2

( ) 2f x x= +

-2

-4

-6

-5 5

g x( ) = x5-2⋅x

x3

-2

neithereven

( ) 2f x x= −

2f x( ) = -2⋅ x2f x( ) = 2⋅x5-3⋅x3

5 3( ) 2 3f x x x= −

-2-2

odd even

Homework: p. 183- 185 # 7-9, 11-20

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I. Scale ChangesA. Example 1: Consider the graph of

3( ) 4f x x x= − 6

4

2

f x( ) = x3-4⋅x

-2

-4

-6

-5 5

B. Now, replace y with y/3Solve the new equation for y:

Graph it

3 43y x x= − 33 12y x x= −

10

g x( ) = 3⋅x3-12⋅x

5

-5

10

f x( ) = x3-4⋅x

C. What happens to the y-coordinates?they are tripled

D. This is called a vertical stretch of magnitude 3.

E. It can be seen as the followingE. It can be seen as the following transformation:

( ): ( , ) ,3T x y x y→

F. Now, replace x with x/3Solve the new equation for y:

Graph it

3

42 2x xy ⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

3

28xy x= −

55

-5

10

g x( ) = x3

8-2⋅x

G. What happens to the x-coordinates?they are doubled

H. This is called a horizontal stretch of magnitude 2.

I. It can be seen as the followingI. It can be seen as the following transformation:

( ): ( , ) 2 ,T x y x y→

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J. Let . Find an equation for g(x), the image of f(x) under

3( ) 4f x x x= −( , ) (2 ,3 )S x y x y→

2

43 2 2y x x⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠2

2y x 23 8y x= −

33 68xy x= −

II. Graph Translation TheoremIn a relation described by a sentence in x and y, the following two processes yield the same graph:

1. replacing (x) by x/2 and (y) by y/2 in the sentence

2 applying the scale change to the graph of the original relation.

Note: If a = b, then you have performed a size change.

A. Example 2: Given the graph below, draw (3 )

2y f x=

9 03 2

1 12 35 2

x y−− −

−

3 23 01 1

1 13 2

32

yx

−− −

8

6

4

2

-2

-4

-6

-8

-10 -5 5 10

f(x)

5 2 323 2

5 13

−

2

-2

-5

B. EX 3: Consider . Find an equation for the function under

Replace (x) with

Replace (y) with

y x=

( , ) , 23xS x y y⎛ ⎞→ −⎜ ⎟

⎝ ⎠3x

2y

−

Now make the new equation:

( )32y x⎛ ⎞− =⎜ ⎟

⎝ ⎠2 3 6 6y x x x= − = − = −

Homework:p. 191-193 # 9-14 (skip 14(c)), 15, 17-20

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3.6: SCALE CHANGES OF DATA

I. Scale Changes of DataA. Example 1: consider the following set of data:

1. Calculate each of the following:{ }5,2,5,9, 1,6,3,2−

Mean Media Mode3 875 4 2 5Mean n Mode

Range

St. Dev.

Variance

3.875

9.27≈10 3.04≈

4 2,5

2. Multiply every term in the set by (a) where a = 3.

Mean Median Mode

Range St. Dev.

Variance

3.875

9.27≈10 3.04≈

4 2,5

{ }15,6,15,27, 3,18,9,6−

C l l t th f ll i

Mean Median Mode

Range St. Dev. Varianc

e

Calculate the following:

11.625

9.27≈10 3.04≈

12 6,15

3. Multiply every term in the set by (a) where a = -3.

Mean Median Mode

Range St. Dev.

Variance

3.875

9.27≈10 3.04≈

4 2,5

{ }15,6, 15, 27,3, 18, 9, 6− − − − −

C l l t th f ll i

Mean Median Mode

Range St. Dev. Varianc

e

Calculate the following:

11.625−

9.27≈10 3.04≈

12− 6, 15− −

II. General Scale Changes:When you multiply a set of data by some constant a:

A. Measures of center (mean, median, mode) are multiplied by a

B. Measures of spread (st. dev, IQR, Range,…) are multiplied by |a|.

C. Variance is multiplied by a2.p y

III. Real-World ExampleA. Example 2: The teachers in a school have a mean salary of $30,000 with a standard deviation of $4,000. If each teacher is given a 5% raise, what will be their new mean salary and the new standard deviation?

( )$30,000 1.06 $31,800=

( )$4000 1.06 $4,240=

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B. Example 3: To give an approximate conversion from miles to kilometers you can multiply the number of miles by 1.61. Suppose data are collected about the number of miles that cars can go on a tank of gas. What will be the effect of changing from miles to kilometers on:

1. the median of the data? multiply by 1 61

2. the variance of the data?

3. the standard deviation of the data?

multiply by 1.61

( )2multiply by 1.61 2.59≈

multiply by 1.61

Homework: p. 198-200 # 4-9, 14-22

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3-7: Composition of Functions

I. Composition of Functions: Definition

A. Suppose f and g are functions. The composite of g with f, written , is the function defined by

B The domain of is the set of values of x in the domain

( )( ) ( ( ))g f x g f x=

g fB. The domain of is the set of values of x in the domain of f for which is in the domain of g.

g f

C. Example 1: Let and . Evaluate:

1.

1( )3 1

g xx

=+

2( )f x x=

( (4))f g =( )

13 4 1

f⎛ ⎞⎜ ⎟⎜ ⎟+⎝ ⎠

113

f ⎛ ⎞= ⎜ ⎟⎝ ⎠

2113⎛ ⎞= ⎜ ⎟⎝ ⎠

1169

=

2. ( )(4)g f =169

( (4))g f( )( )24g= ( )16g=

( )1

3 16 1=

+149

=

D. Is composition communitive? Why or why not?

no, because if you switch the order, the answer changes

f g g f≠

E. Example 2: Using the functions f and g above.

1. Find formulas for

( )( )f g x = ( ( ))f g x = 13 1

fx

⎛ ⎞⎜ ⎟+⎝ ⎠

213 1x

⎛ ⎞= ⎜ ⎟+⎝ ⎠

1( )2

13 1x

=+

( )( )g f x = ( ( ))g f x = ( )2g x ( )21

3 1x=

+ 2

13 1x

=+

2. verify that by graphingf g g f≠

3

2

gof x( ) = 1

3⋅x2+1

4

3

fog x( ) = 1

3⋅x+1( )2

2

1

-1

-4 -2 2 4

2

1

-2 2 4

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3. Give the domains of each of the following:

( )f

( )f x

( )f g x

( )g x1,3

x ≠ −

13 1 03

x x+ ≠ → ≠ −

1,3

x ≠ −( )g f x ( )f g xi. first, use inside function's domain 1,

3x ≠ −

ii. next, use composite function's domain2

2

3 1 013

no restrictions

x

x

+ ≠

≠ −( )23 1 0

13

x

x

+ ≠

≠ −

3

Examples:

(5(1) 3) (2)f f− =2(2) 2(2) 7 15+ + =

( )2(1) 2(1) 7 (10)g g+ + =

5(10) 3 47− =

( ) ( )2

2

2

(5 1) 5 1 2 5 1 7

25 10 1 10 2 7

25 6

f x x x

x x x

x

− = − + − +

= − + + − +

= +

( )(5 3) 5 5 3 325 15 3

25 18

g x xx

x

− = − −

= − −

= −

3.

( )( )

2( ( )) 1 2

1 23

n s x x

xx

= − −

= − −

= −domain of s:

1 01

xx− ≥≥

domain of 3: x −

, 1x ≥

Homework:

p. 205-207 #4-10 even, 11-16, 18-23, 26

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I. Inverses

A. Definition: If a function is written as a set of ordered pairs, its inverse is the set of ordered pairs with the x and y‐coordinates reversed.

B. If a function is written as an equation, its inverse is the equation with the x and y values switched.

C. The inverse of the function f is written f‐1

B. Example 1:1.Use the relation

i. Find the inverse of S.

f

{ }( 1,1),(1,1),( 2,4), (2,4),( 3,9), (3,9)S = − − −

{ }1 (1, 1), (1,1),(4, 2), (4,2),(9, 3), (9,3)S − = − − −

ii. Is S a function?

iii. Is the inverse of S a function? 

yes, no repeated -x values

no, repeated -x values

II. Horizontal Line TestA. The inverse of a function f is itself a function iff no horizontal line intersects the graph of f in more than one point.

4f x( ) = x4f x( ) =

1x

passes: inverse is a function

2

fails: inverse is not a function

2

-2

B. Example 2: Give the inverse of the functions below, then graph both the functions and their inverses on the same set of axes

31. ( )f x x=2

f x( ) = x3

3y x=3x y=2

1

-2

-4

does this pass the HLT?yes

3 x y=1 3( )f x x− =

-2

g x( ) = x13 33 3x y=

42. ( )f x x= 4y x=4x y=

44 4x y± =

4

2

f x( ) = x4 4

2

does this pass the HLT?no

4 x y± =1 4( )f x x− = ±

-2

5

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C. Notes1. The horizontal line test for a function is the same as the VLT for its inverse

4

2

5

f -1(x)

4

2

5

f(x)

2. A function and its inverse are symmetric to the line y=x-2

-2

4

2

-2

5

f -1(x)

f(x)

III. Inverse Function theorem.

A. any two functions f and g are inverses if f(g(x)) = x in the domain of g.

B  This theorem can be used to test if two functions are B. This theorem can be used to test if two functions are inverses of each other. 

C. Example 3: use the inverse function theorem to show that the following functions are inverses of each other. 4( ) 3 4, ( )

3xf x x g x +

= − =( )( )f g x =

4x +⎛ ⎞ 4x +⎛ ⎞ ( )3 443

xf +⎛ ⎞ =⎜ ⎟⎝ ⎠

43 43

x +⎛ ⎞ − =⎜ ⎟⎝ ⎠

( )3 44

3x +

− = ( )4 4x + − = x( )( )g f x =

( )3 4g x − = ( )3 4 43

x − += 3 4 4

3x − +

=3xx= x

these functions are inverses∴

4

2g x( ) = x+43

Note their symmetry to the line y x=

h x( ) = x

3

f x( ) = 3⋅x-4

D. Example 4: Use the Inverse Function Theorem to show that the functions below are not inverses.

4 4( ) , ( )f x x g x x−= =

( ( ))f g x = ( )4f x− = ( )44x− = 16x− x≠they are not inverses∴

note they are notsymmetric to theline y x=

4

2

h x( ) = x

g x( ) = x-4

f x( ) = x4

HW: p. 212‐214 #2‐12 even, 13‐20, 22‐23