Module 1: Answer Key - Open School BCmedia.openschool.bc.ca/ocrmedia/pmath12/pdf/pma12_mod1...Module 1 Section 1, Lesson 1 Answer Key Note: The graphing calculator displays in these

Principles of Mathematics 12 Answer Key, Contents 137

Module 1

Module 1: Answer Key

Section 1: Review of Mathematics 11

Lesson 1 Graphing Calculator Review 139

Lesson 2 Functions and Interval Notation 142

Lesson 3 Inverse Functions 143

Lesson 4 Polynomial Functions and Their Graphs 145

Review 148

Section 2: Transformations

Lesson 1 Translations 151

Lesson 2 Reflections 157

Lesson 3 Absolute Value Functions 163

Lesson 4 Stretches and Compressions 167

Lesson 5 Reciprocal Relations 174

Lesson 6 Composition of Transformations 180

Review 192

138 Answer Key, Contents Principles of Mathematics 12

Module 1

Module 1

Section 1, Lesson 1

Answer Key

Note: The graphing calculator displays in these answers all employ the standardwindow of X[ –10, 10 ] and Y[–10, 10]. In most cases, this is not the best windowsize; the tops and bottoms of the curves extend beyond the viewing area. Giveyourself a pat on the back (and possibly an extra mark on the provincial exam) ifyou used windows with Y-boundaries greater than standard, such as Y[–20, 20].

Although not as important, many of your window X-boundaries could be narrowerthan X [–10, 10]. Furthermore, the two boundary numbers within a pair ofbrackets need not be equal.

1. Equation should be rearranged first: x3 – x2 – 12x + 3 = 0

Display

The graph crosses the x-axis in three places, so there are threesolutions. Here they are labelled #1, #2, and #3.

1. (a) Guess #1: –3.5 (your answer may vary but should liebetween –3 and –4)

(b) Bounds: {–4,–3} (your bounds may vary)

The equation to use at the solve (or eqn:0= prompt) is: X^3 – X^2 – 12X + 3

(c) Answer #1: –3.134 (rounded to 3 decimal places)

2. (a) Guess #2: 0.5(b) Bounds: {0,1}

(c) Answer #2: 0.246

3. (a) Guess #3: 3.5

(b) Bounds: {3,4}

(c) Answer #3: 3.888

SOLUTION: –3.134, 0.246, 3.888

Principles of Mathematics 12 Section 1, Answer Key, Lesson 1 139

2. –x3 + 2x2 – x + 1 = 0

Display

For this equationthere is only onesolution

Guess: 1.5Bounds: {1,2}The equation to use is: –X^3 + 2X^2 – X + 1 (It must start withthe negation sign, not minus.)

Answer: 1.754877666

SOLUTION: 1.755

3. x3 + 6x2 + 3x – 5 = 0

Display

The graph shows thereare three solutions. Herethey are labeled A), B),and C).

A) Guess: –5.5Bounds: {–6,–5}

The equation to use is: X^3 + 6X^2 + 3X – 5 (It must startwith the negation sign, not minus.)Answer: –5.246551365

B) Guess: –1.5Bounds: {–2,–1}Answer: –1.423112476

C) Guess: 0.5Bounds: {0,1}Answer: 0.6696638406

SOLUTION: –5.247, –1.423, 0.670 (to 3 decimal places)

X[–7, 3] Y[–6,20]

140 Section 1, Answer Key, Lesson 1 Principles of Mathematics 12

Module 1

4. Equation should be rearranged first: x3 – 3x2 – 9x + 9 = 0

Display

The graph shows thereare three solutions.Here they are labeledA), B), and C).

A) Guess: –2.5Bounds: {–3,–2}

The equation iskeyed in as:X^3 – 3X^2 – 9X + 9 Answer: –2.377551642

B) Guess: 0.5Bounds: {0,1}Answer: 0.8329448266


SOLUTION: –2.378, 0.833, 4.545 (to 3 decimal places)

5. Display

The graph shows thereare three solutions. Herethey are labeled A), B),and C).

A) Guess: –3.5Bounds: {–4,–3}Answer:–3.783908881

B) Guess: –0.5Bounds: {–1,0}Answer: –0.3449601863


SOLUTION: –3.874, -0.345, 6.129 (to 3 decimal places)


Module 1

Lesson 2

1. a) 3 g) 4x2 – 3x + 4

b) 1 h) –4x2 – x – 2

c) 21 i) 105

d) 1/2 j) 3

e) 4a2 – a + 3 k)

f) 3 – 2b

2. a) y-int: y = –8 x-int: x = –2, 2

b) y-int: x-int:

c) y-int: y = 5 x-int: x = 5

3. i) a) domain: {x|x ∈ ℜ} range: {y|y > –8}

b) domain: range: {y|y > 0}

c) domain: {x|x ∈ ℜ} range: {y|y ∈ ℜ}

ii) a) domain: ℜ range: (–8,∞)

b) domain: range: (0,∞)

c) domain: ℜ range: ℜ

4. a) determine each of the following:

i) p(q(x)) =

ii) p(q(3)) =

iii) q(p(a)) =

b) Find the domain and range of

i) domain (p°q) = (1,∞)

range (p°q) = (0,∞)

ii) domain (q°p) = (4,∞)

range (q°p) = (1,∞)

3 4 1a − +

6

3 3x −

5,

2 − ∞

5{ | }

2x x > −

52

x = −5y =

2

3 24 9 8

bb b

−− +


Module 1

Lesson 3

1.

( ) ( )

1

1 1

1 1

2

1

21 1 2

a) ( )3( ) 3

3( ( )) ( ( )) 3

3 3domain of range of

domain of range of

b) ( ) 2

( ) 2

( ( )) 2 2 ( ( )) 2 2

domain of range of

xf x

f x x

x xf f x x f f x x

f f

f f

f x x

f x x

f f x x x f f x x x

f f

−

− −

− −

−

− −

=

• =

• = = = = • = ℜ = ℜ

= ℜ = ℜ= −

• = +

• = + − = = − + =

• = ℜ ( )( ) ( )

( )

( )

1 1

21

21

1 1

1

2,

domain of 2, range of 0,

c) ( ) 3 2

2( )

3

3 2( ( )) 2

3

2domain of , range of 0,

3

2domain of range of ,

31

d) ( ) , 03

1( ) 3

(

f f

f x x

xf x

xf f x x

f f

f f

f x xx

f xx

f f

− −

−

−

− −

−

= − ∞

= − ∞ = ∞

= −

+• =

+ • = − =

• = ∞ = ∞ = ℜ = ∞

= ≥−

• = +

•

{ } { }{ } { }

1 1

1 1

1 1 1( )) ( ( )) 3

1 1 13 3 3

domain of ( ) | 3 range of ( ) | 0

domain of ( ) | 0 range of ( ) | 3

x x f f x x

x x xf x x f y y

f x x f y y

− −

− −

= = = = + =

+ − −• = ≠ = ≠

= ≠ = ≠


Module 1

2. 1( )

31

31

3(3 ) 1

3 1(3 1) 1

13 1

xf x

xx

yx

yx

yx y yxy y

y x

yx

+=

+=

+=

= +− =− =

=−


Module 1


Module 1

Lesson 4

1.

2. • opens down• opens up• falls from Quadrant II

3. a) v)b) iii)c) ii)d) vi)e) i)f) vii)g) iv)h) viii)

4. a) i) 4x − x3 = x(2 + x)(2 – x) zeros at –2, 0, 2

ii) max no turns = 2

iii) falls from Quadrant II

iv) domain of f = ℜ range of f = ℜ

4

4

3 4 2

4 3 2

5 2

5 2

a) ( ) 8

8 degree 4max no. of turns = 3

b) ( ) 5 2

5 2 degree 4max no. of turns = 3

c) ( ) 3

3 degree 5max no. of turns = 4

f x x x

x x

g x x x x

x x x

h x x x x

x x x

= −− + =

= + −

+ − =

= − − +

− − + =

v)

b) i) zero at x = 2

ii) max no of possible turns = 3, actual number of turns = 1

iii) opens up

iv) domain of g = ℜ range of g = (0,∞)

v)

c) i) x3 + x2 − 6x = x(x2 + x –6) = x(x + 3)(x – 2)zeros at –3, 0, 2

ii) max no of possible turns = 2

iii) rises from Quadrant III

iv) domain of h = ℜ range of h = ℜ

y

x

y

x


Module 1


v) y

x

Module 1


Module 1

Review

Answer Key

1. a) –1.463, –2.473, 1.935

b) –0.922, 0.922

c) 0, 0.809, 1.926

d) –1.539

2. a) 17

b) undefined

c) 62

d)

e) 3a2+27a + 59

f)

g)

h) x

i)

j)

3. a) i) y = 1

ii) domain of f: range of f:

b) i) y = 4

ii) domain of g: range of g:

c) i) y = 32

ii) domain of h: range of h: ( )25.96,− ∞ℜ

2, 2, 4x = −

( )4,− ∞ℜ

2 2, 2 2x = − − − +

( )0,∞1,2

−∞

12

x =

2 12

x −

( )23 15 17 2 5t t t+ + +

2 3 2 1x x+ + +

11

593

4. a)

b)

c)

5. a) domain of f:

range of f:

b) domain of g:

range of g: ℜ

ℜy

x•• • •321–2–3

ℜ

ℜy

x1

–2•

•

1 11

xx x+ = +

3 3 1x + −

1 2( )

5x

f x− +=

Principles of Mathematics 12 Section 1, Answer Key, Review 149

Module 1

150 Section 1, Answer Key, Review Principles of Mathematics 12

Module 1

Notes

Module 1

Section 2, Lesson 1

Answer Key

1. a)

b)

c)

1 2 3 4

1

2

3

−1−1

x

y

4

5

6(1, 6)l

1 2 3 4

1

−1−1

−2

−3

x

y

−4

−5

l

1 2 3 4

1

2

3

−1−1

−2

−3

x

y

5 6

l


d)

e)

f)

1 2

1

2

3

−1−1

x

y

4

5

6l

−2−3−4

(−4, 6)

1 2 3 4

1

2

3

−1−1

−2

−3

x

y

5 6

l−4 (6, −4)

1 2

1

2

3

−1−1

−2

−3

−3−4 −2x

y

−5

l


Module 1

2. a)

b)

c)

1 2

1

2

−1−1−2

x

y

−2

−3

−4

1 2

1

2

−1−1−2

x

y

−2

−3

−4

1 2

1

2

−1−1−2

x

y

−2


Module 1

d)

3.

4.

5.

6. The graph of n(x) is the same as the graph of g(x) except it ismoved three units to the left of g(x).

3 2

3 2

3 2

a) ( ) ( 2) ( 2) 3( 2) ( 2) 6

b) ( ) ( ) 3 ( 3 6) 3

c) ( ) ( 2) 1 ( 2) 3( 2) ( 2) 6 1

g x f x x x x

h x f x x x x

m x f x x x x

= + = + + + − + +

= − = + − + −

= − + = − + − − − + +

a)

b)

c)

g x

g x

g x

+

+

+ −

1

1

1 1

b gb gb g

Question

(a)

(b)

(c )

(d)

Function Domain Range y-intercept x-intercepts

f(x) ℜ [2, ∞) 2

f(x) – 6 ℜ [–4, ∞) –4

f(x + 1) ℜ [2, ∞) 3

f(x – 2) – 3 ℜ [–1, ∞) 3

x2 2 0+ =

∅no solution or

2

2

( 2) 6 0

4 02

x

x

x

+ − =− =

= ±

2

2

( 1) 2 0

( 1) 2

x

x

+ + =

+ = −

∅

2

2

2

( 2) 2 3 0

( 2) 1 0

( 2) 1

2 1

3 or 1

x

x

x

x

x

− + − =

− − =

− =

− = ±

=

1 2

1

2

−1−1−2

x

y

−2

−3

−4

l(2, −1)


Module 1

7.

8. a)

b)

1 2 3 4

1

2

3

−1−1

−2

−3

−2x

y

5 6 7 8l

l

(2, 0)

(8, 1)

1 2 3 4

1

2

3

−1−1

−2

−3

−2x

y

5 6 7 8

l

l

(2, 1)

(8, 2)

x

y

1−1 2−2−3 3

1

2

3

−1

−2

−3


Module 1

c)

1 2 3 4

1

2

3

−1−1

−2

−3

−2x

y

5 6 7 8

l

l

(0, 1)

(6, 2)


Module 1

Lesson 2

Answer Key

1. a)

b)

c)

x

y

1

1

−1

−1

2

x

y

1

1

−1

−1

−2

x

y

1

1

−1 2


Module 1

2. a)

b)

c)

x

y

1

1

−1 2 3 4

x

y

1

1

−1

−1

x

y

1

1

−1


Module 1

3. a)

Steps: 1. Reflect f(x) in the line y = x.2. Reflect the result in x-axis.

b)

Steps: 1. Reflect f(x) in the line y = x.2. Reflect the result in y-axis.

4. a) The same as they were, since a reflection in the x-axis leavesall points on the x-axis unchanged.

b) –3, 5, and 0, since a reflection in the y-axis will reflect anypoint on the right of the y-axis the same distance from the y-axis but to its left, and vice versa.

c) Since f(x) = 0 at the numbers 3, –5, and 0, it follows that f(–x +2) will be equal to zero whenever –x + 2 is equal to 3, –5, or 0.Therefore, the x-intercepts of f(–x + 2) are: –1, 7, and 2.

1 2 3 4

1

2

3

−1−1

−2

−3

−3−4 −2x

y

l

l

1 2 3 4

1

2

3

−1−1

−2

−3

−3−3 −2x

y

l

l


Module 1

To find the x-intercepts from a transformational point ofview you must rewrite f(–x + 2) as f(–(x – 2)) whichrepresents the reflection in the y-axis of the curvefollowed by a translation of two units to the right. Thesewere the steps followed.

Be careful how you interpret the last step usingtransformations. It is quite tricky!

d) Impossible to solve since the translation will shift everypoint one unit up. The x-intercepts are the points of thecurve that originally intersected the line y = –1. You donot have enough information to determine these points.

5. a) Minus 5, since the curve is reflected in the x-axis. Or, youknow that the y-intercept is 5, so f(0) = 5; therefore, –f(0) = –5.

b) Still 5, since a reflection in the y-axis does not move anypoints on the y-axis.

c) 2, since the graph has been lowered by 3 units.d) 6, since the graph is the same as the graph in part (b) but

one unit up. Therefore, the y-intercept of 5, in part (b),becomes 6.

Function Transformation x-intercepts

f(x) none 3, –5, 0

f(–x) reflection in y-axis –3, 5, 0

f(–x + 2) = f(–(x – 2)) a shift of two units to the right –1, 7, 2


Module 1

6.

The graphs are the same. Reflecting f(x) in the y-axis produces thesame graph with interchanged branches.

7. f(x) = f(–x) for: (a) and (d).

f(–x) = –f(x) for: (b) and (c).

1 2 3 4

1

2

3

−1−1

−2

−3

−3−3 −2x

y

f(−x)

1 2 3 4

1

2

3

−1−1

−2

−3

−3−4 −2x

y

f(x)


Module 1

Notes


Module 1

Lesson 3

Answer Key

1. a)

b)

c)

1 2 3 4

1

2

3

−1−1

−2

−3

−3−4 −2x

y

1 2

1

2

3

−1−1

−2

−3

−3−4 −2x

y

−5

1 2 3 4

1

2

3

−1−1

−2

−3

−3−4 −2x

y

12


Module 1

d)

e)

f)

1 2 3 4

1

2

3

−1−1

−2

−3

−3−4 −2x

y

1 2 3 4

1

2

3

−1−1

−2

−3

−3−4 −2x

y

1 2 3 4

1

2

3

−1−1

−2

−3

−3−4 −2x

y


Module 1

The domain and range are easily deduced from the sketches. Tofind the intercepts you may have to solve some equations.

e) Since is undefined there is no y-intercept. Similarly,

has no solution so there is no x-intercept.10

x=

10

x

x

x

3

3

1 0

1 01

− =

− ==

d) y = − =0 1 13

x

x

x x

x

2

2

1 0

1 0

1 1 0

1

− =

− =− + =

= ±b gb g

c) y = − =0 1 12

( ) ( )

3 2 03 23 2 or 3 23 2 or 3 2

1 or 5

xxx xx x

x

+ − =+ =+ = − + =+ = + = −

= − −

b) y = + − =0 3 2 1

2 1 0

2 112

x

x

x

− =

=

=

a) y = − =2 0 1 1b g

Question Domain Range y-intercept x-intercepts

(a) ℜ [0, ∞) 1

(b) ℜ (–2, ∞) 1 –1, –5

(c) ℜ [0, ∞) 1 –1, 1

(d) ℜ [0, ∞) 1 1

(e) {x|x ≠ 0} (0, ∞) ∅ ∅

(f) [0, ∞) [0, ∞) 1 1

12


Module 1

2.

3.

4. –|f(x)|

5. –f(x)

x

y

1−1 2

−1

1

−2

2

−2

1 2 3 4

1

2

3

−1−1

−2

−3

−3−4 −2x

y

x

x

x

x

− =

− =

==

1 0

1 0

11

f) y = − =0 1 1


Module 1

Lesson 4

Answer Key

1. a) Same shape as y = x3 except much steeper due to the verticalstretch by a factor of 5.

b) Same shape as y = x3 except much flatter due to the

vertical compression by a factor of

2. a)

Invariant points:

(–4,0) (0,0) (0,2)

b)

Invariant points:

(0,0)

x

y

1 2−2 −1

1

2

3

4

−1

−2

x

y

1 2−2−4

1

2

3

4

−2

5

6

7

8

−3

−4

−1−1−3

23

.


Module 1

c)

Invariant points:

(–4,0) (0,0) (2,0)

d)

Invariant points:

(0,0)

e)

Invariant points:

(–4,0) (0,0) (2,0)

x

y

(1, − )43

1 2−2−4

1

2

3

4

−3 −1−1

−2

(−2, )83

x

y

−2

−2

2

1

−1−1 21

3

4

x

y

1 2−2−4

1

2

3

4

−1−3−1

−2

−3

−4

−5

−6

−7

−8


Module 1

f)

Invariant points: (0,0)

g)

Invariant points:

(0,0)

x

y

1 2−2−4

1

2

3

4

−2

5

6

7

8

−3

−4

−1−1−3

x

y

1 2−8 −4

1

2

3 4

3

4

−7 −6 −5 −3 −2 −1−1

−2


Module 1

h)

Invariant points:

(0,0)

3. a) 3, 1, –4 (take of each of original x-intercepts)

b) 18, 6, –24 (multiply each x-intercept by 3)

c) –2,

d)

4.

5. a) 1

12 so 1

21

1 12

yx

x x yx

y y yx

=

→ − =

→ − = +

( )( ) ( )

a) 2 b) (5 )

c) 2 3 d) 3

f x f x

f x f x

−

− −

2 4 82 4

2

x

x

x

− = −= −= −

2 4 22 6

3

x

x

x

− ===

2 4 62 10

5

x

x

x

− ===

(take of each x-intercept, change

sign, since this is a reflection in y-axis)

2 8,

3 3−

12

x

y

1

2

3

4

−2

5

6

7

8

−3

−4

−1

23

23

43

− −


Module 1

13

b)

c)

6. a)

b)

x

y

1 2−2−4

1

2

3

4

−3 −1−1

−2

3

−3

−4

x

y

1 2−2−4

1

2

3

4

−3 −1−1

−2

3

−3

−4 l

4 5

( )( )

( )

1 11 so

14 14

4which can be written as

1

y x yx

yx

→ − + =− +

−=+

( )

1

1 1 1so

3 3 23

22

yx

y y yx

x x yx

=

→ − − =−

−→ − =−


Module 1

c)

The domain and range are easily deduced from the sketches.To find the intercepts you may have to solve some equations.

x-intercept = 2 or –2

Therefore, y-intercept is 4.( ) ( )b) 0 2 0 4 40 2 4

2 42 4 or 2 4

g

x

xx x

=− + ==− +== =−

However, is undefined; hencethere is no y-intercept.

Check:

Therefore, x-intercept = 5.

( )a) 0 2 1 4

0 2 1 4

4 2 1

2 14 15

f

x

x

x

x

x

= − −

= − −

= −

= −= −=


(a) [1, ∞) [–4, ∞) ∅ 5

(b) ℜ (–∞, 4] 4 ±2

(c) ℜ (–1, 1] 1 –1, 1

x

y

1

1

−1

−1


Module 1

( )0f

x-intercept = ±1

Therefore, y-intercept is 1.( ) 2

2

2

2

2

2c) 0 1 1

0 12

0 11

21

1

1 2

1

h

x

x

x

x

= − =+

= −+

=+

+ =

=


Module 1

Lesson 5

Answer Key

1. a)

b)

x

y

112

−1

−2

x

y

1 212−


Module 1

c)

d)

e)

x

y

14−

x

y

1 2

14−

−1−2

x

y

1

−1

−2

−3

−4


Module 1

3. a)

b)

x

y

1

1

−1

−1

−2−3−4−5 2 3 4 5

Summary

f(x)y = 1, –1 → fixed0 < y < 1, –1 < y < 0 → largey > 1 or y < –1 → small

x

y

1

1

−1−1

Question Domain Range y-intercept Zeros

(a) {x|x ≠ 2} {y|y ≠ 0} ∅

(b) {x|x ≠ 0} {y|y ≠ –2} ∅

(c) {x|x ≠ –3} {y|y ≠ –4}

(d) {x|x ≠ ±2} ∅

(e) ℜ ∅

− 14

− 14

− 12

12

−113

−114

( )1, 0 0,

4 − ∞

∪

1, 0

4 −


Module 1

c)

d)

e)

x

y

1

−1

12−

x

y

1

x

y

1


Module 1

3. a)

⇓

b)

⇓

x

y

1

1

x

y

1

l l

l l

ll

x

y

1 2 3 4 5

x

y

1 2 3 4 5

l

l

l


Module 1

c)

⇓

All the information above is available from the graphs.

4. 7111

,FHGIKJ

Question Domain Range Zeros

(a) (4, ∞) (0, ∞) ∅

(b) {x|x ≠ 1} (0, ∞) ∅

(c) {x|x ≠ 2} (0, ∞) ∅

x

y

1 3

1

2

3

4

x

y

1

l l

l l

ll

3

1

2

3

4


Module 1

Lesson 6

Answer Key

1. a) Stretch f(x) vertically by a factor of 4.

b) Reflect f(x) in the x-axis.

c) Compress f(x) vertically by a factor of 4.

d) Stretch f(x) vertically by a factor of 3 and reflect theresulting graph in the x-axis.

e) Compress f(x) vertically by a factor of 5 and reflect theresulting graph in the x-axis.

f) Reflect f(x) in the y-axis.

g) Stretch f(x) vertically by a factor of 2 and translate theresulting graph one unit upward.

h) Translate f(x) one unit to the left and stretch theresulting graph vertically by a factor of 4.

i) Translate f(x) two units to the right, stretch the resultinggraph vertically by a factor of 3, followed by a reflectionin the x-axis.

j) Compress f(x) horizontally by a factor of 2 and translatethe resulting graph five units downward.

k) Stretch f(x) horizontally by a factor of 3, reflect theresulting graph in the y-axis, followed by a translation offour units upward.

l) Reflect f(x) in the y-axis and reflect any part of theresulting graph, which is below the x-axis, in the x-axis.

2. a)

⇓

x

y

1

−1

−1

x −12


Module 1

⇓

b)

x

y

1−1

1

2 3

−9

−1 l

y = (x − 2) − 13

x

y

1−1

1

2 3

−8

(x − 2)3

y =

x

y

1−1

1

x −121y =

x

y

1−1

−1

x −12

1y =


Module 1

c)

⇓

⇓

x

y

1 3

1

2

3

4

2 4

y = 2 4 − x

x

y

1 3

1

2

3

4

2 4l

ly = 4 − x

x

y

1 3

1

2

3

4

2 4

y = 4 − x


Module 1

d)

⇓

⇓

x

y

1−1

1

y = |x| − 1

x

y

y =|x| − 1

1−1−1

x

y

y =|x|


Module 1

e)

⇓

⇓

x

y

1

1

−1

l

23

x − 13| |y = − 1

x

y

1

x − 13| |y =

1

x

y

1

−1

x − 13y =


Module 1

Most of the information is available from the sketch. The solutionfor the intercepts is given below. (Note the symbols x and y areused for the intercepts.)

y-intercepts

x-intercepts

d) 0 1

0 1

1

1

= −

= −

=

± =

x

x

x

x

( )

c) 0 2 4

0 4

0 4

4

Check: 2 4 4 2 0 0.

Therefore, 4.

x

x

x

x

x

= −

= −= −=

− = =

=

3

3

b) 0 ( 2) 1

1 ( 2)1 23

x

x

x

x

= − −= −= −=

a)

no solution

01

10 1

2=−

=x

a) b)

c) d)

e) f) is undefined

y y

y y

y

=−

= − = = − − = − − = −

= − = = = − = − =

= − − = − =−

10 1

1 1 0 2 1 8 1 9

2 4 0 2 2 4 0 1 1 1

0 1 1 1 1 01

0 0

2

3

3

b g

b g

Asymptotes

x = ±1, y = 0

Zeros

∅

y-intercept

1

Range

(0, ∞)

Domain

{x|x ≠ ±1}

Question

(a)

none3–9ℜℜ(b)

none44[0, ∞)(–∞, 4](c)

none±11[0, ∞)ℜ(d)

none0[–1, ∞)ℜ(e)

x = 0, y = 0∅∅(0, ∞)(–∞, 0)(f)

0 23,


Module 1

3.

4. a)

b)

c)

f x

f x

f x

−

−−

b gb gb g1

( )( )

( )

1 1, 0 since the original function

1

1 is undefined at 0.

f f x f x xx

x

f xx

= = = ≠

=

e)

or

or

or

0 1 1

1 1

1 1 1 1

2 0

2 0

3

3

3 3

3 3

3

= − −

= −

= − − = −

= =

= =

x

x

x x

x x

x x


Module 1

Enrichment Questions

1. a)

⇓

b)

⇓

x

y

1 3

1

2

3

4

2 4

3 − xy =

x

y

y = x

x

y

|x|y =


Module 1

⇓

c)

⇓

x

y

1 3

1

2

2 4−1

−2

−3

−4−(x − 2)2y =

x

y

1 3

1

2

3

4

2 4

33 − xy = −

x

y

1 3

1

2

3

4

2 4

3

3 − xy =


Module 1

⇓

2.

The sequence of sketches would be:

⇓

x

y

xy =

( )If is the basic function, then 2 2

2 .

y x y x x

x

= = − = − += − −

x

y

1

2

4−1

−2

1 32

y = 1 − (x − 2)2

x

y

1 3

1

2

2 4−1

−2

−3

y = 1 − (x − 2)2


Module 1

⇓

If the square root operation is used on the graph of y = 2 – xthe sequence of graphs is:

⇓

x

y

1 3

1

2

3

4

2 4

y = 2 − x

x

y

1 2

1

2 2

y = −(x − 2)

x

y

1 2

x − 2y =


Module 1

3.

4. You would draw the graph if more information wererequested. However, since only the domain is required, allyou have to do is solve the following inequalities:

The sign diagram is:

Therefore, the solution is: [0, 1).(–∞, 1)µ(1,+∞)

The sign diagram is:

Therefore, the solution is [–1, 3]

−1 3+ −−

( )( ) ( )

( )( )

2b) 4 1 0

2 1 2 1 0

3 1 0

x

x x

x x

− − ≥

− − + − ≥ − + ≥

0 1 ++ −

a)x

x −>

10

x

y

1 3

1

2

2 4−4 −2−3 −1

x

y

1 2

1

2 2

y = 2 − x


Module 1

Review

Answer Key

1. a) b)

c) d)

e) f)

1

2

3

4

y

x−3−4

l l(−5, 3)

−5 −2 −1

(−1, 3)

y

−1

−2

−3

x4 5

l l (5, −3)(1, −3)

321

y

1

2

x3 4 521

x

y

1 2−1−2

−1

−2

−3l l (2, −3)(−2, −3)

x

y

1 2−1−2

1

2

3

4

l l (2, 3)(−2, 3)

y

1

2

−1

−2

x−1−2−3 1−4−5


Module 1

g) h)

i)

j)

k) l)

2 4 22 6

3

x

x

x

− ===

2 4 22 2

1

x

x

x

− = −==

x

y

1

2

−1

−2

l l

32

1 2 352

x

y

1

2

−1

−2

ll

13

23

13

23− −

x

y

1 2−1−2

1

2

−1

−2

ll

3 4 5 6−3−4−5−6

x

y

1 2−1−2

1

2

−1

−2

ll

x

y

1 2−1−2

1

2

−1

−2

ll(2, −1)(−2, −1)x

y

1 2−1−2

1

2 ll (2, 2)(−2, 2)


Module 1

2. a) b)

(Reflect through y = x or take points from original graph, andinterchange x- and y-coordinates.)

3. a)

b) c)

x

y

1−1

1

2

−1

−2

−3

3− 3x

y

1 2−1−2

1

2

−1

−2

x

y

1 2−1−2

1

−1

−2

l

l

3 4 5 6 7 8

−3

−4

−5

−6

l

x

y

1 2−1−2

1

2

−1

−2

l

l

3x

y

1 2−1−2

1

2

−1

−2

l

l


Module 1

The sketches can be used to fill in the table.

4. a)

b)

x

y

1 2−1−2

1

2

−1

−2

x

y

1−1

1

2

−1

−2

−2 12

12− l

l


3 (a) [–1,∞) [–6,∞) –4 8

(b) ℜ ( –∞,1][–6,∞) 1

(c) ( –∞,–3];(–1,∞) –3 3±(–∞, –1);(–1, 1);(1,∞)

1 1,

2 2−


Module 1

c)


5. a) b)

x

y

−3

9

l(−2, 1)

x

y

1 2−1−2

1

2

−1

−2

Question Domain Range Zeros

(a) {x|x ≠ –2} {y|y ≠ 2}

(b) ℜ (0, 1] ∅

(c) {x|x ≠ 2} (0, ∞) ∅

12

x

y

1 2−1−2

1

2

−1

−2


Module 1

c) d)

e)


Question y-intercept x-intercepts VerticalAsymptote

HorizontalAsymptote

(a) 1 ∅ x = 1 y = 0

(b) 9 –3 ∅ ∅

(c) 3 ∅ ∅

(d) 1 ±1 ∅ ∅

(e) 2 ∅ ∅ ∅

−2 3

Domain Range

{x|x ≠ 1} 0 < y < ∞

ℜ ℜ

x ≤ 3 y ≤ 0

ℜ y ≥ 0

ℜ y ≥ 0

x

y

1 2−1−2

1

2

−1

−2

l

x

y

1 2−1−2

1

2

−1

−2

x

y

1 2 3l

3−2


Module 1

6. ( )( )( ) ( )( ) ( )

( )( )( )( )

( )( )( )

1

1

1)

1)

1

a) 3 2

b) 4 6

c) 3 1; 1 2

d) 2 3; 3 2 or, sincein all cases,

2 2

e) 5 0

f) 4 4, reasoning similar to (d)

−

−

−

−

−

=−

=−

= =

− = =−=

− =−

=

=

f

g

f g

f ff f a af f

f

g g


Module 1

Documents

Module 1: Answer Key - Open School BCmedia.openschool.bc.ca/ocrmedia/pmath12/pdf/pma12_mod1...Module 1 Section 1, Lesson 1 Answer Key Note: The graphing calculator displays in these