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Answer Key
Algebra II Regents Course Workbook
with Exam Questions
2019-20 Edition Donny Brusca
© 2019 Donald Brusca. All rights reserved.
www.CourseWorkbooks.com
3
Ch. Unit Page 2. RATIONAL EXPRESSIONS AND EQUATIONS ...................................................... 4
3. RADICALS ........................................................................................................ 19
4. COMPLEX NUMBERS ...................................................................................... 27
5. QUADRATIC FUNCTIONS AND CIRCLES .......................................................... 29
6. EXPONENTIAL FUNCTIONS ............................................................................. 36
7. LOGARITHMS .................................................................................................. 43
8. POLYNOMIAL FUNCTIONS .............................................................................. 51
9. TRIGONOMETRIC FUNCTIONS ........................................................................ 65
10. COMPARE FUNCTIONS ................................................................................... 77
11. SYSTEMS ......................................................................................................... 82
12. SEQUENCES AND SERIES ................................................................................. 88
13. PROBABILITY ................................................................................................... 93
14. STATISTICS ...................................................................................................... 97
Notation A code next to each Regents Question number states from which Algebra II Common Core Regents exam the question came. For example, AUG ’18 [25] means the question appeared on the August 2018 as question 25.
4
2. RATIONAL EXPRESSIONS AND EQUATIONS
2.1 Undefined Expressions
Practice Problems
1. 𝑥 + 2 = 0, so 𝑥 = −2 2. 3𝑥 + 1 = 0, so 𝑥 = −1
3
3. 𝑥2 − 4 = 0 𝑥2 = 4
𝑥 = ±√4 = ±2
4. 𝑥2 − 4𝑥 − 12 = 0
(𝑥 + 2)(𝑥 − 6) = 0
𝑥 = {−2,6}
5. 9 − 𝑥2 = 0 −𝑥2 = −9 𝑥2 = 9 𝑥 = ±3
6. 𝑥2 + 5𝑥 − 6 = 0
(𝑥 + 6)(𝑥 − 1) = 0
𝑥 = {−6,1}
7. 𝑥 + 2 = 0 or 𝑥 − 1 = 0 𝑥 = {−2,1}
8. 2𝑥2 + 1 = 0 2𝑥2 = −1
𝑥2 = −1
2
𝑥 = ±√−1
2 (not real numbers)
The expression is defined for all possible real values of x.
Regents Questions
1) AUG ’17 [1] Ans: 1
5
2.2 Negative Exponents
Practice Problems
1. 7
1
p 2.
( )2 2
1 1
255 xx=
3. 1
2 9
11 1
3+ =
4. 2−2 − 20 + 22 =1
4− 1 + 4 = 3
1
4
5. 3
5
2m
n−
6. 5𝑥3
7. 3
5
bc− 8.
3 3
3
5 125
2 8
y y
x x
=
9. (3
4)
2∙ 42 = 9 10.
6
5
a
b
11. ( )
( )32
3 2
23 8
932
−
−
−= = −
−
12. 4
2 3 4 3
1 25
5
b
a b a− −=
13. 6
3
y
x 14.
4 5
3
x y
15. 7
22
y
x
16.
( )
( )
( )
25 3 74 5
2 43 7
5 6 14 6 9
4 4
2
9
3 23
2
3 4 12
12
y x yx y
xx y
y x y x y
x x
x
y
−−
−−
−−
= =
= =
6
2.3 Simplify Rational Expressions
Practice Problems
1. 𝑥 + 2 2. 2𝑥2 + 3𝑥 + 1
3. 3𝑥 − 9𝑥3 4. 𝑥4 − 9𝑥2 + 1
5. 3𝑎2𝑏2 − 6𝑎
6. ( )2 6x x −
6x −2x=
7. ( )25 5x −
( ) ( )5 5x x+ −
25
5x=
+ 8.
( )3x + ( )
( )
5
3
x
x x
−
+
5x
x
−=
9. ( )5x + ( )
( )
1
5
x
x
+
+ ( )
1
55
x
xx
+=
−− 10.
( )3 3 5x x y−
( ) ( )3 5 3 5x y x y+ −
3
3 5
x
x y=
+
11. ( )3x + ( )
( )
5
3
x
x x
−
+
5x
x
−= 12.
( ) ( )2 3x x+ −
( ) ( )2 3x x− −
2
2
x
x
+=
−
13.
( )
( )
( )22 5 14 2 7
4 7
x x x
x
+ − +=
+
( )
( )
2
4 7
x
x
−
+
2
2
x −=
14.
( )
( )33
2 3
xx
x
− −−=
− ( )2 3x −
1
2= −
15.
( )( )
( )x yy x
x y x y
− −−=
+ − ( ) ( )x y x y+ −
1
x y
−=
+
16.
( )( )
9
5 9
xx x
x x
−=
−
( )9x −
5 x− ( )9x −
1
5= −
17.
( )
( )
( )2
3 43 4
4
y yy y
y y
−−=
− ( )2 4y y− −
3
y= −
18.
( )( ) ( ) ( )3 33 3
3
xy xyxy xy
xy
+ −+ −=
− ( )3xy− −
( )3 3xy xy
=
− + = − −
19.
( )2 58 15
5
xx x
x
++ +=
+
( )3
5
x
x
+
+3x= +
20. Base = 14𝑥 + 21 − 2(4𝑥 + 6) = 6𝑥 + 9
( )3 2 36 9
14 21
xx
x
++=
+ ( )7 2 3x +
3
7=
7
Regents Questions
1) JUN ’17 [23] Ans: 4 2) JAN ’18 [18] Ans: 3
3) JUN ’18 [3] Ans: 3 4) JAN ’19 [21] Ans: 4
8
2.4 Multiply and Divide Rational Expressions
Practice Problems
1. 27 9 3 3
3 14 1 2 2
x x x
x = = 2.
2 3
2 4 2 2
4 21 1 3 3
17 20 5 5
x y y y
y x x x = =
3.
( )
2 21 4
1
1
x x
x x
x
− =
+
+ ( ) 21 4
1
x x
x x
−
+
( ) 24 11 44 4
1 1 1
x xx xx x
=
−− = = −
4.
24 1
1 3 3
4
1
x x
x x
x
x
− =
− +
−
( )1x +
( )1x −
( )3 1x +
4
3
x=
5.
( )
2 1 3
1 3 3
1
x x
x x
x
− + =
+ −
+ ( )1x −
1x + ( )
3
3 1
x
x
+
−
3
3
x +=
6.
2
2 4 20
2 6 82
x x
x xx
+ + =
+ +
+ ( )
( ) ( )
4 5
2 4 2
x
x x
+
+ +
( )2 5 2 10
4 4
x x
x x
=
+ +=
+ +
7.
22 5 4
3 3 2 4
2
x x x
x x
x
+ + + =
+ +
+
( )3 1x +
( ) ( )4 1x x+ +
( )2 2x +
4
6
x +=
8.
( )
2
2 2
9
9 18 3
3
x x
x x x x
x
− =
+ + −
+ ( )3x −
( ) ( )6 3x x+ + ( )3
x
x x
−
1
6x=
+
9.
2
2
3 9
3 3 39
3
x x x x
x x xx
x
x
− = =
+ +−
+
( )3x +
( )3 3
3 3
x x
x
− −=
10.
2
2
2 16
4 4 216
4
x x x x
x x xx
x
x
− = =
+ +−
+
( )4x +
( )4 4
2 2
x x
x
− −=
9
11.
( )
2
2 2
2 2
2
2
9 12
12 36 6
9 6
1212 36
9
6
x x
x x x x
x x x
xx x
x
x
=+ + +
+ =
+ +
+ ( )
( )6
6
x x
x
+
+
( )
2 2
12
3 3
4 6 4 24
x
x x
x x
=
=+ +
12.
( )
2
2
3 6 4 3 6 3
4 12 3 4 12 4
3 2
x x x x
x x x x
x
+ − + + = =
+ + + −
+
( )4 3x +
3x +( )2x + ( )
( )
2
3 3
4 2 4 8
x
x x
=−
=− −
13.
( )
2 2
2 2
2 8 42 3
6 9
2 3
x x x x
x x
x
− − − =
−
+ ( ) ( )2
7 3
6
x x x
x
− −( )3x + ( )3x −
7
3
x
x
=
−
14.
( )
2 2
2 2
3 9 6
5 6 9
3 3
x x x x
x x x
x x
+ − − =
+ + −
+
( )3x + ( )2x +
( )2x +
( )3x −
( ) ( )3 3x x+ −
3
3
x
x
=
+
15.
( )
2 2
2
9 14 56
3 649
7
x x x x
xx
x
+ + + − =
+−
+ ( )2x +
( )7x + ( )7x −
( ) ( )8 7x x+ −
( )3 2x +
8
3
x
=
+
16.
( )
2
2
2 6 2
2 4 2 15
2 3
x x x
x x x
x
− + =
+ + −
−
( )2 2x +
( )2x x +( ) ( )5 3x x+ − 5
x
x=
+
17.
( )
2 2
2 2
2 15 5 36
4 45 12
5
x x x x
x x x x
x
+ − − − =
− − + −
+ ( )3x −
( )5x + ( )9x −
( )4x +
( )9x −
( )4x + ( )3x −1=
18.
( )
2 2 3 2
2 2 2
3
4 3 2 4 4 4
2 10 3 3 2x
x x x x x x
x x x x x x+
+ + + + + =
− − + + +
( )1x +
( ) ( )2 5 2x x− +
( )
( )
22 1 2
3
x x
x x
+
+
( )2x +
( )2x +
( )2x + ( )1x +
( ) 22 2 1 4 2
2 5 2 5
x x x x
x x
=
+ +=
− −
10
2.5 Add and Subtract Rational Expressions
Practice Problems
1. 2
8
1x +
2. 24
2
x
x −
3. 12
2 4
x
x
+
+ 4.
4
5x
5.
( )2 52 10
5
yy
y
++=
+ 5y +2=
6.
( ) ( )
2 26 5 6 5
1 1 1
5 1
x x x x
x x x
x x
+ + ++ = =
+ + +
+ +
1x +5x= +
7. LCM is 12
5 2 3 10 3 13
6 2 4 3 12 12 12
x x x x x + = + =
8. LCM is 5𝑥
3 5 2 15 2 15 2
5 5 5 5 5
x x x
x x x x x
− − = − =
9. LCM is 21𝑛
3 3 7 7 9 49 40
7 3 3 7 21 21 21n n n n n
− = − = −
10. LCM is 2𝑥
2 2 2
2 2 2 2 2
a b a b a b
x x x x x
+ + = + =
11. LCM is 3𝑏
3 1 3 3
3 3 3 3 3
a b a b a b
b b b b b
− − = − =
12. LCM is 12𝑥2
2 2 2
2
7 2 7 2
12 26 12 127 2
12
x y x y
x x x x xx y
x
− = − =
−
13.
2
6 5
5 25
56
5
y
y y
y
y
+− =
− −
+−
− ( )5y + ( )5
6 1 5
5 5 5
y
y y y
=−
− =− − −
14.
6 ( 1)+x x
( 1)+x
( 1)( 1)
3 ( 2)( 2)
6 ( 1)( 1)
( 2) 3 ( 2)
+ ++ =
−−
+ ++
− −
x x
x xx
x x x
x x x
.
LCM is 3𝑥(𝑥 − 2)
2 2
2 2
2
6 3 ( 1)( 1)
( 2) 3 3 ( 2)
18 2 1
3 ( 2) 3 ( 2)
19 2 1 19 2 1
3 ( 2) 3 6
+ + + =
− −
+ ++ =
− −
+ + + +=
− −
x x x x
x x x x
x x x
x x x x
x x x x
x x x x
11
15.
2
2
5 3 5 2 3
1 2 1 2 2
( 5)( 2) 3 3 10 3
2 2
3 7
2
− − ++ = +
+ + +
− + + − − += =
+ +
− −=
+
y y y
y y y
y y y y
y y
y y
y
16.
3 3 3 3
1 1 1 10
01
a a a a
a
+ = − =− − − −
=−
17.
( )
3 9 3 9
2 6 6 2 2 6 2 6
3 33 9
2 6
x x
x x x x
xx
x
+ = − =− − − −
−−=
− ( )2 3x −
3
2=
18.
( )
( ) ( )
( ) ( ) ( )
1 1
1 2 2 1 2 1
1 2 1
1 2 1 1 2 2 1
2 1 2 1 2 1
2 1 2 1 2 1 2 2
x x
x x x x
x x
x x x x
x x x
x x x x
− = − =− − − −
+ = + =
− − − −
+ ++ = =
− − − −
12
2.6 Solve Rational Equations
Practice Problems
1.
1 116 16 16
16 4 24 8
4
x
x
x
+ =
+ =
=
2.
( )6 6 6 22 6
3 12
4 12
3
x x
x x
x
x
+ =
+ =
=
=
3.
( )3 2
5 5 5 45 5
3 2 20
3 18
6
x
x
x
x
+ =
+ =
=
=
4.
( ) ( )3 5
4 4 2 4 4 64 4
3 8 5 24
32 2
16
x x
x x
x
x
+ = −
+ = −
=
=
5.
( )3 1
12 12 12 54 3
9 4 60
5 60
12
x x
x x
x
x
= +
= +
=
=
6.
5 1 36 6 6
2 630 3 3
3 27
9
n n nn n
n
n
n
− =
− =
− = −
=
7.
2 1 7 215 15 15
5 3 156 5 7 2
7
x x
x x
x
− + =
+ = −
=
8.
( )2 26
3
2 3 26
3 24
8
x x xx x
x
x
x
− =
− =
− =
= −
9.
( )2
6 6 6 53 6
4 30
5 30
6
x x
x x
x
x
+ =
+ =
=
=
10.
1 2 15 321 21 21
7 3 213 14 15 3
6
x x
x x
x
− + =
+ = −
=
13
11.
( )
( )
16 6 6
3 2
2 3 1 6
2 3 3 6
5 3 6
3
x xx
x x x
x x x
x x
x
+ + =
+ + =
+ + =
+ =
=
12.
( )( )
2
2
8 1 112 12 12
3 12 6
32 2
30 0
5 6 0
{ 5,6}
xx x x
x x
x x
x x
x x
− − =
− + =
− − =
+ − =
−
13.
( ) ( )
( )
34 3 4 9
43 3 36
3 9 36
3 279
x
x
x
xx
+ =
+ =
+ =
=
=
14.
( ) ( )
( ) ( )
35 2 5 4
53 2 5 4
3 6 5 20
26 213
x x
x x
x x
xx
+ = −
+ = −
+ = −
=
=
15.
( )( )
( ) ( )
3 110 10 10 2 3
5 2
2 15 1 20 3
2 15 15 20 6017 15 20 60
45 3
15
mmm
m m m
m m mm m
m
m
− + = −
+ − = −
+ − = −
− = −
=
=
16.
( )
( )( )
2 2 22
2
2
6 11
6
6 0
2 3 0
{ 2, 3}
x x xxx
x x
x x
x x
− =
− =
− − =
+ − =
−
17.
( )
( ) ( )
( )( )
2
2 2
2
43
4 122 1
2
4 12 2 2
4 12 24 2
3 14 24 0
3 4 6 0
,6
xx x
x x
x x x x
x x x x
x x
x x
+ − = +
− + = +
− − = +
− − =
+ − =
−
18.
( )
( ) ( )
2 4 73 1
3 1
2 1 12 1 21
2 2 12 12 2114 14 21
14 7
2
x xx x x
x x x
x x xx x
x
x
+ + = +
+ + + =
+ + + =
+ =
=
=
14
19.
( )( )
( ) ( ) ( )( )
( )
( )( )
2
2
2
54
3 2 43 3 4
3 4 3
9 4 6 3 4 3 4
9 36 6 18 4 12
15 18 4 4 48
4 19 30 0
4 5 6 0
,6
x xx x
x x x x
x x x x
x x x
x x
x x
+ − + = + −
− + + = + −
− + + = − −
− = − −
− − =
+ − =
−
20.
( )( )( )( )
( ) ( )
( )( )
2
2
9 505 5
5 5 5 5
5 9 5 50
5 9 45 50
4 5 0
5 1 0
5
xx x
x x x x
x x x
x x x
x x
x x
+ − + =
+ − + −
− + + =
− + + =
+ − =
+ − =
− ,1
–5 is an extraneous root, since 𝑥 + 5 = (−5) + 5 = 0
21.
( )( )( )( )
( ) ( )
( )( )
2
2
1 283 4
4 3 3 4
3 4 28
3 4 28
2 24 0
6 4 0
6, 4
xx x
x x x x
x x x
x x x
x x
x x
+ − − =
− + + −
+ − − =
+ − + =
+ − =
+ − =
−
4 is an extraneous root.
22.
( )
( ) ( )
( )( )
( )
2
2
2
4 31 7
1
4 1 3 7 1
4 4 3 7 7
7 6 4 0
6 6 4 7 4
2 7
6 148 6 2 37
14 14
3 37
7
x xx x
x x x x
x x x x
x x
x
x
x
+ − = +
+ − = +
+ − = +
+ − =
− − −=
− − = =
− =
15
Regents Questions
1) FALL ’15 [1] Ans: 4 2) AUG ’16 [17] Ans: 3 3) JAN ’17 [17] Ans: 1 4) JUN ’17 [19] Ans: 1 5) JUN ’18 [9] Ans: 4 6) JAN ’19 [15] Ans: 3 7) JUN ’16 [25]
1 1 13
3 33 1
4
xx x
x
x
− = −
− = −
=
8) AUG ’17 [33]
( )( )
( ) ( ) ( )
( )( )
2 2
2
2
3 23 5
5 3 3
3 3 2 5 5
3 9 2 10 5
2 12 10 0
6 5 0
5 1 0
5, 1
p pp p
p p p
p p p p p
p p p p p
p p
p p
p p
+ − − =
− + +
+ − − = −
+ − + = −
+ + =
+ + =
+ + =
− −
9) AUG ’18 [29]
( )
( ) ( ) ( )
( )
2
2
3 1 16 3
3 2 6 2
18 3 3 3 3 3
18 3 9 3 3 9
3 0
3 0
0,3
xx
x
x x x x
x x x x
x x
x x
− + + = − +
− + + = + − +
− + + = + − −
− =
− =
16
2.7 Model Rational Expressions and Equations
Practice Problems
1.
3 7 3
7 44( 4) 3( 7)
4 16 3 21
5
x
xx x
x x
x
− +=
++ = +
+ = +
=
Original fraction is 25
.
2.
2
2
6416
6416
64 16
16 64 0
( 8)( 8) 0
8
xx
x xx
x x
x x
x x
x
+ =
+ =
+ =
− + =
− − =
=
3.
12
5 73
2
5 72 3
2
10 7 6
3 6
x x
xx x
x
x
x
= +
= +
= +
=
=
4.
( )
( ) ( )
( )( )
2
2
6 71
2
6 72 1
2
6 2 7 2
13 12 2
11 12 0
1 12 0
1
x x
x xx x
x x x x
x x x
x x
x x
+ =+
+ + = +
+ + = +
+ = +
− − =
+ − =
− ,12
Reject –1. Numbers are 12 and 14.
5.
125
1 1 1
6 41 1 1
126 4
2 3 12
5 12
2.4hrs
x
xx
x x
x
x
+ =
+ =
+ =
=
= =
6.
152
1 1 1
2 51 1 1
102 5
10 5 2
7.5hrs
c c
cc c
c
c
+ =
+ =
+ =
= =
17
7.
4 16
5 54(5 ) 16(5 )
20 4 80 16
20 60
3mph
c cc c
c c
c
c
=− ++ = −
+ = −
=
=
8.
1656 1 3168
12 2 12
3312 3168
12 123312( 12) 3168( 12)
3312 39744 3168 38016
144 77760
540 mph
=
+ −
=+ −− = +
− = +
=
=
x x
x xx x
x x
x
x
9.
36 42
22
36 4 44 2
2 8
4
k
k
k k
k
k
+=
+
+ = +
=
=
10.
Written as a fraction, 914 4
2.25 2= = .
Therefore, 94
1 1 4
9TR= = .
( )
( ) ( )
( ) ( )( )
( )
2
2
2
1 1 49 3
3 9
9 3 9 4 3
9 27 9 4 12
4 6 27 0
6 6 4 4 27
2 4
6 468
8
x xx x
x x x x
x x x x
x x
x
x
+ + = +
+ + = +
+ + = +
− − =
− − −=
=
Positive solution 𝑥 ≈ 3.5.
Regents Questions
1) JUN ’16 [2] Ans: 3 2) JUN ’17 [22] Ans: 3 3) JUN ’18 [24] Ans: 3
4) JAN ’18 [27]
247
1 1 1
8 6
1 1 124 24
8 6
3 4 24
7 24
3.4
b
b bb
b b
b
b
t
t tt
t t
t
t
+ =
+ =
+ =
=
=
18
2.8 Graphs of Rational Functions
Practice Problems
1. 𝑥 = −2 and 𝑥 = −9 2. 𝑦 = 1 Leading coefficients of both the
numerator and denominator are 1.
3. 𝑦 − 0 The degree of the numerator (1) is
smaller than the degree of the denominator (2).
4. There is no horizontal asymptote because the degree of the numerator (2) is greater than the degree of the denominator (1).
5. Vertical asymptote at 𝑥 = −1
2.
Horizontal asymptote at 𝑦 = 2.
6.
( )3 12 2 3 5
31 1 1 1
x x
x x x x
+ ++ = + =
+ + + +
Vertical asymptote at 𝑥 = −1. Horizontal asymptote at 𝑦 = 3.
7. (2) Vertical asymptotes at –3 and 3, which are the roots of 𝑥2 − 9 = 0, and a horizontal asymptote at 2.
19
3. RADICALS
3.1 Operations with Radicals
Practice Problems
1. 5√3 + √3 = 6√3 2. 3√3 + 2√3 = 5√3
3. 4√3 − 2√3 = 2√3 4. 5√6 + 2√6 = 7√6
5. 3√2 + 2√2 = 5√2 6. 5√2 − 4√2 = √2
7. 6√2 − 3√2 = 3√2 8. 10√2 − √2 = 9√2
9. 5√7 + 6√7 = 11√7 10. 30√2 + 6√2 = 36√2
11. 5 − 2√3 + 3√3 + 6 = 11 + √3 12. 𝑦√3 − 4√2 − 3𝑦√3 = −2𝑦√3 − 4√2
13. √90 = 3√10 14. √3600 − √144 = 60 − 12 = 48
15. 6√100 − 21√20 = 60 − 42√5 16. 3√98 + 12√392 = 21√2 + 168√2 =
189√2
17. 9 − 5 = 4 18. 5√2 − 20 + 2 − 4√2 = −18 + √2
19. 10√18𝑥7 = 10√9𝑥6√2𝑥 = 30𝑥3√2𝑥 20. 15√2𝑥6𝑦2 = 15𝑥3𝑦√2
21. 9𝑥 − 18√𝑥 + 9 22.
𝑐2 = (𝑥 + √2)2
+ (𝑥 − √2)2
𝑐2 = 𝑥2 + 2𝑥√2 + 2 + 𝑥2 − 2𝑥√2 + 2 𝑐2 = 2𝑥2 + 4
𝑐 = √2𝑥2 + 4
23. √13 24.
28 2 7
72 2
= =
25. 5√50 = 25√2 26.
15 3 3 3 18 36 3
3 3
+= =
27.
4 3 15 3 10 3 3
13 3
− + −= = −
28.
3 3 5 3 8 3
42 3 2 3
+= =
29.
16 21
5 12 8 3 10 3 2 32 7
− = − = −
30.
√18𝑥4𝑦3 = 3𝑥2𝑦√2𝑦
20
3.2 Rationalize Denominators
Practice Problems
1.
3 7 3 7
77 7
=
2.
2 2 2 2
222 2
= =
3.
3 5 10 3 50 15 2 3 2
2 10 20 42 10 10
= = =
4.
3 8 3 3 3 24
33 3
3 3 2 6
3
− −=
−=
5.
1 1
5 1 1 25 5
5 55 5
+ +
= =
6.
5 3 2 5(3 2)
9 23 2 3 2
15 5 2
7
+ += −− +
+=
7.
5 4 11 20 5 11
16 114 11 4 11
20 5 114 11
5
+ += −− +
+= = +
8.
4 5 13 20 4 13
25 135 13 5 13
20 4 13 5 13
12 3
+ += −− +
+ += =
9.
1 3 7 3 7
9 73 7 3 7
3 7
2
+ += −− +
+=
10.
5 7 5 7 5 5
49 57 5 7 5
7 5 5
44
+ += −− +
+=
11.
2 14 2 28 2 2
14 414 2 14 2
2 7 2 2 7 2
10 5
− −= −+ −
− −= =
12.
3 3 5 3 5 3
3 253 5 3 5
3 5 3 3 5 3 5 3 3
22 22 22
− −= −+ −
− − −= = − =
−
21
13.
2 2 2 2 4 4 2 2
4 22 2 2 2
6 4 23 2 2
2
− − − += −+ −
−= = −
14.
3 5 3 5 3 10 3 25
3 253 5 3 5
28 10 3 14 5 3
22 11
+ + + += −− +
+ += = −
−
22
3.3 Solve Radical Equations
Practice Problems
1. 𝑥 − 4 = 49 𝑥 = 53
2. 2𝑥 − 1 = 9 𝑥 = 5
3. 3 + 𝑥 = 9 𝑥 = 6
4. (√𝑥 − 𝑎)2
= 𝑏2
𝑥 − 𝑎 = 𝑏2 𝑥 = 𝑏2 + 𝑎
5. 𝑥2 − 3𝑥 + 3 = 1
𝑥2 − 3𝑥 + 2 = 0 (𝑥 − 1)(𝑥 − 2) = 0 {1,2}
6. 𝑥 + 1 = (𝑥 + 1)2 𝑥 + 1 = 𝑥2 + 2𝑥 + 1 𝑥2 + 𝑥 = 0 𝑥(𝑥 + 1) = 0 {0, −1}
7. 2𝑥 − 4 = (𝑥 − 2)2 2𝑥 − 4 = 𝑥2 − 4𝑥 + 4 𝑥2 − 6𝑥 + 8 = 0 (𝑥 − 2)(𝑥 − 4) = 0 {2,4}
8. 𝑥2 = 4(2𝑥 − 3) 𝑥2 − 8𝑥 + 12 = 0 (𝑥 − 2)(𝑥 − 6) = 0 {2,6}
9. 9𝑥 + 10 = 𝑥2 𝑥2 − 9𝑥 − 10 = 0 (𝑥 + 1)(𝑥 − 10) = 0 {−1, 10} –1 is extraneous
10. 5𝑥 + 29 = (𝑥 + 3)2 5𝑥 + 29 = 𝑥2 + 6𝑥 + 9 𝑥2 + 𝑥 − 20 = 0 (𝑥 + 5)(𝑥 − 4) = 0 {−5, 4} –5 is extraneous
11.
√𝑥 + 4 = √𝑥 − 3 + 1
(√𝑥 + 4)2
= (√𝑥 − 3 + 1)2
𝑥 + 4 = 𝑥 − 3 + 2√𝑥 − 3 + 1
2√𝑥 − 3 = 6
√𝑥 − 3 = 3
(√𝑥 − 3)2
= 32
𝑥 − 3 = 9 𝑥 = 12
12.
(√𝑥 + 1 − √𝑥 − 4)2
= 𝑥 − 7
𝑥 + 1 − 2√𝑥 + 1√𝑥 − 4 + 𝑥 − 4 = 𝑥 − 7
2√𝑥 + 1√𝑥 − 4 = 𝑥 + 4
2√𝑥2 − 3𝑥 − 4 = 𝑥 + 4 4(𝑥2 − 3𝑥 − 4) = (𝑥 + 4)2 4𝑥2 − 12𝑥 − 16 = 𝑥2 + 8𝑥 + 16 3𝑥2 − 20𝑥 − 32 = 0 (3𝑥 + 4)(𝑥 − 8) = 0
{−4
3, 8}
−4
3 is extraneous
23
Regents Questions
1) JUN ’16 [5] Ans: 3 2) AUG ’17 [4] Ans: 2 3) JAN ’18 [2] Ans: 3 4) AUG ’18 [7] Ans: 3 5) SPR ’15 [8]
√𝑥 − 5 = −𝑥 + 7 𝑥 − 5 = 𝑥2 − 14𝑥 + 49 𝑥2 − 15𝑥 + 54 = 0 (𝑥 − 6)(𝑥 − 9) = 0 𝑥 = {6, 9} Reject 𝑥 = 9 because
√9 − 5 + 9 ≠ 7. 𝑥 = 6
6) AUG ’16 [35]
(√2𝑥 − 7)2
= (5 − 𝑥)2
2𝑥 − 7 = 25 − 10𝑥 + 𝑥2 𝑥2 − 12𝑥 + 32 = 0 (𝑥 − 4)(𝑥 − 8) = 0 𝑥 = {4, 8} Reject 𝑥 = 8 because
√2(8) − 7 + 8 ≠ 5. 𝑥 = 4
7) JAN ’17 [37]
0 = √𝑡 − 2𝑡 + 6
√𝑡 = 2𝑡 − 6
(√𝑡)2
= (2𝑡 − 6)2
𝑡 = 4𝑡2 − 24𝑡 + 36 4𝑡2 − 25𝑡 + 36 = 0 (4𝑡 − 9)(𝑡 − 4) = 0
𝑡 = {9
4, 4}
Reject 𝑡 =9
4 because
2 (9
4) − 6 < 0.
√1 − 2(1) + 6 = 5
√3 − 2(3) + 6 = √3
5 − √3 ≈ 3.268 327 mph
8) JUN ’17 [30]
√𝑥 − 4 = −𝑥 + 6 𝑥 − 4 = 𝑥2 − 12𝑥 + 36 𝑥2 − 13𝑥 + 40 = 0 (𝑥 − 5)(𝑥 − 8) = 0 𝑥 = {5, 8} 8 is extraneous because
√8 − 4 ≠ −8 + 6 9) JUN ’18 [33]
√6 − 2𝑥 + 𝑥 = 2𝑥 + 30 − 9
√6 − 2𝑥 = 𝑥 + 21 6 − 2𝑥 = 𝑥2 + 42𝑥 + 441 𝑥2 + 44𝑥 + 435 = 0 (𝑥 + 29)(𝑥 + 15) = 0 𝑥 = {−29, −15} –29 is extraneous because
√6 − 2(−29) − 29 ≠ 2(−14) − 9.
10) JAN ’19 [36]
3√𝑥 − 2𝑥 = −5
3√𝑥 = 2𝑥 − 5 9𝑥 = 4𝑥2 − 20𝑥 + 25 4𝑥2 − 29𝑥 + 25 = 0 (4𝑥 − 25)(𝑥 − 1) = 0
𝑥 = {25
4, 1}
1 is extraneous because
3√1 − 2(1) ≠ −5.
24
3.4 nth Roots
Practice Problems
1. 11 2. 6
3. 2.91 4.
49 2
5. 310 10 6.
3 64 4=
7.
( )
( )
23 3 3
3 23
93 3 81 81
9 39 9
= =
8.
( )
( )
( )( )
4
4 4
3 34 4
34 4
4 34
6 8 6
16 2
2 6 263 2 3 8
22 2
=
= = =
9.
( )3 33 7x =
𝑥 = 343
10.
5 5 5 243x =
𝑥 = 3
Regents Questions
1) AUG ’17 [30] 8.75 = 1.25𝑥49 7 = 𝑥49 49 7 1.04x= 4 % growth
25
3.5 Rational Exponents
Practice Problems
1. 45x 2.
74x
3. 153x
− 4. 3x
5.
3 2
1
x
6.
2 23 3
227 64 4 16
64 27 3 9
−
= = =
7.
1817 1799 3
1 19 9
33 323 17x x
x x x
x x
= = = =
8.
113
2 233 9
23
43 2927
27 3 3x
x y xy x y
xy−
= = =
9.
( )122 6
32 6
1 19
39x y
xyx y
−= =
10.
2132
6 43 4
3
yx y x y
x
−− −
= =
11.
( )1 1
12 22
5 94 2
9 5
x xx x
x x
−
−
= = =
12.
( )
( )
23
23
6
2 4 2 66 2
1 1 1m
m m m mm m
−
= = =
13.
2 43 13 66 2
1 16 6
3 2
6
x x xx x x
x x x
= = = = =
14.
5 52 2
7 7 73 3 3
15 91 26 36 3
14 76 3
2 5 3 3 3
3 7
3 26
x x x x x x
x x x x
x xx x x x
x x
+ += = + =
+ = + = +
26
15.
523 6
3352 22
3 6
54
y x
y x
y x
=
=
=
16.
13
13
13
2
2
2
4
6
12
V a b
Vb
a
Vb
a
Vb
a
=
=
=
=
Regents Questions
1) JUN ’16 [1] Ans: 4 2) JAN ’17 [7] Ans: 2 3) JUN ’17 [16] Ans: 4 4) AUG ’17 [23] Ans: 4 5) JAN ’18 [11] Ans: 4 6) JUN ’18 [20] Ans: 2 7) AUG ’18 [12] Ans: 3 8) SPR ’15 [5]
83
43
43
43
y
y
xx
x
x x
y
=
=
=
9) AUG ’16 [26]
( )15
2 25 53 3 9
= =
10) JAN ’17 [30]
6 65 55 53 6
2
x y
x y
=
=
11) JUN ’17 [31] 3 51 21
3 6 6 62x x x x x
= =
12) AUG ’17 [25]
( )43
4 43( 8) 8 ( 2) 16− = − = − =
13) JAN ’18 [32] The denominator of the rational exponent represents the index of a root, and the 4th root of 81 is 3 and 33 is 27.
14) AUG ’18 [26] 3
122
22
2
2
xx x
x
= =
15) JAN ’19 [25]
5 52 8213 3 312312
3 394 312
2 53
3 44
x y x y x yx y
x y x y x y
−
= = =
27
4. COMPLEX NUMBERS
4.1 Imaginary and Complex Numbers
Practice Problems
1. √100√−1√3 = 10𝑖√3 2. −8 +3
4√16√−1√5 = −8 + 3𝑖√5
3. √36√𝑥16√−1√5 = 6𝑥8𝑖√5 4. −27𝑖
28
4.2 Operations with Complex Numbers
Practice Problems
1. 8 − 2𝑖 2. (3 + 2𝑖)(2 − 𝑖) = 6 − 3𝑖 + 4𝑖 − 2(−1) = 8 + 𝑖
3. (3 − 7𝑖)(3 − 7𝑖) = 9 − 21𝑖 − 21𝑖 + 49(−1) = −40 − 42𝑖
4. (2√2 + 5𝑖)(5√2 − 2𝑖) =
20 − 4𝑖√2 + 25𝑖√2 − 10(−1) =
30 + 21𝑖√2
5. (−1 + 𝑖)(−1 + 𝑖)(−1 + 𝑖) = (1 − 𝑖 − 𝑖 − 1)(−1 + 𝑖) = (−2𝑖)(−1 + 𝑖) = 2𝑖 − 2(−1) = 2 + 2𝑖
6. (3𝑖)(2𝑖)2(2 + 𝑖) = (3𝑖)(−4)(2 + 𝑖) = (−12𝑖)(2 + 𝑖) = −24𝑖 − 12(−1) = 12 − 24𝑖
7. (𝑥 + 𝑖)(𝑥 + 𝑖) − (𝑥 − 𝑖)(𝑥 − 𝑖) = (𝑥2 + 2𝑥𝑖 − 1) − (𝑥2 − 2𝑥𝑖 − 1) = 4𝑥𝑖
8. 2𝑥𝑖(𝑖 − 4𝑖2) = 2𝑥𝑖(𝑖 + 4) = −2𝑥 + 8𝑥𝑖
Regents Questions
1) JUN ’16 [3] Ans: 2 2) JUN ’17 [4] Ans: 2 3) AUG ’17 [2] Ans: 3 4) JUN ’18 [5] Ans: 3 5) AUG ’18 [15] Ans: 3 6) JAN ’19 [11] Ans: 1 7) SPR ’15 [6]
(4 − 3𝑖)(5 + 2𝑦𝑖 − 5 + 2𝑦𝑖) (4 − 3𝑖)(4𝑦𝑖) 16𝑦𝑖 − 12𝑦𝑖2 12𝑦 + 16𝑦𝑖
8) AUG ’16 [27] 𝑥𝑖(−6𝑖)2 𝑥𝑖(36𝑖2) −36𝑥𝑖
9) JAN ’17 [25] (1 − 𝑖)(1 − 𝑖)(1 − 𝑖) (1 − 2𝑖 + 𝑖2)(1 − 𝑖) −2𝑖(1 − 𝑖) −2𝑖 + 2𝑖2 −2 − 2𝑖
10) JAN ’18 [25] 𝑖2 = −1, not 1. 6 + 10𝑖 − 4(−1) 10 + 10𝑖
29
5. QUADRATIC FUNCTIONS AND CIRCLES
5.1 Solve Quadratic Equations
Practice Problems
1. 𝑥 = ±√−25 = ±√25√−1 = ±5𝑖 {−5𝑖, 5𝑖}
2. 𝑥 = ±√−16 = ±√16√−1 = ±4𝑖 {−4𝑖, 4𝑖}
3. 𝑥2 =3
2(−18) = −27
𝑥 = ±√−27 = ±3𝑖√3
{−3𝑖√3, 3𝑖√3}
4. 𝑥 + 2 = ±√−9 𝑥 + 2 = ±3𝑖 𝑥 = −2 ± 3𝑖
5. 𝑥2 − 4𝑥 = −9 𝑥2 − 4𝑥 + 4 = −9 + 4 (𝑥 − 2)2 = −5
𝑥 − 2 = ±√−5
𝑥 = 2 ± 𝑖√5
6. 𝑥2 + 8𝑥 = −25 𝑥2 + 8𝑥 + 16 = −25 + 16 (𝑥 + 4)2 = −9
𝑥 + 4 = ±√−9 𝑥 = −4 ± 3𝑖
7.
( ) ( ) ( )( )
( )
23 3 4 1 7
2 1
3 19 3 19
2 2 2
i
− − − −=
−=
8.
( ) ( ) ( )( )
( )
212 12 4 4 25
2 4
12 256 12 16 32
8 8 2
ii
− − − −=
− = =
9. 3𝑥2 − 4𝑥 + 5 = 0
( ) ( ) ( )( )
( )
24 4 4 3 5
2 3
4 44 4 2 11 2 11
6 6 3 3
i i
− − − −=
− = =
10. −3𝑥2 + 2𝑥 − 2 = 0
( ) ( ) ( )( )
( )
22 2 4 3 2
2 3
2 20 2 2 5 1 5
6 6 3 3
i i
− − − −=
−
− − − = =
− −
11.
3 33 2
3
x xx
x
+ + + =
𝑥2 + 3𝑥 + 3𝑥 + 9 = 6𝑥 𝑥2 + 9 = 0 𝑥2 = −9 𝑥 = ±3𝑖
12.
( )5 2
xx x + =
𝑥2 + 5 = 2𝑥 𝑥2 − 2𝑥 = −5 𝑥2 − 2𝑥 + 1 = −5 + 1 (𝑥 − 1)2 = −4
𝑥 − 1 = ±√−4 𝑥 − 1 = ±2𝑖 𝑥 = 1 ± 2𝑖
30
13.
( )82
xx x = −
𝑥2 = 2𝑥 − 8 𝑥2 − 2𝑥 = −8 𝑥2 − 2𝑥 + 1 = −8 + 1 (𝑥 − 1)2 = −7
𝑥 − 1 = ±√−7
𝑥 − 1 = ±𝑖√7
𝑥 = 1 ± 𝑖√7
14.
( )32 2
xx x + = −
2𝑥2 + 3 = −2𝑥 2𝑥2 + 2𝑥 + 3 = 0
𝑥2 + 𝑥 +3
2= 0
𝑥2 + 𝑥 = −3
2
𝑥2 + 𝑥 +1
4= −
3
2+
1
4
(𝑥 +1
2)
2= −
5
4
𝑥 +1
2= ±√−
5
4
𝑥 = −1
2±
𝑖√5
2
15.
872 0
8 9
xx
x
+ =
9𝑥2 + 64 = 0
𝑥2 = −64
9
𝑥 = ±√−64
9
8
3
ix =
16.
22
5 62x
xx
+ =
2𝑥2 + 5 = 6𝑥 2𝑥2 − 6𝑥 = −5
𝑥2 − 3𝑥 = −5
2
𝑥2 − 3𝑥 +9
4= −
5
2+
9
4
(𝑥 −3
2)
2= −
1
4
𝑥 −3
2= ±√−
1
4
𝑥 −3
2= ±
1
2𝑖
𝑥 =3
2±
1
2𝑖 or
3
2
ix
=
17. −2 − 3𝑖 18. (𝑎 + 𝑏𝑖)(𝑎 − 𝑏𝑖) = 𝑎2 − 𝑎𝑏𝑖 + 𝑎𝑏𝑖 − (𝑏𝑖)2 = 𝑎2 + 𝑏2
31
Regents Questions
1) FALL ’15 [4] Ans: 3 2) JUN ’16 [12] Ans: 1 3) AUG ’16 [1] Ans: 4 4) JAN ’17 [11] Ans: 4 5) JUN ’17 [7] Ans: 4 6) AUG ’17 [3] Ans: 3 7) JAN’18 [12] Ans: 1 8) JUN ’18 [8] Ans: 1 9) AUG ’18 [9] Ans: 3
10) JAN ’19 [5] Ans: 2 11) JAN ’19 [9] Ans: 4 12) JUN ’18 [27]
( )( )
( )
25 5 4 2 8
2 2
5 39
4 4
x
i
− −=
= −
32
5.2 Graphs of Quadratic Functions
Practice Problems
1.
( 8)
22( 2)
x− −
= = −−
22( 2) 8( 2) 3 11y = − − − − + = 2x =− and (–2,11)
2.
2
( 2)1
2( 1)
( 1) 2( 1) 1 2
x
y
− −= = −
−
= − − − − + =
1x =− and ( 1,2)−
3.
4.
5.
6.
33
7.
2 26
92 2
b = =
2
2
2
2
2
6 10
10 6
1 6 9
1 ( 3)
( 3) 1
y x x
y x x
y x x
y x
y x
= + +
− = +
− = + +
− = +
= + +
vertex: (–3,1)
8.
2 210
252 2
b = =
2
2
2
2
2
10 21
21 10
4 10 25
4 ( 5)
( 5) 4
y x x
y x x
y x x
y x
y x
= + +
− = +
+ = + +
+ = +
= + −
vertex: (–5,–4)
Regents Questions
1) JAN ’19 [22] Ans: 4
34
5.3 Focus and Directrix
Practice Problems
1. Vertex is (2, −4)
𝑝 =1
4𝑎=
1
4÷
1
16= 4
Since 𝑝 > 0, the directrix is below the vertex. Directrix is 𝑦 = −4 − 4 = −8
2. ℎ = 0 and 𝑘 =3+5
2= 4
𝑝 = 3 − 4 = −1
𝑎 =1
4𝑝= −
1
4
𝑦 = −1
4𝑥2 + 4
3. ℎ = 4 and 𝑘 =2−4
2= −1
𝑝 = 2 − (−1) = 3
𝑎 =1
4𝑝=
1
12
𝑦 =1
12(𝑥 − 4)2 − 1
4. 𝑥2 + 6𝑥 = −4𝑦 − 5 𝑥2 + 6𝑥 + 9 = −4𝑦 − 5 + 9 (𝑥 + 3)2 = −4𝑦 + 4 (𝑥 + 3)2 − 4 = −4𝑦
𝑦 = −1
4(𝑥 + 3)2 + 1
Vertex is (−3,1)
𝑝 =1
4𝑎=
1
4÷ (−
1
4) = −1
Since 𝑝 < 0, the focus is 1 unit below the vertex, at (−3,0), and the directrix is 1 unit above the vertex, at 𝑦 = 2.
5. Vertex is (−1,2). 𝑝 = 8 ÷ 4 = 2
The equation of the directrix is 𝑦 = 𝑘 − 𝑝, so 𝑦 = 2 − 2, or 𝑦 = 0
6. Vertex is (–3,1) 𝑝 = −20 ÷ 4 = −5
The equation of the directrix is 𝑦 = 𝑘 − 𝑝, or 𝑦 = 6.
7. 𝑝 = 1 − (−3) = 4 (𝑥 − 2)2 = 16(𝑦 − 1)
8. Vertex is (2,1).
4𝑝 = 2, so 𝑝 =1
2.
Focus is 1
2 unit above the vertex, at (2,
3
2).
Directrix is 𝑦 = 𝑘 − 𝑝, or 𝑦 =1
2.
Regents Questions
1) SPR ’15 [2] Ans: 4 2) AUG ’16 [19] Ans: 4 3) JUN ’17 [17] Ans: 4 4) AUG ’17 [6] Ans: 2 5) JAN ’18 [16] Ans: 1 6) JUN ’18 [21] Ans: 4 7) AUG ’18 [23] Ans: 4 8) JAN ’19 [14] Ans: 3
9) JUN ’16 [30] The vertex is (4, −3) and 𝑝 = 12 ÷ 4 = 3. The x-coordinates of the focus and the vertex are the same. Since 𝑝 = 3, the focus is 3 units up from the vertex, so the coordinates of the focus are (4,0).
35
5.4 Equations of Circles
Practice Problems
1. 𝑥2 − 6𝑥 + 𝑦2 + 14𝑦 + 42 = 0 𝑥2 − 6𝑥 + 𝑦2 + 14𝑦 = −42 (𝑥2 − 6𝑥 + 9) + 𝑦2 + 14𝑦 = −42 + 9 (𝑥 − 3)2 + 𝑦2 + 14𝑦 = −33 (𝑥 − 3)2 + (𝑦2 + 14𝑦 + 49) = −33 + 49 (𝑥 − 3)2 + (𝑦 + 7)2 = 16
Center is (3, −7) and radius is √16 = 4.
2. 2𝑥2 + 4𝑥 + 2𝑦2 − 20𝑦 − 46 = 0 Divide by 2: 𝑥2 + 2𝑥 + 𝑦2 − 10𝑦 − 23 = 0 𝑥2 + 2𝑥 + 𝑦2 − 10𝑦 = 23 (𝑥2 + 2𝑥 + 1) + 𝑦2 − 10𝑦 = 23 + 1 (𝑥 + 1)2 + 𝑦2 − 10𝑦 = 24 (𝑥 + 1)2 + (𝑦2 − 10𝑦 + 25) = 24 + 25 (𝑥 + 1)2 + (𝑦 − 5)2 = 49
Center is (−1,5) and radius is √49 = 7.
Regents Questions
1) JUN ’16 [19] Ans: 4
36
6. EXPONENTIAL FUNCTIONS
6.1 Solve Simple Exponential Equations
Practice Problems
1. 4𝑥 = 64 4𝑥 = 43 𝑥 = 3
2. 3𝑥 = 81 3𝑥 = 34 𝑥 = 4
3. 5𝑥 = 25 5𝑥 = 52 𝑥 = 2
4. (1
2)
𝑥=
1
8
(1
2)
𝑥= (
1
2)
3
𝑥 = 3
5. 2𝑥+1 = 23 𝑥 + 1 = 3 𝑥 = 2
6. 32𝑥−2 = 34 2𝑥 − 2 = 4 𝑥 = 3
7. 3𝑥−3 = 30 𝑥 − 3 = 0 𝑥 = 3
8. 43𝑥+5 = 42 3𝑥 + 5 = 2 𝑥 = −1
9. 3𝑥+1 = 27 3𝑥+1 = 33 𝑥 + 1 = 3 𝑥 = 2
10. (22)4 = 23𝑥−1 28 = 23𝑥−1 8 = 3𝑥 − 1 𝑥 = 3
11. 2𝑥 = 𝑥 + 4 𝑥 = 4
12. 2𝑥 = 22(𝑥+1) 𝑥 = 2(𝑥 + 1) 𝑥 = 2𝑥 + 2 𝑥 = −2
13. 22𝑥 = 23𝑥+1 2𝑥 = 3𝑥 + 1 𝑥 = −1
14. 26𝑥 = 2𝑥+5 6𝑥 = 𝑥 + 5 𝑥 = 1
15. 3𝑥−5 = 32(𝑥−3) 𝑥 − 5 = 2𝑥 − 6 𝑥 = 1
16. 23(𝑥−2) = 2𝑥 3𝑥 − 6 = 𝑥 𝑥 = 3
17. 21 = 22𝑥+1 1 = 2𝑥 + 1 𝑥 = 0
18. 23(𝑥−2) = 22𝑥 3𝑥 − 6 = 2𝑥 𝑥 = 6
37
6.2 Rewrite Exponential Expressions
Practice Problems
1. 52𝑥 = (52)𝑥 = 25𝑥 2. 10(1.1)5𝑥 = 10(1.15)𝑥 = 10(1.61051)𝑥
3. 23𝑥+2 = (23)𝑥 ∙ 22 = 4(8)𝑥 4. 4(3)𝑥+1 = 4(3)𝑥(31) = 12(3)𝑥
5.
( )2
2 33
3 93
273
xx
x− = =
6. ( ) ( )
( )4 5 4 5
4 5 4 54
2 22 2 32
16 2
y yy y
y y
+ ++ −
= = = =
Regents Questions
1) AUG ’17 [10] Ans: 3 2) JAN ’18 [8] Ans: 4
38
6.3 Graphs of Exponential Functions
Practice Problems
1.
2.
3.
4.
39
5.
6.
7.
8.
Regents Questions
1) JAN ’18 [23] Ans: 4 2) JUN ’18 [2] Ans: 2
3) JUN ’17 [29]
40
6.4 Exponential Regression
Practice Problems
1. 0.1(4)xy = 2. ( )14
xy =
Regents Questions
1) JAN ’17 [13] Ans: 3 2) JUN ’18 [4] Ans: 2 3) JAN ’18 [26]
𝐷 = 1.223(2.652)𝐴
41
6.5 Periodic Growth or Decay
Practice Problems
1. (3) because the base of the exponent is less than 1.
2. (1 +0.033
12)
12𝑡= (1.00275)𝑚
3. 𝑓(𝑡) = 300 (1.021
12)12𝑡
≈
300(1.0017)12𝑡
4. a) 100(1.001)365 ≈ 144 (wow!) b) 1.001365 ≈ 1.44,
so 𝑓(𝑡) = 100(1.44)𝑡
5. a) 𝑓(𝑡) = 2000(1.02)4𝑡 ≈2000(1.0824)𝑡
b) 2000(1.02)4(1.5) ≈ 2252
6. a) 1000(1.04)5 = $1,216.65
b) ( )( )
( )( )
51 200 200 1.04
1 1.041 1
$1,083.26
ntrn
rn
c c− + −=
−− +
=
c) 1216.65 + 1083.26 = $2,299.91
Regents Questions
1) SPR ’15 [4] Ans: 2 2) JUN ’16 [15] Ans: 4 3) JUN ’16 [21] Ans: 3 4) AUG ’16 [13] Ans: 1 5) AUG ’16 [22] Ans: 4 6) JAN ’17 [2] Ans: 1 7) JUN ’17 [13] Ans: 3 8) JUN ’17 [24] Ans: 2 9) JAN ’18 [5] Ans: 4 10) JUN ’18 [23] Ans: 4 11) AUG ’18 [8] Ans: 2 12) JAN ’19 [6] Ans: 3 13) JAN ’19 [20] Ans: 3
14) FALL ’15 [17]
𝐴(𝑡) = 100(0.5)𝑡
63 t is time in years, and 𝐴(𝑡) is the amount of titanium-44 left after t years.
( ) ( )10 0 89.58132 100
10 0 10
1.041868
A A− −=
−
= −
The estimated mass at 𝑡 = 40 is 100 − 40(−1.041868) ≈ 58.3. The actual mass is
𝐴(40) = 100(0.5)40
63 ≈ 64.3976 The estimated mass is less than the actual mass.
42
6.6 Continuous Growth or Decay
Practice Problems
1. a) 𝐴 = 1000(1.025)3
b) 𝐴 = 1000 (1 +0.025
12)
(12)(3)
c) 𝐴 = 1000𝑒(0.025)(3)
2. 𝐴 = 5000𝑒(0.03)(4) = 5637.48
3. 𝐴 = 𝑃𝑒𝑟𝑡
30,000 = 𝑃𝑒(0.05)(6)
0.3
30,000$22,225P
e=
Regents Questions
1) JUN ’17 [18] Ans: 2 2) JAN ’19 [18] Ans: 4
43
7. LOGARITHMS
7.1 Introduction to Logarithms
Practice Problems
1. 4𝑥 = 64, so 𝑥 = 3 2. 5𝑥 =1
125, so 𝑥 = −3
3. 6𝑥 = 1, so 𝑥 = 0 4. 𝑘 = log2 1 means 2𝑘 = 1, so 𝑘 = 0
5. 𝑥 = 34 = 81 6. 𝑥 = 52 = 25 √𝑥 = 5
7. 𝑥 + 1 = 23 𝑥 + 1 = 8 𝑥 = 7
8. log10(𝑥 + 2) = 4 𝑥 + 2 = 104 𝑥 + 2 = 10,000 𝑥 = 9,998
9. –2, –1, 0, 1, 2 10. 2𝑥 = 𝑥 + 4, so 𝑥 = 4
11.
−
= −
=
=
3 32
3 32
3
3
2log 10 log 20
log 10 log 20
10log
20log 5
12.
+ −
= + −
=
3 2
3
2
3log log 2log
log log log
log
b b b
b b b
b
x y z
x y z
x y
z
44
7.2 Common Logarithms
Practice Problems
1. log 2𝑥 = log 5 𝑥 log 2 = log 5
log5
2.32log2
x =
2. log 2𝑥 = log3
16
𝑥 log 2 = log3
16
316
log2.42
log2x = −
3. log 23𝑥 = log 7 3𝑥 log 2 = log 7
log73
log2
log70.94
3 log2
x
x
=
=
4. log 32𝑥−1 = log 20 (2𝑥 − 1) log 3 = log 20
log202 1 2.7268
log3
2 3.7268
1.863
x
x
x
− =
5. 3(5𝑥+1) = 125
5𝑥+1 =125
3
log 5𝑥+1 = log125
3
(𝑥 + 1) log 5 = log125
3
1253
log1 2.317
log5x + =
𝑥 ≈ 1.317
6. 8,000 (1 +0.06
12)
12𝑡= 10,000
(1 +0.06
12)
12𝑡= 1.25
(1.005)12𝑡 = 1.25 log(1.005)12𝑡 = log 1.25 12𝑡 log 1.005 = log 1.25
log1.25
3.712log1.005
t = years
45
Regents Questions
1) JAN ’17 [15] Ans: 3 2) AUG ’16 [34]
( )
( )
8.02
8.02
7 20 0.5
0.35 0.5
log0.5log0.35
8.02
8.02log0.3512
log0.5
t
t
t
t
=
=
=
=
3) AUG ’16 [37]
(a) 𝐴(𝑛) = 5,000(1.045)𝑛
( )4
0.0465,000 1
4
n
B n
= +
(b) 𝐵(6) − 𝐴(6) ≈ 6578.87 − 6511.30 ≈ 67.57
40.046
10,000 5,000 14
n
= +
(c) 2 = 1.01154𝑛 log 2 = 4𝑛 log 1.0115
log215.2
4log1.0115n= yrs
4) JUN ’17 [37]
(a) 100 = 140 (1
2)
5
ℎ
(b) log (100
140) = log (
1
2)
5
ℎ
log5
7=
5
ℎlog
1
2
12
57
5log10.3002
logh=
(c) 40 = 140 (1
2)
𝑡
10.3002
log2
7= log (
1
2)
𝑡
10.3002
12 27
27
12
loglog
10.3002
10.3002log18.6
log
t
t
=
=
5) AUG ’17 [36] (a) 𝑦 = 4.168(3.981)𝑥 (b) 100 = 4.168(3.981)𝑥
1004.168
1004.168
log log3.981
log2.25
log3.981
x
x
=
=
6) JUN ’18 [35]
( )12
0.025563,000 1
12
t
C t
= +
( )
( )
12
12 10063
10063
0.025563,000 1 100,000
12
1.002125
12 log 1.002125 log
18.14
t
t
t
t
+ =
=
=
7) JAN ’19 [37] (a) 𝐴(𝑡) = 318,000(1.07)𝑡 (b)
(c) 1,000,000 = 318,000(1.07)𝑡 1000
318= 1.07𝑡
log1000
318= 𝑡 log 1.07
1000318
log17
log1.07t = yrs
(c) The graph nearly intersects the point (17,1000000)
46
7.3 Natural Logarithms
Practice Problems
1. ln 𝑒𝑥 = ln 15 𝑥 = ln 15 ≈ 2.71
2. ln 𝑒2𝑥 = ln 13 2𝑥 = ln 13
ln13
1.282
x =
3. 42050
1500xe=
420501500
ln ln xe=
20501500
ln 4x=
20501500
ln0.078
4x =
4. Use the power rule: 𝑎 ln 𝑥 = ln 𝑥𝑎. ln 𝑥2 = ln 16 𝑥2 = 16 𝑥 = ±4 𝑓(𝑥) = ln 𝑥 is restricted to 𝑥 > 0, so reject –4. 𝑥 = 4
5. 𝐴 = 𝑃𝑒𝑟𝑡
𝐴
𝑃= 𝑒𝑟𝑡
1.25 = 𝑒3𝑟 ln 1.25 = ln 𝑒3𝑟 ln 1.25 = 3𝑟
ln1.25
7.4%3
r =
6. 𝐴 = 𝑃𝑒𝑟𝑡 3000 = 2500𝑒5𝑟 1.2 = 𝑒5𝑟 ln 1.2 = ln 𝑒5𝑟 ln 1.2 = 5𝑟
ln1.2
3.6%5
r =
47
Regents Questions
1) JAN ’17 [23] Ans: 1 2) JUN ’17 [2] Ans: 1 3) JAN ’18 [13] Ans: 3 4) JAN ’18 [10] Ans: 1 5) JUN ’18 [18] Ans: 1 6) AUG ’18 [1] Ans: 4 7) AUG ’18 [19] Ans: 3 8) FALL ’15 [13]
(a) 100 = 325 + (68 − 325)𝑒−2𝑘 −225 = −257𝑒−2𝑘
ln225
257= −2𝑘
225257
ln0.066
2k =
−
(b) 𝑇(7) = 325 − 257𝑒−0.066(7) ≈ 163
9) JUN ’16 [32] 𝐴 = 𝑃𝑒𝑟𝑡 135,000 = 100,000𝑒5𝑟 1.35 = 𝑒5𝑟 ln 1.35 = 5𝑟
ln1.350.06
5r = or 6%
10) JAN ’17 [28] 12
ln
1590 is negative, so 𝑀(𝑡)
represents decay. 11) JAN ’18 [29]
20𝑒0.05𝑡 = 30𝑒0.03𝑡 2
3𝑒0.05𝑡 = 𝑒0.03𝑡
0.0323 0.05
t
t
e
e=
2
3= 𝑒−0.02𝑡
ln2
3= −0.02𝑡
23
ln20.3
0.02t =
− months
12) AUG ’18 [35] 2 = 𝑒0.0375𝑡 ln 2 = 0.0375𝑡
ln218.5
0.0375t =
48
7.4 Graphs of Parent Log Functions
Practice Problems
1. a) (3) b) (2) c) (1)) d) (4)
Regents Questions
1) JUN ’16 [18] Ans: 1 2) AUG ’18 [16] Ans: 2
49
7.5 Evaluate Loan Formulas
Practice Problems
1. ( )
( )
360
360
0.0025 1.0025300,000
1.0025 1M
= −
𝑀 ≈ $1,265
( )
( )
360
360
0.0025 1.0025275,000
1.0025 1M
= −
𝑀 ≈ $1,159 She can reduce her payments by $106.
2.
300
300
0.025 0.0251
12 12650,000
0.0251 1
12
+
= + −
M
( )( )
( )
300
300
0.00208 1.00208650,000
1.00208 1=
−M
𝑀 ≈ $2,915
50
Regents Questions
1) SPR ’15 [9]
0.048 0.048120,000 1
12 12720
0.0481 1
12
+
= + −
n
n
( )
( )
480 1.004720
1.004 1=
−
n
n
720(1.004)𝑛 − 720 =480(1.004)𝑛 240(1.004)𝑛 = 720 1.004𝑛 = 3 𝑛 log 1.004 = log 3
log3275.2
log1.004n= months
275.2
12≈ 23 years
2) JAN ’17 [36] (a)
601 1.00625
0.0062520,000 PMT
−− =
60
0.0062520,000
1 1.00625PMT
−
=
−
𝑃𝑀𝑇 ≈ $ 400.76 (b)
601 1.00625
0.0062521,000 300x
−− − =
𝑥 ≈ $ 6,028
3) JUN ’17 [34] 𝑁 = 12 × 15 = 180 (a)
( )180
180
0.00305 1.00305172,600
1.00305 1M = •
−
𝑀 ≈ $ 1,247 (b)
( )( )180
180
0.00305 1.003051100 172,600
1.00305 1x= −
−
1100 ≈ (172,600 − 𝑥)(0.007228) 152,193 ≈ 172,600 − 𝑥 𝑥 ≈ $ 20,407
4) JUN ’18 [31] 360
360
0.036 0.036137,250 1
12 12
0.0361 1
12
M
+
+ −
=
𝑀 ≈ $ 624
51
8. POLYNOMIAL FUNCTIONS
8.1 Factor Polynomials
Practice Problems
1. 𝑥2(𝑥 + 3) + 2(𝑥 + 3) (𝑥2 + 2)(𝑥 + 3)
2. 2𝑥2(2𝑥 + 5) − 5(2𝑥 + 5) (2𝑥2 − 5)(2𝑥 + 5)
3. 𝑥2(𝑥 + 3) − 4(𝑥 + 3) (𝑥2 − 4)(𝑥 + 3) (𝑥 + 2)(𝑥 − 2)(𝑥 + 3)
4. 𝑥2(𝑥 − 2) − 9(𝑥 − 2) (𝑥2 − 9)(𝑥 − 2) (𝑥 + 3)(𝑥 − 3)(𝑥 − 2)
5. 𝑥2(𝑥 + 2) − (𝑥 + 2) (𝑥2 − 1)(𝑥 + 2) (𝑥 + 1)(𝑥 − 1)(𝑥 + 2)
6. 𝑥2(3𝑥 − 5) − 16(3𝑥 − 5) (𝑥2 − 16)(3𝑥 − 5) (𝑥 + 4)(𝑥 − 4)(3𝑥 − 5)
7. 𝑎(𝑎 + 𝑏) + 𝑐(𝑎 + 𝑏) (𝑎 + 𝑐)(𝑎 + 𝑏)
8. 𝑎3 + 𝑎2 − 𝑎𝑏 − 𝑏 𝑎2(𝑎 + 1) − 𝑏(𝑎 + 1) (𝑎2 − 𝑏)(𝑎 + 1)
9. Split the middle term: 𝑎4 − 𝑎2𝑏2 − 4𝑎2𝑏2 + 4𝑏4 𝑎2(𝑎2 − 𝑏2) − 4𝑏2(𝑎2 − 𝑏2) (𝑎2 − 4𝑏2)(𝑎2 − 𝑏2) (𝑎 + 2𝑏)(𝑎 − 2𝑏)(𝑎 + 𝑏)(𝑎 − 𝑏)
10. Split the middle term: 2𝑥4 − 2𝑥2𝑦2 + 𝑥2𝑦2 − 𝑦4 2𝑥2(𝑥2 − 𝑦2) + 𝑦2(𝑥2 − 𝑦2) (2𝑥2 + 𝑦2)(𝑥2 − 𝑦2) (2𝑥2 + 𝑦2)(𝑥 + 𝑦)(𝑥 − 𝑦)
11. (𝑥 + 5)(𝑥2 − 5𝑥 + 25) 12. (𝑏 + 4)(𝑏2 − 4𝑏 + 16)
13. (𝑦 − 6)(𝑦2 + 6𝑦 + 36) 14. (3𝑥 − 2)(9𝑥2 + 6𝑥 + 4)
15. Let 𝑢 = 𝑥2 𝑢2 + 𝑢 − 12 (𝑢 + 4)(𝑢 − 3) (𝑥2 + 4)(𝑥2 − 3)
16. Let 𝑢 = 2𝑥2 2𝑢2 − 3𝑢 − 2 (2𝑢 + 1)(𝑢 − 2) (4𝑥2 + 1)(2𝑥2 − 2) 2(4𝑥2 + 1)(𝑥2 − 1) 2(4𝑥2 + 1)(𝑥 + 1)(𝑥 − 1)
17. Let 𝑢 = 𝑥2 and 𝑣 = 𝑥 − 6 𝑢2 − 𝑣2 (𝑢 + 𝑣)(𝑢 − 𝑣) (𝑥2 + 𝑥 − 6)(𝑥2 − 𝑥 + 6) (𝑥 + 3)(𝑥 − 2)(𝑥2 − 𝑥 + 6)
18. We can factor out the GCF of 2 from the middle terms:
(𝑥5 + 2𝑥)2 + 2(𝑥5 + 2𝑥) + 1 Now let 𝑢 = 𝑥5 + 2𝑥 𝑢2 + 2𝑢 + 1 (𝑢 + 1)(𝑢 + 1) (𝑥5 + 2𝑥 + 1)(𝑥5 + 2𝑥 + 1)
52
19. Factor by grouping: 𝑥4(2𝑥 − 1) + 5𝑥2(2𝑥 − 1) + 4(2𝑥 − 1) (𝑥4 + 5𝑥2 + 4)(2𝑥 − 1) Let 𝑢 = 𝑥2 𝑢2 + 5𝑢 + 4 (𝑢 + 4)(𝑢 + 1) (𝑥2 + 4)(𝑥2 + 1)(2𝑥 − 1)
Regents Questions
1) FALL ’15 [5] Ans: 4 2) AUG ’16 [5] Ans: 3 3) AUG ’16 [15] Ans: 3 4) JAN ’17 [3] Ans: 4 5) AUG ’17 [15] Ans: 1 6) AUG ’18 [14] Ans: 4 7) JAN ’19 [3] Ans: 1 8) FALL ’15 [12]
The expression is of the form 𝑢2 −5𝑢 − 6 or (𝑢 − 6)(𝑢 + 1). Let 𝑢 = 4𝑥2 + 5𝑥. (4𝑥2 + 5𝑥 − 6)(4𝑥2 + 5𝑥 + 1) (4𝑥 − 3)(𝑥 + 2)(4𝑥 + 1)(𝑥 + 1)
9) JUN ’17 [27] 𝑥2(4𝑥 − 1) + 4(4𝑥 − 1) (𝑥2 + 4)(4𝑥 − 1)
10) JAN ’18 [28] 3𝑥3 + 𝑥2 + 3𝑥𝑦 + 𝑦 𝑥2(3𝑥 + 1) + 𝑦(3𝑥 + 1) (𝑥2 + 𝑦)(3𝑥 + 1)
11) AUG ’18 [25] (𝑥2 − 6)(𝑥2 + 2)
53
8.2 Solve Polynomial Equations
Practice Problems
1. 𝑥(𝑥2 + 𝑥 − 2) = 0 𝑥(𝑥 + 2)(𝑥 − 1) = 0 {−2,0,1}
2. 𝑥2(2𝑥 − 1) − 4(2𝑥 − 1) = 0 (𝑥2 − 4)(2𝑥 − 1) = 0 (𝑥 + 2)(𝑥 − 2)(2𝑥 − 1) = 0
{1
2, ±2}
3. 4𝑥2(2𝑥 + 1) − 9(2𝑥 + 1) = 0 (4𝑥2 − 9)(2𝑥 + 1) = 0 (2𝑥 + 3)(2𝑥 − 3)(2𝑥 + 1) = 0
{±3
2, −
1
2}
4. 𝑥3 + 5𝑥2 − 4𝑥 − 20 = 0 𝑥2(𝑥 + 5) − 4(𝑥 + 5) = 0 (𝑥2 − 4)(𝑥 + 5) = 0 {±2, −5}
5. (𝑥2 + 1)(𝑥2 − 4) = 0 𝑥2 + 1 = 0 or 𝑥2 − 4 = 0 {±2, ±𝑖}
6. 3𝑥(𝑥4 − 16) = 0 3𝑥(𝑥2 + 4)(𝑥2 − 4) = 0 3𝑥(𝑥2 + 4)(𝑥 + 2)(𝑥 − 2) = 0 {0, ±2, ±2𝑖}
7. 𝑥(𝑥3 + 4𝑥2 + 4𝑥 + 16) = 0
𝑥(𝑥2(𝑥 + 4) + 4(𝑥 + 4)) = 0
𝑥(𝑥2 + 4)(𝑥 + 4) = 0 {0, −4, ±2𝑖}
8. 9𝑥2(𝑥 − 10) + 64(𝑥 − 10) = 0 (9𝑥2 + 64)(𝑥 − 10) = 0
For the first factor, 𝑥2 = −64
9,
so 𝑥 = ±8
3𝑖
{10, ±8
3𝑖}
Regents Questions
1) JUN ’16 [6] Ans: 1 2) AUG ’17 [8] Ans: 4 3) AUG ’18 [21] Ans: 4
54
8.3 Operations with Functions
Practice Problems
1. 𝑠(𝑥) = 13𝑥 − 4 𝑝(𝑥) = 36𝑥2 − 11𝑥 − 5
2. 𝑑(𝑥) = 3𝑥2 + 9𝑥 𝑞(𝑥) = 𝑥 + 4
3. (a) 𝑃(𝑥) = 6𝑥 − 170 (b) 6𝑥 − 170 > 0 6𝑥 > 170 𝑥 > 28.33 Need to sell 29 headphones.
Regents Questions
1) JUN ’16 [8] Ans: 4 2) JAN ’17 [10] Ans: 3 3) JUN ’17 [9] Ans: 2 4) JUN ’18 [13] Ans: 1 5) AUG ’18 [3] Ans: 4
6) JAN ’18 [33] (2𝑥2 + 𝑥 − 3)(𝑥 − 1) − [(2𝑥2 + 𝑥 − 3) + (𝑥 − 1)] = (2𝑥3 − 2𝑥2 + 𝑥2 − 𝑥 − 3𝑥 + 3) − (2𝑥2 + 2𝑥 − 4) = 2𝑥3 − 3𝑥2 − 6𝑥 + 7
55
8.4 Long Division
Practice Problems
1.
( )
( )
2
2
5 2
1 5 3 2
5 5
2 2
2 2
0
x
x x x
x x
x
x
−
+ + −
− +
− −
− − −
2.
( )
( )
2 3 2
3 2
2
2
2
6 3 4 12
6
2 2 12
2 2 12
0
x
x x x x x
x x x
x x
x x
+
+ − + − −
− + −
+ −
− + −
3.
( )
( )
( )
2
3 2
3 2
2
2
2 2
2 3 2 7 2 9
2 3
4 2
4 6
4 9
4 6
15
x x
x x x x
x x
x x
x x
x
x
+ −
+ + + +
− +
+
− +
− +
− − −
2 152 2
2 3x x
x+ − +
+
4.
( )
( )
( )
2
3 2
3 2
2
2
3
3 2 0 4
3
0
3
3 4
3 9
5
x x
x x x x
x x
x x
x x
x
x
+ +
− − + −
− −
+
− −
−
− −
2 53
3x x
x+ + +
−
56
5.
( )
( )
( )
3 2
3 6 5 3 2
6 3
5 3 2
5 2
3
3
2 3
2 2 5 4 6
2
2 3 4
2 4
3 6
3 6
0
x x
x x x x x
x x
x x x
x x
x
x
+ +
+ + + + +
− +
+ +
− +
+
− +
6.
( )
( )
2 3 2
3 2
2
2
2 3
4 1 2 5 10
2 8 2
3 10
3 12 3
11 13
x
x x x x x
x x x
x x
x x
x
+
− + − + −
− − +
− −
− − +
−
2
11 132 3
4 1
xx
x x
−+ +
− +
Regents Questions
1) FALL ’15 [3] Ans: 1 2) JUN ’16 [14] Ans: 2 3) AUG ’17 [13] Ans: 1 4) JAN ’18 [9] Ans: 4 5) AUG ’18 [5] Ans: 3 6) JUN ’18 [29]
( )
( )
( )
2
3 2
3 2
2
2
2 5 2
3 2 6 11 4 9
6 4
15 4
15 10
6 9
6 4
5
a a
a a a a
a a
a a
a a
a
a
+ +
− + − −
− −
−
− −
−
− −
−
2 52 5 2
3 2a a
a+ + −
−
7) JAN ’19 [34]
( )
( )
3
4 3
4 3
4
2 2 4 10
2
4 10
4 8
18
x
x x x x
x x
x
x
+
+ + + −
− +
−
− +
−
3 184
2x
x+ −
+
57
8.5 Synthetic Division
Practice Problems
1.
1 0 4 4 8 2
2 4 0 8
1 2 0 4 | 0
− −
−
−
2.
2 5 11 4 4
8 12 4
2 3 1 | 0
− − −
Regents Questions
1) AUG ’16 [11] Ans: 2 2) JAN ’17 [32]
3 7 20 2
6 26
3 13 | 6
−
6
3 132
xx
+ +−
58
8.6 Remainder Theorem
Practice Problems
1. 𝑓(−2) = (−2)3 + 2(−2)2 + (−2) + 6 𝑓(−2) = 4
2. 𝑎(𝑥)
𝑥−2= 3𝑥 + 13 +
6
𝑥−2
𝑎(𝑥) = 3𝑥(𝑥 − 2) + 13(𝑥 − 2) + 6 𝑎(𝑥) = 3𝑥2 − 6𝑥 + 13𝑥 − 26 + 6 𝑎(𝑥) = 3𝑥2 + 7𝑥 − 20
3. (a) (−4)3 + 3(−4)2 + (−4)𝑘 − 24 = 0 −64 + 48 − 4𝑘 − 24 = 0 −40 − 4𝑘 = 0 𝑘 = −10
(b)
( )
( )
( )
3 2
3
2
2
2
2
3 24
4
10
6
4
4
6 24
6 2
0
1
4
0
x x
x x x x
x x
x x
x x
x
x
+ −
− +
− −
− − −
− −
− −
− −
+ −
−
(𝑥2 − 𝑥 − 6)(𝑥 + 4) = 0 (𝑥 − 3)(𝑥 + 2)(𝑥 + 4) = 0 {−4, −2,3}
59
Regents Questions
1) AUG ’16 [21] Ans: 3 2) JAN ’17 [20] Ans: 2 3) JUN ’17 [11] Ans: 1 4) AUG ’17 [20] Ans: 2 5) JAN ’18 [19] Ans: 4 6) JUN ’18 [12] Ans: 3 7) SPR ’15 [7]
By the Factor Theorem, 𝑥 − 𝑎 is a factor of 𝑓(𝑥) only when 𝑓(𝑎) = 0. 𝑓(4) = 2(4)3 − 5(4)2 − 11(4) − 4 𝑓(4) = 0
8) FALL ’15 [15] 0 = 6(−5)3 + 𝑏(−5)2 − 52(−5) + 15 0 = −750 + 25𝑏 + 260 + 15 475 = 25𝑏 𝑏 = 19 𝑧(𝑥) = 6𝑥3 + 19𝑥2 − 52𝑥 + 15 By the Factor Theorem, since 𝑧(−5) = 0, then 𝑥 + 5 is a factor of 𝑧(𝑥).
6 19 52 15 5
30 55 15
6 11 3 | 0
− −
− −
−
(𝑥 + 5)(6𝑥2 − 11𝑥 + 3) = 0 (𝑥 + 5)(3𝑥 − 1)(2𝑥 − 3) = 0
𝑥 = {−5,1
3,
3
2}
9) JUN ’16 [27] By the Factor Theorem, 𝑥 − 𝑎 is a factor of 𝑓(𝑥) only when 𝑓(𝑎) = 0. 2(5)3 − 4(5)2 − 7(5) − 10 = 105, so 𝑥 − 5 is not a factor.
10) JUN ’17 [25] 𝑟(2) = (2)3 − 4(2)2 + 4(2) − 6 𝑟(2) = −6 By the Factor Theorem, 𝑥 − 𝑎 is a factor of 𝑓(𝑥) only when 𝑓(𝑎) = 0. Since 𝑟(2) = −6, 𝑥 − 2 is not a factor of 𝑟(𝑥).
11) AUG ’18 [34] 𝑗(−1) = 2(−1)4 − (−1)3 −35(−1)2 + 16(−1) + 48 = 0 By the Factor Theorem, since 𝑗(−1) = 0, then 𝑥 + 1 is a factor of 𝑗(𝑥).
2 1 35 16 48 1
2 3 32 48
2 3 32 48 | 0
− − −
− −
− −
Factor by grouping: 2𝑥3 − 3𝑥2 − 32𝑥 + 48 𝑥2(2𝑥 − 3) − 16(2𝑥 − 3) (𝑥2 − 16)(2𝑥 − 3) (𝑥 + 4)(𝑥 − 4)(2𝑥 − 3) (𝑥 + 1)(𝑥 + 4)(𝑥 − 4)(2𝑥 − 3) = 0
𝑥 = {−1, −4, 4,3
2}
60
8.7 Properties of Graphs
Practice Problems
1. Degree = 5, leading coefficient = –3.
As 𝑥 → −∞, 𝑓(𝑥) → ∞, and as 𝑥 → ∞, 𝑓(𝑥) → −∞
2. Degree = 4, leading coefficient = 5.
As 𝑥 → −∞, 𝑓(𝑥) → ∞, and as 𝑥 → ∞, 𝑓(𝑥) → ∞
3. 𝑥 = {−2.31, −0.76, 0.57} y-intercept at (0,2)
4. 𝑥 = {−0.41, 2.41} y-intercept at (0,–1) 2𝑥 + 1 = 0
𝑥 =1
2
Regents Questions
1) SPR ’15 [1] Ans: 1 2) JUN ’16 [4] Ans: 3 3) JUN ’16 [20] Ans: 2 4) JUN ’17 [1] Ans: 1 5) AUG ’17 [12] Ans: 4
6) JAN ’18 [17] Ans: 3 7) JUN ’18 [16] Ans: 2 8) AUG ’18 [4] Ans: 1 9) JAN ’19 [8] Ans: 1 10) JAN ’19 [19] Ans: 1
61
8.8 Graph Polynomial Functions
Practice Problems
1. x-intercepts at –2, 0, and 3. y-intercept at 𝑓(0) = 0. Degree of 3 (odd), leading coefficient of 1
(positive), so end behavior of
2. x-intercepts at –2, –1, and 1, with –1 as a double root.
y-intercept at 𝑓(0) = −4. Degree of 4 (even), leading coefficient of
2 (positive), so end behavior of
62
Regents Questions
1) AUG ’16 [33] 0 = 𝑥2(𝑥 + 1) − 4(𝑥 + 1) 0 = (𝑥2 − 4)(𝑥 + 1) 0 = (𝑥 + 2)(𝑥 − 2)(𝑥 + 1) 𝑥 = {−2, −1,2}
2) JAN ’17 [29]
Various answers, such as
or
3) AUG ’17 [32] Various answers, such as
4) JAN ’18 [31]
Various answers, such as
5) JUN ’18 [26]
63
6) AUG ’18 [37] 𝑃(𝑥) = 𝑅(𝑥) − 𝐶(𝑥) = −330𝑥3 + 9,000𝑥2 − 67,000𝑥 +167,000
Least profitable at year 5 because there is a minimum at 𝑃(5). Most profitable at year 13 because there is a maximum at 𝑃(13).
7) JAN ’19 [26]
64
8.9 Polynomial Identities
Practice Problems
1. (𝑥 + 𝑎)(𝑥 + 𝑏) = 𝑥2 + 𝑎𝑥 + 𝑏𝑥 + 𝑎𝑏 Multiply = 𝑥2 + (𝑎 + 𝑏)𝑥 + 𝑎𝑏 Distributive Pr.
2. (𝑥2 − 1)2 + (2𝑥)2 = 𝑥4 − 2𝑥2 + 1 + (2𝑥)2 Expand square of binomial = 𝑥4 − 2𝑥2 + 1 + 4𝑥2 Power Rule = 𝑥4 + 2𝑥2 + 1 Combine terms = (𝑥2 + 1)2 Rewrite as square of binomial
Regents Questions
1) AUG ’16 [20] Ans: 4 2) JAN ’18 [6] Ans: 2 3) JUN ’18 [22] Ans: 4 4) FALL ’15 [11]
Let x and 𝑥 + 1 represent the integers. (𝑥 + 1)2 − 𝑥2 = 𝑥2 + 2𝑥 + 1 − 𝑥2 = 2𝑥 + 1 2𝑥 is an even integer, so 2𝑥 + 1 is an odd integer.
5) JUN ’16 [31]
3
3
3 3
3
3
11
8
8 1
8 8
9
8
x
x
x x
x
x
++
+= +
+ +
+=
+
6) JAN ’17 [33] (1) 2𝑥3 − 10𝑥2 + 11𝑥 − 7 = (𝑥 − 4)(2𝑥2 + ℎ𝑥 + 3) + 𝑘 (2) 2𝑥3 − 10𝑥2 + 11𝑥 − 7 = 2𝑥3 + ℎ𝑥2 + 3𝑥 − 8𝑥2 − 4ℎ𝑥 −12 + 𝑘 (3) −2𝑥2 + 8𝑥 + 5 = ℎ𝑥2 − 4ℎ𝑥 + 𝑘 (4) So, ℎ = −2 and 𝑘 = 5
7) AUG ’17 [27] (𝑥2 − 𝑦2)2 + (2𝑥𝑦)2 = 𝑥4 − 2𝑥2𝑦2 + 𝑦4 + (2𝑥𝑦)2 = 𝑥4 − 2𝑥2𝑦2 + 𝑦4 + 4𝑥2𝑦2 = 𝑥4 + 2𝑥2𝑦2 + 𝑥4 = (𝑥2 + 𝑦2)2
8) JAN ’19 [27] (𝑎 + 𝑏)3 = 𝑎3 + 3𝑎2𝑏 + 3𝑎𝑏2 + 𝑏3 = 𝑎3 + 𝑏3 + 3𝑎𝑏(𝑎 + 𝑏) No. Erin’s shortcut only works if 3𝑎𝑏(𝑎 + 𝑏) = 0; that is, only if 𝑎 = 0, 𝑏 = 0, or 𝑎 = −𝑏.
65
9. TRIGONOMETRIC FUNCTIONS
9.1 Trigonometric Ratios
1. 1 1 3
cot60tan60 33
= = =
2. 1 1 2 2 3
csc60sin60 33 3
2
= = = =
3. 25
csc7
hypA
opp= =
4. cos2 𝑥
5.
coscot sin cos
1csc
sin
xx x xx
x
= = 6. 2
cossin
cot sin sin cos1sec
cos
xx
x x x xx
x
= =
9.2 Radians
Practice Problems
1.
(a) 45180 4
= rad
(b) 3
270180 2
= rad
(c) 5
150180 6
= rad
2.
(a) 180
306
= (d)
5 180100
9
=
(b) 5 180
2254
= (e)
8 180288
5
=
(c) 3 180
1085
=
3. 12 3 9.44
L
= = inches 4. 𝐿 = 2 ∙ 4 = 8 inches
5. 8 10 = 4
5
= rad
6. 65 5r= r = 13 feet
66
9.3 Unit Circle and Reference Angles
Practice Problems
1. (4) 2. (2)
3. (4) 4. (2) (cos 30° , sin 30°) = (√3
2,
1
2)
5.
− cos 70°
6.
− sin 50°
7. tan 50° 8. sin 135° = sin 45° =1
√2
9. sin4𝜋
3= − sin
𝜋
3= −
√3
2 10. sin
3𝜋
2+ cos
2𝜋
3= −1 + (−
1
2) = −
3
2
11. cos 120° = − cos 60° = −1
2 12. cos 150° = − cos 30° = −
√3
2
13. (−√3
2, −
1
2) 14. cos−1 (
√2
2) = 45°
15. cos−1 (√3
2) = 30°
180° + 30° = 210°
16. cos−1(0.6) ≈ 53° 𝜃 ≈ 360° − 53° ≈ 307°
Regents Questions
1) AUG ’16 [16] Ans: 1 2) AUG ’17 [7] Ans: 4 3) JAN ’18 [15] Ans: 1
4) JAN ’17 [27]
csc 𝜃 =1
sin 𝜃, and sin 𝜃 on a unit
circle represents the y value of a point on the unit circle. Since
𝑦 = sin 𝜃. csc 𝜃 =1
𝑦.
67
9.4 Trig Functions on Coordinate Grids
Practice Problems
1. 𝑟 = √𝑥2 + 𝑦2 = √25 = 5
sin 𝜃 =𝑦
𝑟=
4
5
2. cot 𝜃 =𝑥
𝑦=
−3
−4=
3
4
3. 𝑟 = √𝑥2 + 𝑦2 = √16 = 4
sec 𝜃 =𝑟
𝑥=
4
−4= −1
4. 𝑟 = √𝑥2 + 𝑦2 = √13
csc 𝜃 =𝑟
𝑦=
√13
2
5. 𝑟 = √89
sec 𝜃 =𝑟
𝑥= −
√89
8
6. (𝑥, 𝑦) = (𝑟 cos 𝜃 , 𝑟 sin 𝜃)
2 cos𝜋
6= 2 ∙
√3
2= √3
2 sin𝜋
6= 2 ∙
1
2= 1
𝑃(√3, 1)
Regents Questions
1) SPR ’15 [3] Ans: 1 2) JUN ’16 [17] Ans: 1
3) JUN ’18 [32]
𝑟 = √22 + 12 = √5
sin 𝜃 =𝑦
𝑟= −
1
√5
68
9.5 Pythagorean Identity
Practice Problems
1. sin2 𝜃 + cos2 𝜃 = 1 𝑎2 + cos2 𝜃 = 1 cos2 𝜃 = 1 − 𝑎2
cos 𝜃 = √1 − 𝑎2
2. (sin 𝑥)(cos 𝑥)(tan 𝑥) =
( )( ) 2sinsin cos sin
cos
xx x x
x =
3.
2 2
2 2
1 sin cos1
cos cos
x x
x x
−= =
4.
2 2
22 2
sin cos 1sec
1 sin cos
x xx
x x
+= =
−
5.
( )22
2
2 1 sin2 2sin
cos cos
2cos2cos
cos
xx
x x
xx
x
−−= =
=
6. cos 𝑥 (cos 𝑥 + 1) + sin2 𝑥 = cos2 𝑥 + cos 𝑥 + sin2 𝑥 = cos 𝑥 + 1
7. cos 𝑥 (sec 𝑥 − cos 𝑥) = cos 𝑥 ∙ sec 𝑥 − cos2 𝑥 =
21cos cos
cosx x
x − =
1 − cos2 𝑥 = sin2 𝑥
Regents Questions
1) JAN ’17 [4] Ans: 1 2) JUN ’17 [12] Ans: 2 3) AUG ’18 [11] Ans: 2 4) AUG ’16 [28]
sin2 𝜃 + (−0.7)2 = 1 sin2 𝜃 = 0.51
sin 𝜃 = ±√0.51 Since 𝜃 is in Quadrant II,
sin 𝜃 = √0.51 and
tan 𝜃 =sin 𝜃
cos 𝜃=
√0.51
−0.7≈ −1.02
5) JAN ’19 [31] sin2 𝜃 + cos2 𝜃 = 1
(4
7)
2+ 𝑡2 = 1
16
49+ 𝑡2 = 1
𝑡2 =33
49
𝑡 = ±√33
7
Since 𝜃 is in Quadrant II,
𝑡 = −√33
7
69
9.6 Graphs of Parent Trig Functions
Practice Problems
1. (4) 2. (3)
3. (3) 4. (4)
5. (4)
70
9.7 Transform Trigonometric Graphs
Practice Problems
1. amplitude is 2; period is 2𝜋
2. 2 3. amplitude is 3; frequency is 2; each period has a length of 𝜋
4. (3) 5. (2)
6. amplitude is 2
3; frequency is 4;
each period is 𝜋
2 (or 90°) long
7. midline is 𝑦 = 4; frequency is 2 ∙ 3 = 6;
each period is 𝜋
3 (or 60°) long
8. 3; There is no vertical shift and the amplitude is 3.
9. −1
3; There is no vertical shift and the
amplitude is 1
3.
10. 6; The amplitude is 2 plus there is a vertical shift of 4.
11. –5; The amplitude is 2 and there is a vertical shift of 3 units down.
(The negation of a does not affect the amplitude; it merely reflects the graph over the midline.]
12. (2) 13. If the sine graph is shifted left by 𝜋
2, then
it will coincide with the cosine graph, so
ℎ =𝜋
2.
Regents Questions
1) FALL ’15 [6] Ans: 4 2) JAN ’17 [1] Ans: 2 3) JAN ’17 [22] Ans: 3 4) JUN ’17 [6] Ans: 4 5) JUN ’17 [8] Ans: 1
6) AUG ’17 [5] Ans: 3 7) AUG ’17 [18] Ans: 4 8) AUG ’18 [20] Ans: 1 9) JAN ’19 [13] Ans: 1
71
9.8 Model Trigonometric Functions
Practice Problems
1. (4) 2. 30 + 27 = 57 feet.
3.
3
26
= seconds 4. 1 cycle is
5
210
= secs, so a 40-sec ride
would complete 4 revolutions. Radius is 20 (the amplitude). The base is 24 − 20 = 4 feet off the
ground.
Regents Questions
1) JUN ’16 [13] Ans: 3 2) JUN ’16 [24] Ans: 4 3) AUG ’16 [10] Ans: 2 4) JUN ’17 [15] Ans: 4 5) JAN ’18 [4] Ans: 2 6) JUN ’18 [10] Ans: 4 7) AUG ’18 [22] Ans: 2 8) JAN ’19 [7] Ans: 1 9) AUG ’16 [25]
Amplitude, because the height of the graph shows the volume of the air.
10) JUN ’17 [28]
Period is 2
3. The wheel rotates once
every 2
3 second.
11) JUN ’18 [36]
( ) ( )2 112
2 1
h h−= −
−
Use the calculator’s Calc → zero feature to find where ℎ(𝑡) = 0, or:
12 cos (𝜋
3𝑡) + 8 = 0
cos (𝜋
3𝑡) = −
2
3
Reference ∡ is cos−1 (2
3) ≈ 0.8411
cos is negative in Q2 and Q3, so 𝜋
3𝑡 = 𝜋 − 0.8411 ≈ 2.30 and
𝜋
3𝑡 = 𝜋 + 0.8411 ≈ 3.98
𝜋
3𝑡 = 2.30 gives us 𝑡 =
3(2.30)
𝜋≈ 2.2
𝜋
3𝑡 = 3.98 gives us 𝑡 =
3(3.98)
𝜋≈ 3.8
ℎ(𝑡) = 0 at 𝑡 ≈ {2.2, 3.8}.
72
9.9 Graph Trigonometric Functions
Practice Problems
1.
2.
𝑦 = 3 sin 2𝑥
3.
2 points of intersection.
𝑔 (𝜋
6) = −2 cos
𝜋
6= −2 ∙
√3
2= −√3
4.
Solve for x: 4 cos 𝑥 ≥ 2
cos 𝑥 ≥1
2
𝑥 ≥ cos−1 1
2
−𝜋
3≤ 𝑥 ≤
𝜋
3
73
5.
𝑑(𝑡) = sin 𝑡
6. The minimum value of the cosine function is –1. 10(−1) + 20 = 10, so the minimum price is $10. The price is $10 when
10 cos (𝜋
5𝑡) + 20 = 10
10 cos (𝜋
5𝑡) = −10
cos (𝜋
5𝑡) = −1
cos−1(−1) = 𝜋,
so 𝜋
5𝑡 = 𝜋 or 𝑡 = 5 (year 1985).
Each period is
5
210
= ,
so the price cycles every 10 years. The minimum price of $10 occurs in 1985, 1995, and 2005.
74
7. 8 cos 𝑡 + 78 = 80 8 cos 𝑡 = 2
cos 𝑡 =1
4
cos−1 1
4≈ 1.3 (left of the interval)
Since the period is 2𝜋, the graph intersects 𝑦 = 80 at: 𝑥 = 2𝜋 + 1.3 ≈ 7.6 𝑥 = 4𝜋 − 1.3 ≈ 11.3 𝑥 = 4𝜋 + 1.3 ≈ 13.9 (See below)
75
Regents Questions
1) JAN ’17 [19] Ans: 2 2) SPR ’15 [14]
Since the water level has a minimum of –12 and a maximum of 12, the amplitude is 12. The value of A is –12 since at 8:30 it is low tide. The time from low to high tide is 6.5 hours, so the period
of the function is 13 hours. 2𝜋
𝐵=
13, so 𝐵 =2𝜋
13. Therefore, 𝑓(𝑡) =
−12 cos (2𝜋
13𝑡).
Since the function is increasing from 𝑡 = 13 to 𝑡 = 19.5 (which corresponds to 9:30 pm to 4:00 am), fishing at 10:30 pm is recommended.
3) JUN ’16 [28]
4) AUG ’17 [35]
Part a sketch is shifted
𝜋
3 units
right. 5) JAN ’18 [30]
q has the smaller minimum value for the domain [−2,2]. h’s minimum is 2(−1) + 1 = −1 and q’s minimum is −8.
6) JAN ’18 [37]
The period of P is
2
3, which means
the patient’s blood pressure
reaches a high every 2
3 second and
a low every 2
3 second. The patient’s
blood pressure is high because 144 over 96 is greater than 120 over 80.
76
7) AUG ’18 [30]
77
10. COMPARE FUNCTIONS
10.1 Average Rate of Change
Practice Problems
1. (a) 5.06 3.91 1.15
0.0961999 1987 12
−=
− (b)
7.50 5.06 2.440.244
2009 1999 10
−=
−
2. 𝑓(5) = 52 + 2 = 27 𝑓(15) = 152 + 2 = 227
227 27 200
2015 5 10
R−
= = =−
3. 𝑓(−3) = (−3)2 + 10(−3) + 16 = −5 𝑓(3) = 32 + 10(3) + 16 = 55
( )( )
55 5 6010
3 3 6R
− −= = =
− −
4. 𝑓(−1) = (−1)4 + 2(−1)3 = −1 𝑓(1) = 14 + 2(13) = 3
( )( )
3 1 42
1 1 2R
− −= = =
− −
5.
𝑓 (𝜋
2) = 2 ∙ 1 = 2
𝑓 (3𝜋
2) = 2 ∙ (−1) = −2
32 2
2 2 4R
− −= = −
−
Regents Questions
1) JAN ’17 [21] Ans: 4 2) JAN ’17 [24] Ans: 1 3) JUN ’17 [21] Ans: 3 4) AUG ’17 [9] Ans: 3 5) JUN ’18 [7] Ans: 1 6) JUN ’16 [36]
( ) ( )( )
( ) ( )( )
( )
4 2 80 1.2513.125
4 2 6
4 2 179 4938
4 2 6
f f
g g
− − −= =
− −
− − − −= =
− −
𝑔(𝑥) has a greater rate of change 7) AUG ’16 [31]
156.25 56.25 1507.5
70 50 20
−= =
−
Between 50-70 mph, each additional mph in speed requires 7.5 more feet to stop.
8) AUG ’18 [27]
( ) ( )8 448.78
8 4
p p−
− 9) JAN ’19 [30]
( ) ( )11 810.1
11 8
B B− −
−
The average monthly high temperature decreases 10.1 degrees each month from August to November.
78
10.2 Even and Odd Functions
Practice Problems
1. even 2. neither
3. odd
( ) ( )1 1
f x f xx x
− = = − = −−
4. even
( ) ( )1 1
f x f xx x
− = = =−
5. neither
𝑓(−𝑥) = 2−𝑥 − 1 =1
2𝑥 − 1
6. odd
( )( )
( ) ( )
( )( )
2 2
3 3
2 2
33
4 4
4 4
x xf x
x xx x
x xf x
x xx x
− + +− = = =
− +− − −
+ += − = −
−− −
Regents Questions
1) FALL ’15 [2] Ans: 3 2) AUG ’16 [14] Ans: 1 3) JUN ’18 [6] Ans: 2
4) AUG ’17 [31] 𝑗(−𝑥) = (−𝑥)4 − 3(−𝑥)2 − 4 = 𝑥4 − 3𝑥2 − 4 Since 𝑗(𝑥) = 𝑗(−𝑥), the function is even.
79
10.3 Inverse Functions
Practice Problems
1. 𝑥 = 5𝑦 + 2 𝑥 − 2 = 5𝑦
2
5
xy
−=
( )1 2
5
xf x− −
=
2. 2 5
3
yx
+=
3𝑥 = 2𝑦 + 5 3𝑥 − 5 = 2𝑦
3 5
2
xy
−=
( )1 3 5
2
xf x− −
=
3. 𝑥 = −2
3𝑦
−3
2𝑥 = 𝑦
𝑓−1(𝑥) = −3
2𝑥
4. 𝑥 =1
3𝑦 + 2
𝑥 − 2 =1
3𝑦
3(𝑥 − 2) = 𝑦 3𝑥 − 6 = 𝑦 𝑓−1(𝑥) = 3𝑥 − 6
5. 𝑥 = (𝑦 − 2)3
√𝑥3
= 𝑦 − 2
√𝑥3
+ 2 = 𝑦
𝑓−1(𝑥) = √𝑥3
+ 2
6. 𝑥 = log(𝑦 − 2) + 3 𝑥 − 3 = log (𝑦 − 2)
10(𝑥−3) = 𝑦 − 2
10(𝑥−3) + 2 = 𝑦
𝑓−1(𝑥) = 10(𝑥−3) + 2
7. 2
yx
y=
+
𝑥(𝑦 + 2) = 𝑦 𝑥𝑦 + 2𝑥 = 𝑦 2𝑥 = 𝑦 − 𝑥𝑦 2𝑥 = 𝑦(𝑥 − 1)
2
1
xy
x=
−
( )1 2
1
xf x
x− =
−
8. a) 𝑥 = √2 b) 𝑥 = 𝑦2 − 2 𝑥 + 2 = 𝑦2
√𝑥 + 2 = 𝑦
𝑓−1(𝑥) = √𝑥 + 2 c)
80
Regents Questions
1) JUN ’16 [16] Ans: 2 2) JAN ’17 [8] Ans: 3 3) AUG ’17 [14] Ans: 2 4) JAN ’18 [21] Ans: 2 5) JUN ’18 [15] Ans: 3 6) AUG ’18 [6] Ans: 2 7) JAN ’19 [17] Ans: 3
8) FALL ’15 [9] 𝑥 = (𝑦 − 3)3 + 1 𝑥 − 1 = (𝑦 − 3)3
√𝑥 − 13
= 𝑦 − 3
√𝑥 − 13
+ 3 = 𝑦
𝑓−1(𝑥) = √𝑥 − 13
+ 3
81
10.4 Transformations of Functions
Practice Problems
1. (1) 2. (2)
3. (1) 4. The graph shifts 2 units to the right and 3 units up.
Regents Questions
1) JUN ’18 [19] Ans: 4 2) AUG ’18 [17] Ans: 4 3) JAN ’19 [2] Ans: 1 4) AUG ’16 [30]
0 = log10(𝑥 − 4) 100 = 𝑥 − 4 1 = 𝑥 − 4 𝑥 = 5 The x-intercept of ℎ(𝑥) is (2,0). f has the larger value.
5) JUN ’17 [35]
As . As .
6) JAN ’18 [36] 𝑓(𝑥) = 𝑥3(𝑥 + 4)(𝑥 − 3) 𝑔(𝑥) = 𝑓(𝑥 + 2) = (𝑥 + 2)3(𝑥 + 6)(𝑥 − 1)
7) JAN ’19 [32]
82
11. SYSTEMS
11.1 Linear Systems in Three Variables
Practice Problems
1. The first variable, x, is already isolated, so find the others by substitution:
5(3) + 4𝑦 = −9 Eq. 2 15 + 4𝑦 = −9
4𝑦 = −24 𝑦 = −6 −(3) + 4(−6) − 2𝑧 = −25 Eq. 3 −27 − 2𝑧 = −25 −2𝑧 = 2 𝑧 = −1 𝑥 = 3, 𝑦 = −6, 𝑧 = −1
2. 𝑥 + 3𝑦 + 𝑧 = 10 Eq. 1 −𝑥 − 𝑦 − 𝑧 = −2 Eq. 2 × –1
2𝑦 = 8 𝑦 = 4 We have already isolated y, so substitute: (4) − 2𝑧 = 2 Eq. 3 −2𝑧 = −2 𝑧 = 1 𝑥 + (4) + (1) = 2 Eq. 2 𝑥 = −3 𝑥 = −3, 𝑦 = 4, 𝑧 = 1
3. 2𝑥 − 4𝑦 + 5𝑧 = −33 Eq. 1 −2𝑥 + 2𝑦 − 3𝑧 = 19 Eq. 3
−2𝑦 + 2𝑧 = −14 −𝑦 + 𝑧 = −7 Result A 4𝑥 − 𝑦 = −5 Eq. 2 −4𝑥 + 4𝑦 − 6𝑧 = 38 Eq. 3 × 2
3𝑦 − 6𝑧 = 33 Result B −3𝑦 + 3𝑧 = −21 Result A × 3 3𝑦 − 6𝑧 = 33 Result B
−3𝑧 = 12 𝑧 = −4 −𝑦 + (−4) = −7 Result A 𝑦 = 3 4𝑥 − (3) = −5 Eq. 2
𝑥 = −1
2
𝑥 = −1
2, 𝑦 = 3, 𝑧 = −4
4. 𝑥 − 2𝑦 + 3𝑧 = 7 Eq. 1 4𝑥 + 2𝑦 + 2𝑧 = 8 Eq. 2 × 2
5𝑥 + 5𝑧 = 15 𝑥 + 𝑧 = 3 Result A 𝑥 − 2𝑦 + 3𝑧 = 7 Eq. 1 −3𝑥 + 2𝑦 − 2𝑧 = −10 Eq. 3
−2𝑥 + 𝑧 = −3 Result B 2𝑥 + 2𝑧 = 6 Result A × 2 −2𝑥 + 𝑧 = −3 Result B
3𝑧 = 3 𝑧 = 1 𝑥 + (1) = 3 Result A 𝑥 = 2 2(2) + 𝑦 + (1) = 4 Eq. 2 𝑦 = −1 𝑥 = 2, 𝑦 = −1, 𝑧 = 1
83
Regents Questions
1) AUG ’16 [23] Ans: 2 2) JAN ’18 [3] Ans: 4 3) SPR ’15 [10]
7𝑥 + 7𝑧 = 35 (1) + (2) 𝑥 + 𝑧 = 5 4𝑥 + 10𝑧 = 62 (2) + (3) −4(𝑥 + 𝑧 = 5) 4𝑥 + 10𝑧 = 62 6𝑧 = 42 𝑧 = 7 𝑥 + (7) = 5 𝑥 = −2 (−2) + 3𝑦 + 5(7) = 45 3𝑦 = 12 𝑦 = 4
4) JUN ’17 [33] 4𝑦 − 4𝑧 = 12 (1) + (3) 𝑦 − 𝑧 = 3 −2(𝑥 + 𝑦 + 𝑧 = 1) 2𝑥 + 4𝑦 + 6𝑧 = 2 2𝑦 + 4𝑧 = 0 𝑦 + 2𝑧 = 0 𝑦 = −2𝑧 (−2𝑧) − 𝑧 = 3 𝑧 = −1 𝑦 − (−1) = 3 𝑦 = 2 𝑥 + (2) + (−1) = 1 𝑥 = 0
5) AUG ’18 [33] 2(2𝑥 + 3𝑦 − 4𝑧 = −1) −4𝑥 + 𝑦 + 𝑧 = 16 7𝑦 + 7𝑧 = 14 𝑦 − 𝑧 = 2 𝑦 = 𝑧 + 2 4(𝑥 − 2𝑦 + 5𝑧 = 3) −4𝑥 + 𝑦 + 𝑧 = 16 −7𝑦 + 21𝑧 = 28 𝑦 − 3𝑧 = −4 𝑦 = 3𝑧 − 4 𝑧 + 2 = 3𝑧 − 4 2𝑧 = 6 𝑧 = 3 𝑦 = (3) + 2 = 5 −4𝑥 + (5) + (3) = 16 −4𝑥 = 8 𝑥 = −2
6) JAN ’19 [33] Rewrite (3) as −𝑎 + 6𝑏 + 2𝑐 = 14 2𝑏 + 5𝑐 = 21 (1) – (2) 8𝑏 + 3𝑐 = 16 (2) + (3) 4(2𝑏 + 5𝑐 = 21) 8𝑏 + 3𝑐 = 16 17𝑐 = 68 𝑐 = 4 2𝑏 + 5(4) = 21 2𝑏 = 1
𝑏 =1
2
𝑎 + 4 (1
2) + 6(4) = 23
𝑎 + 26 = 23 𝑎 = −3
84
11.2 Nonlinear Systems
Practice Problems
1.
( )( )
2
2
2 1 3 5
6 0
3 2 0
{3, 2}
x x x
x x
x x
x
+ − = +
− − =
− + =
= −
( )3 3 5 14y = + =
( )3 2 5 1y = − + = −
(3,14) and (–2,–1)
2. 3 1 3 1y x y x+ = → = − +
( )( )
2
2
7 22 3 1
10 21 0
7 3 0
{ 7, 3}
x x x
x x
x x
x
+ + = − +
+ + =
+ + =
= − −
( )3 7 1 22y = − − + =
( )3 3 1 10y = − − + =
(–7,22) and (–3,10)
3. 𝑥2 + 2𝑥 − 7 = 2𝑥 + 1 𝑥2 = 8
𝑥 = ±√8 = ±2√2
𝑦 = 2(2√2) + 1 = 1 + 4√2
𝑦 = 2(−2√2) + 1 = 1 − 4√2
(2√2, 1 + 4√2) and (−2√2, 1 − 4√2)
4. 𝑥3 − 6𝑥 + 1 = −2𝑥 + 1 𝑥3 − 4𝑥 = 0 𝑥(𝑥2 − 4) = 0 𝑥(𝑥 + 2)(𝑥 − 2) = 0 𝑥 = {0, −2,2} 𝑦 = −2(0) + 1 = 1 𝑦 = −2(−2) + 1 = 5 𝑦 = −2(2) + 1 = −3 (0,1), (−2,5), (2, −3)
5. Solve by the addition method: 2𝑥2 + 2𝑦2 = 68 Eq. 1 × 2 𝑥2 − 2𝑦2 = 7 Eq. 2
3𝑥2 = 75 𝑥2 = 25 𝑥 = ±5 𝑥2 + (5)2 = 34 Eq. 1 𝑥2 = 9 𝑥 = ±3 𝑥2 + (−5)2 = 34 Eq. 1 𝑥 = ±3 (5,3), (5, −3), (−5,3), (−5, −3)
6. 𝑦 + 2 = 3𝑥 → 𝑦 = 3𝑥 − 2 𝑥2 − 3𝑥 + 9 = 3𝑥 − 2 𝑥2 − 6𝑥 + 11 = 0 𝑥2 − 6𝑥 = −11 𝑥2 − 6𝑥 + 9 = −11 + 9 (𝑥 − 3)2 = −2
𝑥 − 3 = ±√−2
𝑥 = 3 ± 𝑖√2
𝑦 = 3(3 + 𝑖√2) − 2
𝑦 = 7 + 3𝑖√2
𝑦 = 3(3 − 𝑖√2) − 2
𝑦 = 7 − 3𝑖√2
(3 + 𝑖√2, 7 + 3𝑖√2) and
(3 − 𝑖√2, 7 − 3𝑖√2)
85
7.
(−4,1), (2,1)
8.
(−2,5), (0,1), (2, −3)
9.
(−0.58), (1, −1)
10.
(−4.89,0.03), (0.82,1.76)
86
Regents Questions
1) JUN ’16 [22] Ans: 4 2) AUG ’16 [3] Ans: 2 3) AUG ’16 [6] Ans: 3 4) JAN ’17 [5] Ans: 4 5) JAN ’17 [12] Ans: 2 6) JAN ’17 [16] Ans: 2 7) JUN ’17 [5] Ans: 2 8) AUG ’17 [19] Ans: 1 9) JAN ’18 [14] Ans: 1 10) JUN ’18 [1] Ans: 2 11) JUN ’18 [14] Ans: 1 12) JAN ’19 [24] Ans: 4 13) FALL ’15 [7]
−2𝑥 + 1 = −2𝑥2 + 3𝑥 + 1 2𝑥2 − 5𝑥 = 0 𝑥(2𝑥 − 5) = 0
𝑥 = {0,5
2}
14) FALL ’15 [10] Using the intersect function on the calculator, {(–5.17,1.38), (–1.13,4.21), (1.75,3.77)}
15) JUN ’16 [33]
𝑦 = −𝑥 + 5 (𝑥 − 3)2 + (−𝑥 + 5 + 2)2 = 16 𝑥2 − 6𝑥 + 9 + 𝑥2 − 14𝑥 + 49 = 16 2𝑥2 − 20𝑥 + 42 = 0 𝑥2 − 10𝑥 + 21 = 0 (𝑥 − 3)(𝑥 − 7) = 0 𝑥 = {3,7} 𝑦 = −(3) + 5 = 2 𝑦 = −(7) + 5 = −2 (3,2) and (7, −2)
16) JUN ’16 [37] (a) 𝐴(𝑡) = 800𝑒−0.347𝑡 𝐵(𝑡) = 400𝑒−0.231𝑡
(b) 800𝑒−0.347𝑡 = 400𝑒−0.231𝑡 2𝑒−0.347𝑡 = 𝑒−0.231𝑡 ln 2𝑒−0.347𝑡 = ln 𝑒−0.231𝑡 ln 2 + ln 𝑒−0.347𝑡 = ln 𝑒−0.231𝑡 ln 2 − 0.347𝑡 = −0.231𝑡 ln 2 = 0.116𝑡 𝑡 ≈ 6 (c) 0.15 = 𝑒−0.347𝑡 ln 0.15 = ln 𝑒−0.347𝑡 ln 0.15 = −0.347𝑡 𝑡 ≈ 5.5
17) AUG ’17 [37]
( ) ( )V t Z t= when 1.95t .
At 1.95 years, the value of the car equals the loan balance. Zach can cancel the policy after 6 years because at that point the value of the car is less than $3,000.
87
18) JUN ’18 [37] 𝑃(16) = log(16 − 4) ≈ 1.1
14000
19) AUG ’18 [31] 𝑥2 + (𝑥 − 28)2 = 400 𝑥2 + 𝑥2 − 56𝑥 + 784 = 400 2𝑥2 − 56𝑥 + 384 = 0 𝑥2 − 28𝑥 + 192 = 0 (𝑥 − 12)(𝑥 − 16) = 0 𝑥 = {12,16} 𝑦 = (12) − 28 = −16 𝑦 = (16) − 28 = −12 (12, −16) and (16, −12)
88
12. SEQUENCES AND SERIES
12.1 Arithmetic and Geometric Sequences
Practice Problems
1. 𝑎𝑛 = 𝑎1 + (𝑛 − 1)𝑑 𝑎8 = 21 + (8 − 1) ∙ 9 = 84
2. 𝑑 = 3 𝑎𝑛 = 𝑎1 + (𝑛 − 1)𝑑 𝑎27 = 5 + (27 − 1) ∙ 3 = 83
3. 𝑟 = −2 𝑎𝑛 = 𝑎1𝑟𝑛−1 𝑎𝑛 = −(−2)𝑛−1
4. 𝑎𝑛 = 𝑎1𝑟𝑛−1
𝑎7 = 6 (−1
2)
7−1= 6 (
1
26) =6
64=
3
32
(Using the calculator, 𝑎7 = 0.09375.)
5. 𝑎𝑛 = 4(2.5)𝑛−1 6. 𝑎𝑛 = 5(−2)𝑛−1 𝑎15 = 5(−2)14 = 81,920
7. (6,10) and (21,55)
55 10 45
321 6 15
d−
= = =−
𝑎𝑛 = 𝑎1 + (𝑛 − 1)𝑑 10 = 𝑎1 + (6 − 1) ∙ 3 10 = 𝑎1 + 15 𝑎1 = −5 𝑎𝑛 = 𝑎1 + (𝑛 − 1)𝑑 𝑎𝑛 = −5 + (𝑛 − 1) ∙ 3 𝑎𝑛 = −5 + 3𝑛 − 3 𝑎𝑛 = 3𝑛 − 8
8. (4, −23) and (22,49)
( )49 23 72
422 4 18
d− −
= = =−
𝑎𝑛 = 𝑎1 + (𝑛 − 1)𝑑 −23 = 𝑎1 + (4 − 1) ∙ 4 −23 = 𝑎1 + 12 𝑎1 = −35 𝑎𝑛 = 𝑎1 + (𝑛 − 1)𝑑 𝑎𝑛 = −35 + (𝑛 − 1) ∙ 4 𝑎𝑛 = −35 + 4𝑛 − 4 𝑎𝑛 = 4𝑛 − 39
Regents Questions
1) JAN ’17 [14] Ans: 1 2) JUN ’17 [20] Ans: 3 3) JAN ’19 [4] Ans: 2
4) AUG ’17 [34] 6.25 2.25 4
$0.2521 5 16
−= =
− fine per
day. 2.25 − 5(0.25) = $1 replacement fee. 𝑎𝑛 = 1.25 + (𝑛 − 1)(0.25) 𝑎60 = $16
89
12.2 Recursively Defines Sequences
Practice Problems
1. 𝑎2 = 3 + 2 = 5 𝑎3 = 5 + 3 = 8 𝑎4 = 8 + 4 = 12 3, 5, 8, 12 Neither; there is no common difference
nor common ratio.
2. 𝑎2 = 2(−3) − 2 = −8 𝑎3 = 2(−8) − 3 = −19 𝑎4 = 2(−19) − 4 = −42 −3, −8, −19, −42
3. 𝑎2 = −2(3) − 1 = −7 𝑎3 = −2(−7) − 1 = 13
4. (a) 𝑎1 = 40 𝑎2 = 8
𝑎𝑛 =1
2(𝑎𝑛−2 + 𝑎𝑛−1), for 𝑛 > 2
(b) 40, 8, 24, 16, 20, 18, 19, … First odd term is 𝑎7.
Regents Questions
1) JUN ’16 [10] Ans: 4 2) JUN ’16 [23] Ans: 3 3) AUG ’16 [8] Ans: 1 4) AUG ’16 [18] Ans: 3 5) AUG ’16 [24] Ans: 4 6) AUG ’17 [24] Ans: 3 7) JAN ’18 [24] Ans: 3 8) AUG ’18 [10] Ans: 4 9) JAN ’19 [16] Ans: 4 10) SPR ’15 [11]
𝑎1 = 𝑥 + 1 𝑎2 = 𝑥(𝑥 + 1) 𝑎3 = 𝑥2(𝑥 + 1) 𝑎𝑛 = 𝑥𝑛−1(𝑥 + 1) 𝑥𝑛−1 = 0 or 𝑥 + 1 = 0 𝑥 = {0, −1}
11) JAN ’17 [34] Jillian’s plan, because distance increases by one mile each week. 𝑎1 = 10 𝑎𝑛 = 𝑎𝑛−1 + 1 𝑎𝑛 = 𝑛 + 12
12) AUG ’17 [29] : 𝑎1 = 4 𝑎𝑛 = 2𝑎𝑛−1 + 1 𝑎8 = 639
13) JUN ’18 [30] 𝑎1 = 3, 𝑎2 = 7, 𝑎3 = 15, 𝑎4 = 31 No, because there is no common
ratio: 7
3≠
15
7
90
12.3 Evaluate a Summation in Sigma Notation
Practice Problems
1. 3(2) + 3(3) + 3(4) + 3(5) = 6 + 9 + 12 + 15 = 42
2. (22 − 1) + (32 − 1) + (42 − 1) +
(52 − 1) = 3 + 8 + 15 + 24 = 50
3. −2(1) + 100 − 2(2) + 100 − 2(3) +
100 − 2(4) + 100 − 2(5) + 100 = −30 + 500 = 470
4. (23 + 2) + (24 + 2) + (25 + 2) = 10 + 18 + 34 = 62
5. (12 + 1) + (22 + 2) + (32 + 3) +
(42 + 4) + (52 + 5) = 2 + 6 + 12 + 20 + 30 = 70
6. (2 − 1)2 + (2 − 2)2 + (2 − 3)2 = 1 + 0 + 1 = 2
7. 3(2)0 + 3(2)1 + 3(2)2 = 3 + 6 + 12 = 21
8. (−32 + 3) + (−42 + 4) + (−52 + 5) = −6 − 12 − 20 = −38
9. (−14 − 1) + (−24 − 2) + (−34 − 3) = −2 − 18 − 84 = −104
10. (2 ∙ 1 + 1)0 + (2 ∙ 2 + 1)1 +
(2 ∙ 3 + 1)2 = 1 + 5 + 49 = 55
11. (4) 12. (2)
13. (4) 14. 13567(0) 294 294+ =
13567(1) 294 13861+ =
13567(2) 294 27428+ =
Sum is $41,583
91
12.4 Write a Series in Sigma Notation
Practice Problems
1.
( )
( )
11
5
1
2 1
n
k
k
dk a d
k
=
=
+ −
+
2.
( )
( )
11
6
1
4 21
n
k
k
dk a d
k
=
=
+ −
+
3.
( )
( )
( )
11
1
20
1
2 1
39 2 1
20
2 1
n
k
n
k
k
dk a d
k
k
k
k
=
=
=
+ −
−
= −
=
−
4.
( )
( )
( )
11
1
20
1
2 3
43 2 3
20
2 3
n
k
n
k
k
dk a d
k
k
k
k
=
=
=
+ −
+
= +
=
+
5.
( )11
1
15
1
7
105 7
15
7
n
k
n
k
k
dk a d
k
k
k
k
=
=
=
+ −
=
=
6.
( )
( )
( )
11
1
14
1
8 10
122 8 10
14
8 10
n
k
n
k
k
dk a d
k
k
k
k
=
=
=
+ −
+
= +
=
+
7.
( )6
1
1
2n
n
−
=
− or ( )5
0
2n
n=
−
8. (1)
92
12.5 Find the Partial Sum of a Series
Practice Problems
1. 𝑎19 = 3 + (19 − 1)(7) = 129
( )
1919 3 129
1,2542
S+
= =
2. 𝑎20 = 5 + (20 − 1)(9) = 176
( )
2020 5 176
1,8102
S+
= =
3. 𝑎30 = 15 + (30 − 1)(2) = 73
( )
3030 15 73
1,3202
S+
= =
4. 𝑎21 = 18 + (21 − 1)(2) = 58
( )
2121 18 58
7982
S+
= =
5.
( )( )
8
83 3 4
39,3211 4
S− −
= = −− −
6. This is a finite geometric series:
When 𝑛 = 1, ( )1 1
1 9 3 9a−
= − = , then
each term is –3 times the previous.
( )( )
7
79 9 3
4,9231 3
S− −
= =− −
Regents Questions
1) AUG ’16 [9] Ans: 1 2) AUG ’17 [21] Ans: 4 3) JAN ’18 [22] Ans: 2 4) AUG ’18 [13] Ans: 1 5) JUN ’16 [34]
( )33,000 33,000 1.04
1 1.04
n
nS−
=−
( )15
1533,000 33,000 1.04
1 1.04
660,778.39
S−
=−
6) JAN ’19 [29]
( )10
1015 15 1.03
171.9581 1.03
S−
= −
93
13. PROBABILITY
13.1 Theoretical and Empirical Probability
Practice Problems
1. 1
4 2.
6
10=
3
5
3. 6
22=
3
11 4.
3
6=
1
2
5. 1
6 6.
23
29
7. 6
20=
3
10 8.
13
52=
1
4
9. 5
8 10. 𝑃(𝑟𝑒𝑑) =
30
90
𝑃(𝑤ℎ𝑖𝑡𝑒) =31
90
𝑃(𝑏𝑙𝑢𝑒) =29
90
White is the most likely to be picked.
11. 2,000
80,000=
1
40 12.
8
20=
2
5
Regents Questions
1) AUG ’17 [28] Since there are six flavors, each flavor can be assigned a number, 1-6. Use the simulation to see the number of times the same number is rolled 4 times in a row. Note: the fact that there are hundreds of jelly beans in the bag minimizes the effect of selecting without replacement.
94
13.2 Probability Involving And or Or
Practice Problems
1. 6
11 2.
4
5
3. 3
8+
2
8=
5
8 4. 𝑃(𝑝𝑒𝑛) + 𝑃(𝑟𝑒𝑑) − 𝑃(𝑟𝑒𝑑 𝑝𝑒𝑛) =
6
14+
9
14−
4
14=
11
14
5. (1) 𝑃(𝐴 𝑎𝑛𝑑 𝐵) = 𝑃(𝐴 ∩ 𝐵) =
𝑃({5, 8}) =2
10=
1
5
(2) 𝑃(𝐴 𝑜𝑟 𝐵) = 𝑃(𝐴 ∪ 𝐵) =
𝑃({2, 3, 4, 5, 7, 8, 9}) =7
10
6. 𝑃(𝐴 𝑜𝑟 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 𝑎𝑛𝑑 𝐵) [add 𝑃(𝐴 𝑎𝑛𝑑 𝐵) to both sides] 𝑃(𝐴 𝑜𝑟 𝐵) + 𝑃(𝐴 𝑎𝑛𝑑 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) [subtract 𝑃(𝐴 𝑜𝑟 𝐵) from both sides] 𝑃(𝐴 𝑎𝑛𝑑 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 𝑜𝑟 𝐵)
Regents Questions
1) JUN ’16 [29] 𝑃(𝑆 𝑜𝑟 𝑀) = 𝑃(𝑆) + 𝑃(𝑀) −𝑃(𝑆 𝑎𝑛𝑑 𝑀) 974
1376=
649
1376+
433
1376− x
𝑥 =649
1376+
433
1376−
974
1376=
108
1376
95
13.3 Conditional Probability
Practice Problems
1. 1
3×
3
7=
1
7 2. (
1
2)
5=
1
32
3. 3
4×
1
2=
3
8 4. 0.95 × 0.93 × 0.98 ≈ 87%
5. 1
20×
1
19=
1
380 6.
5
7×
2
6=
5
21
7. ( )( )
( )
31035
15 1|
30 2
P S and MP M S
P S= = = =
8. If A and B are independent, then 𝑃(𝐴 𝑎𝑛𝑑 𝐵) = 𝑃(𝐴) ∙ 𝑃(𝐵). Therefore,
( )( )
( )
( )
|P Aand B
P B AP A
P A
= =
( )
( )
P B
P A
( )P B=
9. (a) 𝑃(𝐹|𝐶) =24
80=
3
10.
(b) For any pair of events A and B, we see that 𝑃(𝐴|𝐵) = 𝑃(𝐴). For example,
𝑃(𝐹|𝐶) =24
80=
3
10 and 𝑃(𝐹) =
72
240=
3
10. Therefore, these appear to be independent
events.
96
Regents Questions
1) JUN ’16 [11] Ans: 1 2) AUG ’16 [7] Ans: 1 3) JUN ’17 [14] Ans: 2 4) JUN ’18 [11] Ans: 4 5) AUG ’18 [18] Ans: 2 6) AUG ’18 [24] Ans: 4 7) JAN ’19 [12] Ans: 2 8) SPR ’15 [13]
This scenario can be modeled with a Venn Diagram:
Since 𝑃(𝑆 𝑜𝑟 𝐼̅̅ ̅̅ ̅̅ ̅) = 0.2, 𝑃(𝑆 𝑜𝑟 𝐼) = 0.8. 𝑃(𝑆 𝑎𝑛𝑑 𝐼) = 𝑃(𝑆) + 𝑃(𝐼) −𝑃(𝑆 𝑜𝑟 𝐼) = 0.5 + 0.7 − 0.8 = 0.4 If S and I are independent, then the Product Rule must be satisfied. However, (0.5)(0.7) ≠ 0.4. Therefore, salary and insurance have not been treated independently.
9) FALL ’15 [8] Based on these data, the two events do not appear to be independent.
𝑃(𝐹) =106
200= 0.53, while
𝑃(𝐹|𝑇) =54
90= 0.6,
𝑃(𝐹|𝑅) =25
65= 0.39, and
𝑃(𝐹|𝐶) =27
45= 0.6.
The probability of being female are not the same as the conditional probabilities. This suggests that the events are not independent.
10) AUG ’16 [32] 𝑃(𝐴 or 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 and 𝐵) 0.8 = 0.6 + 0.5 − 𝑃(𝐴 and 𝐵) 𝑃(𝐴 and 𝐵) = 0.3 A and B are independent since 𝑃(𝐴 and 𝐵) = 𝑃(𝐴) × 𝑃(𝐵) 0.3 = 0.6 × 0.5
11) JAN ’17 [31] No, because 𝑃(𝑀|𝑅) ≠ 𝑃(𝑀) 70
180≠
230
490
0.38 ≠ 0.47 12) JAN ’17 [35]
𝑃(𝑃|𝐾) =𝑃(𝑃 𝑎𝑛𝑑 𝐾)
𝑃(𝐾)=
0.019
0.023≈ 82.6%
A key club member has an 82.6% probability of being enrolled in AP Physics.
13) JUN ’17 [32] A student is more likely to jog if both siblings jog.
one jogs: 416
2239≈ 0.19.
both jog: 400
1780≈ 0.22.
14) AUG ’17 [26]
𝑃(𝑊|𝐷) =𝑃(𝑊 and 𝐷)
𝑃(𝐷)=
0.4
0.5≈ 0.8
15) JAN ’18 [34] 47
108=
1
4+
116
459− 𝑃(𝑀 and 𝐽)
𝑃(𝑀 and 𝐽) =31
459
No, because 31
459≠
1
4×
116
459
16) JUN ’18 [25] 103
110+103=
103
213
17) JAN ’19 [28] 𝑃(𝐴 𝑎𝑛𝑑 𝐵) = 𝑃(𝐴) ∙ 𝑃(𝐵|𝐴) =(0.8)(0.85) = 0.68
97
14. STATISTICS
14.1 Observational Studies and Experiments
Practice Problems
1. (3) 2. (1)
3. (4)
Regents Questions
1) JAN ’17 [6] Ans: 3 2) AUG ’17 [17] Ans: 2 3) AUG ’18 [2] Ans: 2 4) JUN ’16 [26]
Randomly assign participants to two groups. One group uses the toothpaste with ingredient X and the other group uses the toothpaste without ingredient X.
5) JAN ’17 [26] sample: pails of oranges; population: truckload of oranges. It is likely that about 5% of all the oranges are unsatisfactory.
98
14.2 Statistical Bias
Practice Problems
1. (3) Seniors or physics students may be
biased by aspects of class scheduling specific to their groups. Selecting only students from the cafeteria would omit students who have already chosen not to eat there.
2. (4) Allowing subjects to self-select their
participation can lead to bias. Honors calculus students may tend to
spend more (or less) time on homework due to the nature of their courses.
Surveying only teenagers at a movie theater would omit other age groups as well as people who don’t like to go to movie theaters.
3. (4) People who attend a football game are
more likely to prefer an increase in the sports budget since they are sports fans.
4. (4)
Regents Questions
1) AUG ’16 [2] Ans: 1 2) JUN ’17 [3] Ans: 3 3) JAN ‘18 [1] Ans: 4 4) JAN ’19 [10] Ans: 2
5) JUN ’18 [28] Self selection is a cause of bias because people with more free time are more likely to respond.
99
14.3 Normal Distribution
Practice Problems
1. 74 + 6 = 80 2. 85 − 2(4) = 77
3. 1
2(80 − 50) = 15 4.
1
4(92 − 78) = 3.5
5. 1
4(69 − 63) = 1.5 6. 𝑆𝐷 =
1
3(81 − 57) = 8
57 + 8 = 65 Mean = 65
(check: 81 − 2(8) = 65 ✓)
7. normalcdf(54.3, 63.5, 54.3, 4.6) ≈ 48%
You could also have used the fact that 63.5 is 2 SD above the mean, and 95% ÷ 2 ≈ 48%.
8. Interval is within 1 SD of the mean, representing about 68% of the homes.
75 × 68% = 51 homes.
9. From 56 − 2(5) to 56 + 2(5), or between 46 and 66.
10. $125,000 to $175,000
11. normalcdf(74, 82, 80, 4) ≈ 0.62 12. normalcdf(12.5, 99, 11, 1.5) ≈ 0.16 (Upper boundary can be any number
higher than the largest possible size.)
100
Regents Questions
1) JUN ’16 [9] Ans: 2 2) AUG ’16 [4] Ans: 3 3) JAN ’17 [18] Ans: 4 4) AUG ’17 [11] Ans: 1 5) JAN ’18 [7] Ans: 3 6) JUN ’18 [17] Ans: 2 7) JAN ’19 [1] Ans: 2 8) JAN ’19 [23] Ans: 1 9) FALL ’15 [16]
normalcdf(510, 540, 480, 24) ≈ 0.0994. Use z-scores to compare the two sets of data. Joanne’s score range corresponds to 1.25 to 2.5 SD above the mean:
𝑧 =510−480
24= 1.25
𝑧 =540−480
24= 2.5
Calculating equivalent scores,
1.25 =𝑥−510
20
𝑥 = 535
2.5 =𝑥−510
20
𝑥 = 560 Maria must score in the interval 535–560.
10) JUN ’17 [26] normalcdf(0, 8.25, 8, 0.5) ≈ 69%
11) AUG ’18 [28] normalcdf(200, 245, 225, 18) ≈ 0.784 1200 × 0.784 ≈ 941
101
14.4 Confidence Interval
Practice Problems
1. 5.6 − 2(0.2) and 5.6 + 2(0.2), or between 5.2 and 6.0.
Yes, 5.9 is within the margin of error.
2. a) 50% b) 68% c) 2.5%
3. a) 1
2(68%) +
1
2(95%) = 81.5%
b) 1
2(100% − 99.7%) = 0.15%
c) 0.15% × 424 = 0.636 ≈ 1 student
d) 1
2(68%) +
1
2(99.7%) = 83.85%
83.85% × 424 = 355.5 ≈ 356 students
4. We can calculate the margin of error based on the sample data:
10.2
2 2 2.04100
sME
n
= = =
Therefore, the population mean is very likely between 44.25 − 2.04 and 44.25 + 2.04, or between 42.21 and 46.29.
5.
( )( )
( )( )
ˆ ˆ12
0.40 0.602 0.098
100
p pME
n
−= =
0.40 − 0.098 ≤ 𝑝 ≤ 0.40 + 0.098 Population proportion should fall within
the interval 0.302 to 0.498.
6.
180
300= 0.6 = 60%
( )( )
( )( )
ˆ ˆ12
0.60 0.402 0.057
300
p pME
n
−= =
0.60 − 0.057 ≤ 𝑝 ≤ 0.60 + 0.057 Population proportion should fall within
the interval 0.543 to 0.657. 54.3% of 1,800 is 977.4 and 65.7% of 1,800 is 1182.6, so we can
estimate that between 977 and 1,183 students own a pet.
7. 𝐶𝐼 = 11.3095 ± 2(0.7625), so the interval is approximately 9.78 to 12.83. The claim is plausible because 10 is within this interval.
102
Regents Questions
1) JUN ’16 [7] Ans: 3 2) AUG ’16 [12] Ans: 2 3) JUN ’17 [10] Ans: 3 4) AUG ’17 [16] Ans: 2 5) AUG ’17 [22] Ans: 1 6) JAN ’18 [20] Ans: 2 7) SPR ’15 [12]
a) Yes. The margin of error = 2 × 𝑆𝐷 = 2 × 0.062 ≈ 0.12. This indicates that 95% of the observations fall within ±0.12 of 0.25. So, plausible values would fall into the interval 0.25 ± 0.12. 9 out of 50, or 0.18, falls within this interval. b) The company has evidence that the population proportion could be at least 25%. As seen in the dotplot, it can be expected to obtain a sample proportion of 0.18 or less several times, even when the population proportion is 0.25, due to sampling variability. The results do not provide enough evidence to suggest that the true proportion is not at least 0.25, so the development of the product should continue.
8) JUN ’16 [35] 0.602 ± 2(0.066) = 0.47 − 0.73 Since 0.50 falls within the 95% interval, this supports the concern there may be an even split.
9) AUG ’16 [29] Using a 95% level of confidence, �̅� ± 2 standard deviations sets the usual wait time as 150-302 seconds. 360 seconds is unusual.
10) JUN ’17 [36] 0.506 ± 2(0.078) = 0.35 − 0.66 The 32.5% value falls below the 95% confidence level.
11) JAN ’18 [35] 138.905 ± 2(7.95) = 123 − 155 No, since 125 (50% of 250) falls within the 95% interval.
12) AUG ’18 [32] 2(0.042) = 0.084 ≈ 0.08 The percent of users making in-app purchases will be within 4% of 35%.
13) JAN ’19 [35] 29.101 ± 2(0.934) = 27.23 −30.97 Yes, since 30 falls within the 95% interval.
103
14.5 Mean Difference
Practice Problems
1. a) 3.2 − 2.5 = 0.7 b) According to the graph, at least 18 of
the 100 rerandomized groups (18%) showed a mean difference of 1.0 or higher. Therefore, a mean difference of 0.7 is certainly within the 95% interval and is not statistically significant. There is no strong evidence of the success of the drug despite an observed difference between the effects on the groups in the sample.
2.
( )( ) ( )
( )( ) ( )
2 21 2
1 21 2
2 2
2
11.3 9.171.5 82.6 2
25 32
11.1 5.5
s sCI x x
n n= − + =
− +
−
Estimated population mean difference is between –16.6 and –5.6.
104
Regents Questions
1) JAN ’17 [9] Ans: 2 2) FALL ’15 [14] :
a) The mean difference between the students’ final grades in group 1 and group 2 is –3.64. This value indicates that students who met with a tutor had a mean final grade of 3.64 points less than students who used an on-line subscription. b) If the observed difference –3.64 is the result of the assignment of students to groups alone, then a difference of –3.64 or less should be observed fairly regularly in the simulation output. However, a difference of –3 or less occurs in only about 2% of the results. Therefore, it is quite unlikely that the assignment to groups alone accounts for the difference; rather, it is likely that the difference between the interventions themselves accounts for the difference between the two groups’ mean final grades.
3) AUG ’16 [36] Some of the students who did not drink energy drinks read faster than those who did drink energy drinks. 17.7 − 19.1 = −1.4. Differences of −1.4 and less occur 25
232 or about 10% of the time, so
the difference is not unusual. 4) JUN ’18 [34]
23 − 18 = 5 �̅� ± 2𝜎 = −3.07 − 3.13 Yes, a difference of 5 or more occurred three times out of a thousand, which is statistically significant.
5) AUG ’18 [36] John found the means of the scores of the two rooms and subtracted the means. The mean score for the classical room was 7 higher than the rap room (82-75). Yes, only 4 of the 250 results (1.6%) had a mean difference of 7 or more, so there is less than a 5% chance of this difference occurring due to random chance. It is likely the difference was due to the music.
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