Chapter 1
How to use these notes
These notes were prepared for the University of Utah’s Math 1050/1060 refresher course. They asssume
that the user has had the Math 1050/1060 courses College Algebra/Trigonometry (“Precalculus”) or its
equivalent (in high school or college). The notes work like this: in order to help you focus on the areas
where you need the most improvement, you will be taking a series of diagnostic quizzes. After taking a
quiz, you may check your answers against the answers found in the Answers section. You will then grade
yourself. If you have made only a few small mistakes, read the complete worked solutions to those
problems in the Solutions section. If you have made more than a few mistakes, the notes will refer you to
online lectures to brush up on the appropriate topics.
On the websites http://www.math.utah.edu/lectures/math1050.html and
http://www.math.utah.edu/lectures/math1060.html, you will find four links for each section:
1. Streaming video: to watch a video lecture right now
2. Downloadable video: to download a video lecture for watching offline
3. Pre-notes: a PowerPoint file outlining the notes, with gaps
4. Post-notes: the PowerPoint file marked up with all the work done during the lecture
The fastest, but not necessarily best, way to review the material would be to go straight to the post-notes. A
more thorough approach is to print a copy of the pre-notes, then watch the video, and fill them in. What
you choose to do will depend on how much review you need of a particular section.
1
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Chapter 2
Quizzes
2.1
2.1.1
Warmup
Some computations
3
7
1. 4 − 3
4
8
8.
43 − 2 · 5
6−4÷2
9. 24
3 7
2. 4 · 3
4 8
10. 2−4
3. 1.5 · 2.25
11. 92
4. 2.25 ÷ 1.5
1
12. 9 2
5. 6 − 3 + 2
13. 9− 2
6. 6 ÷ 3 · 2
14. −42
7. (14 − 5)2
15. (−4)2
2.1.2
1
Linear equations and inequalities
Solve for x:
1. 3x + 4 = 6 − x
5.
3+x
7
=
5
10
6.
3−x 3
=
x
5
7.
x−2 x+1
=
4
12
4.
3+x
7
≤
5
10
5.
x−2 x+1
≥
4
12
2. −3x + 4 = 6 − x
3.
9
3
=
x 14
4.
x 3
=
7 5
Solve for x, writing your answer in interval notation:
1. 3x + 4 < 6 − x
2. −3x + 4 ≥ 6 − x
3.
x 3
<
7 5
6. −3 < 2x + 5 ≤ 7
3
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Now check your answers with the answers section. Give yourself one point for each correct answer. There
are 28 questions in this section. If you got fewer than 23... Try the ones you missed again (carefully!)
WITHOUT looking at the worked solutions.
Still got fewer than 23? This material is not really 1050 material... Consider jumping over to the Math
1010 refresher, or just enrolling in Math 1010. Your math may be a little rustier than you think.
2.2
Functions
Do the following equations, tables, or graphs represent y as a function of x?
1. y = 2x − 3
2. 3x − 2y2 = −3
3. x2 + y5 = −1
√
4. x = y + 1
5.
x
−2
−1
0
1
2
y
4
1
0
1
4
x
−2
−1
0
1
2
y
4
4
4
4
4
x
−1
−1
0
1
1
y
4
1
0
1
4
8.
6.
7.
9.
Stop for a second, and check out 1.4 from the online notes if you’re getting confused.
For the next problems,
1
f (x) =
, g(x) = x2 + 1
1−x
1. Find the domain of f (x).
2. Find the domain of g(x).
3. Find and simplify f (x)/g(x), and give its
domain.
4. Find and simplify g(x)/g(x), and give its
domain.
5. Find and simplify ( f ◦ g)(x) and give its
domain.
6. Find and simplify (g ◦ f )(x) and give its
domain.
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7. Evaluate f (−2).
11. Why does g not have an inverse?
8. Evaluate g(−2).
9. Evaluate f (1/x) and write the result as a
simple fraction.
10. Find
12. Find the inverse of the function
h(x) =
f −1 (x)
2x − 3
x+5
For more help, see the online 1.4 (for the “is this a function” problems) and 1.6 (for the second group).
2.3
Sequences and Series
Find the missing terms of the following sequences:
1. 1, 4, 7, ,
3. 1, −1/2, 1/4, −1/8, ,
2. 3, 6, 12, ,
Find a formula for the nth term of the following sequences:
1. 1, 4, 7, . . .
3. 1, −1/2, 1/4, −1/8, . . .
2. 3, 6, 12, . . .
Find the sum of the following series:
1. 1 + 4 + 7 + · · · + 31
3. 1 − 1/2 + 1/4 − 1/8 + · · · − 1/512
2. 3 + 6 + 12 + · · · + 192
Evaluate the factorial expressions:
8!
6!
9!
2.
5!4!
1.
3.
(n + 2)!
(n − 1)!
Solve the following counting problems:
1. How many flags are possible to make from red, white, and blue vertical, equal-width stripes.
2. How many ways are there to make four horizontal stripes using four colors chosen at random from a
box of eight crayons?
3. I have six pairs of pants and five shirts. How many combinations of these can I make?
4. How many ways are there to order a three-topping pizza from a menu with ten possible toppings?
There are sixteen problems in this part. I hope you got more than 12 right. If not, check out:
1. 6.1, 6.2, 6.3 for the first three parts.
2. 6.5 for the last two parts.
2.4
Translation and reflection of graphs
Sketch the graphs of the following:
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6
1. y = −(x + 2)2 − 3
4. y = (x − 2)3 − 3
2. y = (x − 3)2 + 5
5. y = −(x + 2)3
3. y = |x − 1| + 4
6. y =
−1
x+3
In the online lectures, this is about halfway through 1.5.
2.5
Polynomials
2.5.1
Multiplying
Multiply out and write in standard form:
1. (x − 1)2 (x + 2)
4. (5x + 3)(x2 − 3x − 4)
2. (x − 2)3
5. (x − y)5 (binomial theorem)
3. 3(x − 2)(x + 3)
6. (x + 2a)4 (binomial theorem)
2.5.2
Factoring
Factor as completely as possible:
1. x2 + 7xy − 144y2
4. 36x3 + 12x2 − 48x
2. 10x2 + 6xy − 25xy − 15y2
5. x3 + 3x2 − x − 3
3. 3x3 y2 − 75xy2
6. x4 − y4
This material is technically 1010. Hope you did OK! Check out sections 5.4-5.5 from the notes at
http://www.math.utah.edu/lectures/math1010.html if you need help.
2.5.3
Polynomial equations
Solve for x:
1. x2 − 5x = 6
2. x2 − 2x = 0
3. x2 −
1
=0
4
9. x4 − 2x2 + 1 = 0
√
10. x − 8 x + 7 = 0
11. x3 − 5x2 = 6x
4. x2 + 16 = 0
12. x4 − 4x2 = 0
5. 3x2 + 6x = 24
13. (x − 2)2 = 7
6. 2x2 − 3x − 2 = 0
14. x2 − 2x − 5 = 0
7. 6x2 = 7x + 3
15. x2 + 7x + 3 = 0
8. 12x2 − 13x + 3 = 0
16. x2 + 2x + 3 = 0
You need to be doing these by factoring or completing the square. For more help with completing the
square, check Extras: How to complete the square.
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Also check out sections 8A-D from the notes at
http://www.math.utah.edu/lectures/math1010.html.
2.5.4
Finding roots of an equation with a given root
In these problems, one root of a polynomial is given. Use it to find all the other roots.
1. 2x3 − 5x2 + x + 2 = 0, x = 2 is a root.
2. 2x3 − 3x2 − 11x + 6 = 0, x = −2 is a root.
3
3. 12x3 + 16x2 − 5x − 3 = 0, x = − is a root.
2
1
4. 3x3 + 7x2 − 22x − 8 = 0, x = − is a root.
3
The technique of synthetic division may be useful. For a quick refresher, look at the explanations in these
notes. For further explanation, see 2.3 in the online lectures.
2.5.5
Long division
Divide these polynomials, writing the remainders.
1. (6x3 + 7x2 + 12x − 5) ÷ (3x − 1)
2.
3x2 − 2x + 5
x−3
3.
4x4 − 4x2 + 6x
x−4
2.5.6
Graphs of polynomial functions
1. Write y = 2(x − 1)2 + 4 in the form
y = ax2 + bx + c.
2. Write y = 2x2 + 4x + 5 in the form
y = a(x − h)2 + k.
5. Find the equation of a parabola with vertex
(−1, 3) with y-intercept (0, −2).
6. Find the equation of a parabola with roots −2
and 3.
3. Find the coordinates of the vertex of the
parabola y = 2(x − 1)2 + 4
7. Graph y = 2(x − 1)2 + 4.
4. Find the coordinates of the vertex of the
parabola y = 2x2 + 4x + 5.
8. Graph y = 2x2 + 4x + 5.
Check out 2.1 in the notes for more about graphing.
2.5.7
Graphs of higher-degree polynomial functions
Match the equations with their graphs.
1. y = −x4 + x2
2. y = x3 − 4x2
3. y = (x − 3)2
4. y = −x3 − x2 + 5x − 3
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1.
3.
2.
4.
2.6
Rational expressions
Simplify the following rational expressions as completely as possible:
2x2 + 5x − 3
x2 + 2x − 3
4.
4x − 48
x2 − 144
−4x3 y2 −4xy3
2.
·
2x2 y3 10xy2
5.
x2 − 9 x2 + 2x + 1
·
2x + 2 (x − 3)(x + 1)
6.
x2 + x − 2 x2 + 4x − 5 x2 + 2x − 15
÷
· 2
3x2 − 9x
6x − 18
x −x−6
1.
3.
3
3x
+
x − 2 2x + 3
If you need help with these, go to the notes at
http://www.math.utah.edu/lectures/math1010.html. The sections are 6.2-6.3.
2.6.1
Complex fractions
It’s not considered good mathematical writing to have fractions living inside of other fractions. Such
expressions are called complex fractions. Have no fear: a fraction bar signifies division, so one fraction
over another can be rewritten as a division of fractions problem.
Write these as simple fractions:
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9
x+5
2
1. 23x
x − 25
6x3
1
4+
2
2.
1 1
+
3 6
5
3
+
3. x − 5 x + 3
5
3
+
x+3 x−5
If you need help with these, go to the notes at
http://www.math.utah.edu/lectures/math1010.html. The section is 6.4.
2.6.2
Graphs of rational functions
Sketch graphs of the following rational functions:
1. y =
x+2
x−3
3. y =
2. y =
x2 − 4x + 3
x2 + x − 2
4. y =
x−1
x2 − x − 2
x2 + 3x + 2
x−1
The section 2.6 of the video lectures has a detailed example and outlines the general procedure.
2.7
Exponential and Logarithmic functions
Evaluate:
2. 8−2/3
3/2
4
6.
9
3. (−8)2/3
7. log2 (1/2)
4. (−8)−2/3
8. log3 27
5. −82/3
9. log9 3
1. 82/3
Look at the notes in sections 3.1-3.2 for help.
Solve:
1. log x = −3
2. log2 x = −
5. log3 9x = 3
1
2
6. e2x − 3ex = 28
3. e2x − 2 = 0
7. log(x + 1) − log(x − 1) = log 3
4. 8 − 3 · 20.5x = −40
8. ln(x2 − 4) − ln(x + 2) = ln(3 − x)
Look at the notes in sections 3.4-3.5 for help.
Combine the logarithms into a single logarithm:
1.
1
log x + log2 y
2 2
Expand the single logarithm as fully as possible:
2. 3 log x − 2 log y − log z
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√ x y
1. log
z
2. log4 4x2 y3
Look at the notes in section 3.3 for help.
2.8
Matrices and Linear Algebra
NOTE: If you’re coming in to this section and you haven’t heard of matrices before, you should start
directly with the video lectures in Chapter 5. This might be the case, for example, if you took the 1050
equivalent in high school, it may not have been covered. If the material is new to you, it is a lot to learn in a
couple of hours. Don’t worry too much: matrices are a useful tool for doing something you probably
already know how to do another way. You may not even really see them again until Calc III.
Multiply the matrices, or indicate why the product cannot be formed.
1.

9 −3
1 −3 0 6


 0 2 1 0   0 −2 
 0 5 
2 0 4 −4
2 0



3.
1 3
4 2
−2 3
1 −2
−2 3
1 −2
4.
2.
3 −2 3
0 1 −1
1 0 −7
6 11 −2
1 3
4 2
Solve the following linear systems (is there a way to save time on the first four?):
1.
5.
2x1 + 3x2 = 1
x+z = 1
x1 + 4x2 = 2
2x + 2y + 4z = 2
2x − 2y + 8z = 6
2.
2x1 + 3x2 = 3
x1 + 4x2 = 4
6.
3x1 + x2 + 4x3 = 0
3.
2x1 + 3x2 = 5
9x1 + 5x2 + 16x3 = 0
x1 + 4x2 = 6
21x1 + 11x2 + 36x3 = 0
4.
2x1 + 3x2 = 7
x1 + 4x2 = 8
Find the inverses of the following matrices, if possible:
1.
3.
4 −3
−3 2
4 −2
−2 1
2.
4.


1 2 3
 2 3 5 
0 1 2


1 0 2
 3 3 −4 
6 3 2
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Compute the following determinants:
1.
3.
0 1
2 4
4.
2.
2.9

1 2 3
 2 3 5 
0 1 2

−1 −5
−3 −12


1 0 2
 3 3 −4 
6 3 2
Measuring and describing angles
1. Convert to radian measure: a) 30◦ ; b) 150◦ .
2. Convert to degree measure: a)
7π
3π
; b) − .
2
6
3. Find a positive and a negative angle coterminal to a) 390◦ ; b)
17π
.
6
π
4. Find a) the complement of 72◦ ; b) the supplement of 72◦ ; c) the complement of ; d) the
12
5π
supplement of
.
6
Check out section 1.1 from the trigonometry lectures on
http://www.math.utah.edu/lectures/1060.html. This will be where all the lectures come from for
the trigonometry sections.
2.10
Trigonometric functions
Evaluate:
1. sin(45◦ )
6. sin π/6
2. cos(60◦ )
7. cos(−5π/3)
3. sec(135◦ )
8. tan(3π/4)
4. tan(300◦ )
9. csc(4π/3)
5. cot(0◦ )
10. sec π
1. θ is acute and sin θ = 0.6. Find a) cos θ , b) tan θ .
2. tan θ =
−15
and sin θ < 0. Find a) sin θ ; b) cos θ .
8
3. sin θ =
4
and θ lies in the second quadrant. Find a) cos θ , b) tan θ .
5
Check the notes from 1.2, 1.4 for help.
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2.11
12
Inverse trigonometric functions
Evaluate (give your answer in radians):
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1. sin−1 1
2. tan−1 1
√
3. cos−1 (− 3/2)
4. sin−1 (−1/2)
√
5. tan−1 (− 3)
√
6. cos−1 (− 2/2)
The notes for this material are section 1.7.
2.12
Trigonometric Equations
Solve the following trigonometric equations, finding solutions between 0 and 2π.
1. sin x = cos x
3. 2 cos2 x + cos x − 1 = 0
2. 3 sin x − 2 = 5 sin x − 1
4. 2 cos2 x + 3 sin x = 0
2.13
Graphs of trigonometric functions
1. What is the amplitude of the function
y = 3 cos x?
3. What is the period of the function
y = tan(x/2)?
2. What is the period of the function
y = sin(πx)?
4. What is the phase shift of the function
y = cos(2x − π/2)?
Sketch the graphs of the following trigonometric functions:
1. y = 2 sin(x − π/2)
3. y = tan(x − π/4)
2. y = cos(2x + π/2)
4. y = 3 cos(πx − π) + 2
Check out 1.5-1.6 for some tips on graphing trig functions.
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Chapter 3
Answers
3.1
Warmup
3.1.1
1.
Some operations
7
8
9. 16
1
16
13
32
3. 3.375
10.
4. 1.5
12. 3
2. 18
11. 81
5. 5
13.
6. 4
1
3
7. 81
14. -16
1 54
8. 13.5 (= 13 = )
2
4
15. 16
3.1.2
Linear equations and inequalities
1.
1
2
2. −1
5.
1
2
14
3.
3
21
4.
5
6.
15
8
7.
7
2
1. x ∈ (−∞, 12 )
4. x ∈ (−∞, 12 ]
2. x ∈ [−1, ∞)
5. x ∈ [ 27 , ∞)
3. x ∈ (−∞, 21
5)
6. x ∈ (−4, 1]
3.2
Functions
15
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1. Yes.
6. Yes.
2. No.
7. No.
3. Yes.
8. No.
4. Yes.
9. Yes.
5. Yes.
For the next problems,
f (x) =
1. x 6= 1
2. all real numbers
3.
1
1
=
, x 6= 1
2
3
1−x+x −x
(1 − x)(x2 + 1)
4. (x2 + 1)(1 − x), x 6= 1
5. −
6.
3.3
1
, x 6= 0
x2
1
, x 6= 1
(1 − x)2
1
, g(x) = x2 + 1
1−x
1
7.
3
8. 5
9.
x
x−1
10.
x−1
x
11. For example, g(1) = g(−1) = 2, so there can
be no inverse at 2.
12.
5x + 3
2−x
Sequences and Series
1. 10, 13.
3. 1/16, −1/32.
2. 24, 48.
1. an = 1 + 3(n − 1).
3. an = − 21
n−1
.
2. an = 3 · 2n−1 .
1. 176
3.
343
512
2. 381
Evaluate the factorial expressions:
1. 56
2. 126
Solve the following counting problems:
1. 6
2. 1680
3. 30
4. 120
3. (n + 2)(n + 1)n
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3.4
17
Translation and reflection of graphs
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18
1.
4.
2.
5.
3.
6.
3.5
Polynomials
3.5.1
Multiplying
1. 5x2 − 12x + 4
3. x3 − 6x2 − 4x − 8
2. x3 − 3x + 2
4. 3x2 + 3x − 18
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5. 5x3 − 12x2 − 29x − 12
7. x5 − 5x4 y + 10x3 y2 − 10x2 y3 + 5xy4 − y5
6. 45 − 108i
8. x4 + 8x3 a + 24x2 a2 + 32xa3 + 16a4
3.5.2
Factoring
1. (x + 16y)(x − 9y)
4. 12x(x − 1)(3x + 4)
2. (2x − 5y)(5x + 3y)
5. (x + 3)(x + 1)(x − 1)
3. 3xy2 (x + 5)(x − 5)
6. (x2 + y2 )(x + y)(x − y)
3.5.3
Polynomial equations
1. -1,6
9. -1,1
2. 0,2
10. 1,49
1
3. ±
2
4. no real solution
11. -1,0,6
12. -2,0,2
√
13. 2 ± 7
√
14. 1 ± 6
√
−7 ± 37
15.
2
5. -4,2
1
6. − , 2
2
1 3
7. − ,
3 2
1 3
8. ,
3 4
3.5.4
16. no real solution
Finding roots of an equation with a given root
1
1. − and 1.
2
2.
1
and 3.
2
1
1
3. − and
3
3
4. −4 and 2
3.5.5
Long division
1. 2x2 + 3x + 5
2. 3x + 7 +
26
x−3
3. 4x3 + 16x2 + 60x + 246 +
3.5.6
984
x−4
Graphs of polynomial functions
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20
1. 2x2 − 4x + 6
2. 2(x + 1)2 + 3
3. (1, 4)
4. (−1, 3)
5. y = −5(x + 1) + 3
6. y = (x + 2)(x − 3) = x2 − x − 6. Any nonzero
multiple of this is also OK.
8.
7.
3.5.7
Graphs of higher-degree polynomial functions
1. Graph 2
2. Graph 3
3. Graph 1
4. Graph 4
3.6
Rational expressions
1.
2x − 1
, x 6= −3
x−1
4.
4
, x 6= 12
x + 12
2.
4x
, x, y, 6= 0
5
5.
(x + 3)
, x 6= −1, 1, 3
2
3.
3(x2 + 3)
(2x + 3)(x − 2)
6.
2
, x 6= −5, −2, 1, 3
x
3.6.1
1.
Complex fractions
2x
, x 6= −5, 0
x−5
2. 9
3.
3.6.2
Graphs of rational functions
x
, x 6= −3, 5
x−2
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21
1.
3.
2.
4.
3.7
Exponential and Logarithmic functions
1. 4
1
2.
4
3. 4
1
4.
4
5. -4
1. 0.001
6.
8
27
7. -1
8. 3
9.
1
2
5. 3
2.
√1
2
6. ln 7
3.
√
1
ln 2 = ln 2
2
7. 2
5
2
4. 8
8.
√
1. log2 y x
2. log
x3
y2 z
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22
1
1. log x + log y − log z
2
3.8
2. 1 + 2 log4 x + 3 log4 y
Matrices and Linear Algebra
1.
3.


21 3
 0 1 
10 14
4.
2. not possible
1 −3
−6 8
10 0
−7 −1
1. x1 = −2/5, x2 = 1
4. x1 = 4/5, x2 = 23/5
2. x1 = 0, x2 = 11/5
5. x = 5, y = −2, z = −1
3. x1 = 2/5, x2 = 17/5
6. x = −2t, y = −6t, z = 3t for any real number t
1.
3.
−2 −3
−3 −4


−1 1 −1
 4 −2 −1 
−2 1
1
2. No inverse.
4. No inverse
1. 2
3. -1
2. -3
4. 0
3.9
Measuring and describing angles
1. a) π/6; b) 5π/6.
2. a) 270◦ ; b) −210◦ .
3. a) 30◦ , −330◦ ; b) 5π/6, −7π/6
4. a) 18◦ ; b) 108◦ ; c) 5π/12; d) π/6.
3.10
Trigonometric functions
1.
√
2/2
6.
1
2
2.
1
2
7.
1
2
√
3. − 2
√
4. − 3
5. not defined
1. a) 0.8; b) 43 .
8. −1
9.
√
2 3
3
10. −1
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2. a) − 15
17 ; b)
8
17 .
3. a) − 35 ; b) − 43 .
3.10.1
Inverse trigonometric functions
1. π/2
4. −π/6
2. π/4
5. −π/3
3. 5π/6
6. 3π/4
3.11
Trigonometric Equations
1. π/4, 5π/4
3. π/3, 5π/3, π
2. π/6, 5π/6
4. 7π/6, 11π/6
3.12
Graphs of trigonometric functions
1. 3
3. 2π
2. 2
4. π/4
1.
3.
2.
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4.
24
Chapter 4
Explanations
4.1
4.1.1
Warmup
Some operations
1. This is how I would do it. First you need a common denominator
3
7
6
7
4 −3 = 4 −3
4
8
8
8
Now exchange a “1” from the 4 for 8 eighths:
=3
14
7
−3
8
8
Now the 3s go, and
14 7 7
− = .
8
8 8
Here, changing to “improper fractions” also works fine, but is more work.
=
21
, be careful: you can’t just multiply the whole number and fraction parts separately.
32
1 1
Don’t believe me? Try that on 1 · 1 and see if your answer seems reasonable. Here, we better use
2 2
improper fractions:
2. If you got 12
3 7
4 ·3
4 8
=
=
=
19 31
·
4 8
19 · 31
4·8
589
32
3. Remember the “rule”? Multiply, ignoring the decimal points, count the number of places total after
the decimal in the question, and place the decimal point that many digits from the right. It works,
because
1.5 · 2.25 =
=
=
25
15 225
·
10 100
15 · 225
10 · 100
3375
= 3.375
1000
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4. This is equivalent to 22.5 ÷ 15. General comment: Many people learn the order of operations by the
mnemonic PEMDAS (“please excuse my dear Aunt Sally”) – parentheses, exponentiation,
multiplication, division, addition, subtraction. If you do this, remember that “MD” are really done
together, left to right, when they occur. You don’t do “M” before “D”, just before “A” and “S”. Same
for “AS”.
5. If you got 1, read what I wrote above.
6. If you got 1, read what I wrote above.
7. Parens tell you what to do before squaring.
8. Evaluate the whole numerator and whole denominator (with proper order of operations) first. Then
reduce the resulting fraction. No cancelling until you’ve done all the addition and subtraction!
9. No comment.
10. Negative exponents don’t make the result negative. They mean to take the reciprocal before applying
the corresponding positive power. This is done to make the rules of exponents consistent (ask me!).
So
1
1
2−4 = 4 =
2
16
11. No comment.
1
12. Fractional exponents correspond to roots: 9 2 =
√
9 = 3.
13. Combination of exponent rules.
14. Exponents have higher precedence than the minus sign! So we raise 4 to the second power first, then
take the opposite of that.
15. Here the parentheses tell us to take the opposite of four and raise that to the second power.
4.1.2
Linear equations and inequalities
General tips: If two things are equal, and we do the same thing to both of them, they stay equal. This
justifies “adding the same thing to both sides”, and so on.
A good strategy, when trying to get all the expressions with x on the same side is to move so that the
coefficient of x is positive. Fewer arithmetic errors are made when dividing by positive numbers.
1.
3x + 4 = 6 − x ⇒ 4x = 2
1
⇒ x=
2
2.
−3x + 4 = 6 − x ⇒ −2 = 2x
⇒ x = −1
3.
3
9
=
x 14
⇒ 9x = 3 · 14
⇒ x=
3 · 14 14
=
9
3
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4.
x 3
=
7 5
⇒ 5x = 3 · 7
⇒ x=
21
5
5.
7
3+x
=
5
10
⇒ 10(3 + x) = 5 · 7
⇒ 30 + 10x = 35
⇒ 10x = 5
1
⇒ x=
2
6.
3−x 3
=
x
5
⇒ 5(3 − x) = 3x
⇒ 15 − 5x = 3x
⇒ 15 = 8x
8
⇒ x=
15
7.
x−2 x+1
=
4
12
⇒ 12(x − 2) = 4(x + 1)
⇒ 12x − 24 = 4x + 4
⇒ 8x = 28
28 7
=
⇒ x=
8
2
1.
3x + 4 < 6 − x ⇒ 4x < 2
1
1
⇒ x < .x ∈ (−∞, )
2
2
2.
−3x + 4 ≥ 6 − x ⇒ −2 ≥ 2x
⇒ −1 ≥ x.x ∈ [−1, ∞)
3.
x 3
<
7 5
⇒ 5x < 21
⇒ x<
21
21
.x ∈ (−∞, )
5
5
4.
3+x
7
≤
5
10
⇒ 10(3 + x) ≤ 5 · 7
⇒ 30 + 10x ≤ 35
⇒ 10x ≤ 5
1
1
⇒ x ≤ .x ∈ (−∞, ]
2
2
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5.
x−2 x+1
≥
4
12
⇒ 12(x − 2) ≥ 4(x + 1)
⇒ 12x − 24 ≥ 4x + 4
⇒ 8x ≥ 28
7
28
⇒ x ≥ .x ∈ [ , ∞)
8
2
6.
−3 < 2x + 5 ≤ 7 ⇒ −8 < 2x ≤ 2
⇒ −4 < x ≤ 1.x ∈ (−4, 1]
4.2
Functions
1. Yes. This is already solved for y, so a value of x uniquely determines y.
2. No. If we try to solve for y, we end up taking a square root, which could have positive or negative
values. So y is not uniquely determined.
3. Yes. Odd degree roots are unique.
4. Yes. Here we can solve for y. We square both sides at some point, but that does not cause a problem.
5. Yes. Each x value appears no more than once.
6. Yes. Each x value appears no more than once.
7. No. x values appear multiple times.
8. No. Fails the vertical line test.
9. Yes. Passes the vertical line test.
4.3
Sequences and Series
1. 1, 4, 7, , . The pattern is that each term is three greater than the previous term. A sequence that has
a common difference between subsequent terms is called an arithmetic sequence. The next two terms
of this sequence are thus 10 and 13.
2. 3, 6, 12, , . The pattern is that each term is twice the previous term. A sequence that has a common
ratio between subsequent terms is called a geometric sequence. The next two terms of this sequence
are thus 24 and 48.
3. 1, −1/2, 1/4, −1/8, , . This sequence is also geometric. Each term is −1/2 of the previous term.
Thus the next two terms are 1/16 and −1/32.
1. This sequence starts from 1, and each term is three greater than the last. So by time we get to the nth
term, we have n − 1 times added three. So an = 1 + 3(n − 1).
2. This sequence starts from 3, and each term is double the last. So by time we get to the nth term, we
have doubled n − 1 times. So an = 3 · 2n−1 .
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29
3. Same idea: we start from 1, and multiply by −1/2 to get to subsequent terms. an = − 12
that this even works for n = 1, since raising a number to the zeroth power yields 1.
n−1
. Note
1. 1 + 4 + 7 + · · · + 31. We need to know how many terms there are here: the first and last terms differ
by 30, so there are 11 terms here (the first term and 30/3 = 10 more). If we write:
1 + 4 + 7+ · · · +31
31 + 28 + 25+ · · · +1
we have two copies of the series. Note that the numbers in columns always add up to 32. So we have
eleven copies of 32 when we take the series twice. So the sum is 11 · 32/2 = 176.
2. The reasoning for the sum of geometric series is somewhat complicated. It results in the formula:
a + ar + ar2 + · · · + arn−1 = a
rn − 1
r−1
Here a is the first term, r is the common ratio, and n is the number of terms. For our sequence, we
get to 192, starting from 3 and doubling. 192/3 = 64, and 64 = 26 , so our sequence has 6 + 1 = 7
terms (remember we didn’t multiply to get the first term). So the sum is:
3
27 − 1
127
=3
= 381
2−1
1
3. 1 − 1/2 + 1/4 − 1/8 + · · · − 1/512 Here the common ratio is −1/2, and since 512 = 29 , there are 10
terms. The sum is:
(−1/2)1 0 − 1 (1/1024) − 1 1023 · 2 343
1
=
=
=
(−1/2) − 1
−3/2
3 · 1024 512
1. The factors 1-6 cancel out:
8! 8 · 7 · 6!
=
= 8 · 7 = 56
6!
6!
2.
9 · 8 · 7 · 6 · 5! 9 · 2 · 7
9!
=
=
126
5!4!
5!4 · 3 · 2 · 1
=
3.
(n + 2)! (n + 2)(n + 1)n(n − 1)!
=
= (n + 2)(n + 1)n
(n − 1)!
(n − 1)!
1. Order matters, three choices for the first, two for the second, one for the third. 6.
2. 8 · 7 · 6 · 5 = 1680
3. 6 · 5 = 30
4. Ten choices for the first, nine choices for the second, eight choices for the third. But order doesn’t
matter, so you need to divide by the number of ways of reordering three toppings:
10 · 9 · 8
= 120
3·2·1
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4.4
30
Translation and reflection of graphs
Since this is a refresher, only the final results will be presented here.
If we start with the basic graph of a function y = f (x), it may be modified:
1. y = f (x) + a is the same graph translated up by a units (the net translation is “down” when a is
negative). Adding a after applying the function adds a to y.
2. y = − f (x) is the same graph reflected over the x-axis. Applying the negative sign after the function
changes all function values (y-values) to their opposites.
3. y = f (x + a) is the graph translated left by a units (the net translation is “right” when a is negative).
The rough idea is that to solve for x, we would undo f , and then subtract a.
4. y = f (−x) is the graph reflected over the y-axis. It is the x-values whose opposites are being taken.
In addition to these basic rules, it is never a bad idea to check your work by plugging a point or two from
your graph into the original equation.
4.5
Polynomials
Certain expressions we meet a lot. Since we meet them a lot, there is a lot of terminology for them.
A polynomial is an expression of the form:
an xn + an−1 xn−1 + · · · + a1 x + a0
where the an are real numbers, and n is a whole number.
Some terminology:
1. n is called the degree of the polynomial (note that n is the highest power occuring on an x if the terms
happen not to be written in descending order, as here).
2. The numbers ai are called coefficients.
3. The number an is called the leading coefficient.
4. The number a0 is called the constant term.
When
• n = 0, the polynomial is called constant.
• n = 1, the polynomial is called linear.
• there is a total of one term, the polynomial is called a monomial.
• there is a total of two terms, the polynomial is called a binomial.
• there is a total of three terms, the polynomials is called a trinomial.
Once we enter the world of polynomial expressions, there is one more thing we can do with expressions:
• factor: write an expression as the product of two other expressions.
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31
Factoring out the greatest monomial factor: here we try to find the factors that are common to every term in
an expression, both those made of variables, and not.
Examples:
1. 2x2 + 6x − 4 = 2(x2 + 3x − 2)
2. x3 y2 − x2 y + x4 y3 = x2 y(xy − 1 + x2 y2 )
3. 4x(x + 2) − 3(x + 2) = (x + 2)(4x − 3)
This is the first thing to try when factoring anything.
4.5.1
Multiplying
1.
(x − 2)(5x − 2) = 5x2 − 2x − 10x + 4 = 5x2 − 12x + 4)
2.
(x − 1)2 (x + 2) = (x2 − 2x + 1)(x + 2)
= x3 + 2x2 − 2x2 − 4x + x + 2
= x3 − 3x + 2
3.
(x − 2)3 = (x2 − 4x + 4)(x − 2)
= x3 − 2x2 − 4x2 − 8x + 4x − 8
= x3 − 6x2 − 4x − 8
4.
3(x − 2)(x + 3) = 3(x2 + 3x − 2x − 6) = 3(x2 + x − 6) = 3x2 + 3x − 18
5.
(5x + 3)(x2 − 3x − 4) = 5x3 + 3x2 − 15x2 − 20x − 9x − 12 = 5x3 − 12x2 − 29x − 12
6.
(9 − 6i)2 = (9 − 6i)(9 − 6i)
= 81 − 54i − 54i + 36i2
= 81 − 108i − 36
= 45 − 108i
7. The binomial theorem tells us that the coefficients of the terms in the expansion of a binomial raised
to the nth power are the numbers in the nth row of Pascal’s triangle. So
(x − y)5 = 1x5 (−y)0 + 5x4 (−y)1 + 10x3 (−y)2 + 10x2 (−y)3 + 5x1 (−y)4 + 1x0 (−y)5
= x5 − 5x4 y + 10x3 y2 − 10x2 y3 + 5xy4 − y5
8.
(x + 2a)4 = 1x4 (2a)0 + 4x3 (2a)1 + 6x2 (2a)2 + 4x1 (2a)3 + 1x0 (2a)4
= x4 + 8x3 a + 24x2 a2 + 32xa3 + 16a4
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4.5.2
32
Factoring
1. 144 = 16 · 9. Factors are probably like (x+?y)...
x2 + 7xy − 144y2 = (x + 16y)(x − 9y)
2.
10x2 + 6xy − 25xy − 15y2 = 2x(5x + 3y) − 5y(5x + 3y)
= (5x + 3y)(2x − 5y)
3.
3x3 y2 − 75xy2 = 3xy2 (x2 − 25) = 3xy2 (x + 5)(x − 5)
4.
36x3 + 12x2 − 48x = 12x(3x2 + x − 4)
= 12x(3x2 − 3x + 4x − 4)
= 12x[3x(x − 1) + 4(x − 1)]
= 12x(x − 1)(3x + 4)
5.
x3 + 3x2 − x − 3 = x2 (x + 3) − 1(x + 3)
= (x + 3)(x2 − 1)
= (x + 3)(x + 1)(x − 1)
6.
x4 − y4 = (x2 )2 − (y2 )2
= (x2 + y2 )(x2 − y2 )
= (x2 + y2 )(x + y)(x − y)
4.5.3
Polynomial equations
When the same variable occurs in the same equation to different powers, there are basically no valid
algebraic operations that will allow you to isolate the variable. As an alternative, in this case, we move all
the terms on to on side of the equation, leaving the other side zero. If the resulting expression is factorable,
we can use the zero property of multiplication: if a number of factors are multiplied and the result is zero,
then one of those factors had to be zero itself.
Some polynomial equations do not require this. Using some clever manipulations (one of which is
completing the square), these equations can be solved by taking roots.
1.
x2 − 5x = 6 ⇒ x− 5x − 6 = 0 ⇒ (x − 6)(x + 1) = 0 ⇒ x = −1, 6
2.
x2 − 2x = 0 ⇒ x(x − 2) = 0 ⇒ x = 0, 2
3.
r
1
1
1
1
2
⇒x=±
x − =0⇒x = ⇒x=±
4
4
4
2
2
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4.
x2 + 16 = 0 ⇒ x2 = −16
No real number squares to a negative. This equation has complex solutions x = ±4i.
5.
3x2 + 6x = 24 ⇒ 3x2 + 6x − 24 = 0
⇒ 3(x2 + 2x − 8) = 0
⇒ 3(x + 4)(x − 2) = 0
⇒ x = −4, 2
6.
2x2 − 3x − 2 = 0 ⇒ 2x2 − 4x + x − 2 = 0
⇒ 2x(x − 2) + 1(x − 2) = 0
⇒ (x − 2)(2x + 1) = 0
1
⇒ x = − ,2
2
7.
6x2 = 7x + 3 ⇒ 6x2 − 7x − 3 = 0
⇒ 6x2 − 9x + 2x − 3 = 0
⇒ 3x(2x − 3) + 1(2x − 3) = 0
⇒ (2x − 3)(3x + 1) = 0
3 1
⇒ x = ,−
2 3
8.
12x2 − 13x + 3 = 0 ⇒ 12x2 − 9x − 4x + 3 = 0
⇒ 3x(4x − 3) − 1(4x − 3) = 0
⇒ (4x − 3)(3x − 1) = 0
3 1
⇒ x= ,
4 3
9.
x4 − 2x2 + 1 = 0 ⇒ (x2 )2 − 2(x2 ) + 1 = 0
⇒ (x2 − 1)2 = 0
⇒ x2 − 1 = 0
⇒ x = ±1
10.
√
√
√
x − 8 x + 7 = 0 ⇒ ( x)2 − 8( x) + 7 = 0
√
√
⇒ ( x − 1)( x − 7) = 0
√
√
⇒
x = 1 or x = 7
⇒ x = 1, 49
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34
11.
x3 − 5x2 = 6x ⇒ x3 − 5x2 − 6x = 0
⇒ x(x2 − 5x − 6) = 0
⇒ x(x − 6)(x + 1) = 0
⇒ x = −1, 0, 6
12.
x4 − 4x2 = 0 ⇒ x2 (x2 − 4) = 0 ⇒ x = −2, 0, 2
13.
√
√
(x − 2)2 = 7 ⇒ x − 2 = ± 7 ⇒ x = 2 ± 7
14. Think: x2 − 2x + 1 = (x − 1)2
x2 − 2x − 5 = 0 ⇒ (x − 1)2 − 6 = 0
⇒ (x − 1)2 = 6
√
⇒ x−1 = ± 6
√
⇒ x = 1± 6
15. Quadratic formula
√
−7 ± 72 − 4 · 1 · 3
x + 7x + 3 = 0 ⇒ x =
√2 · 1
−7 ± 37
⇒ x=
2
2
16. For x2 + 2x + 3 = 0, the discriminant 22 − 4 · 1 · 3 = −8 < 0, so the quadratic equation gives no real
solutions.
4.5.4
Finding roots of an equation with a given root
1.
2
2
2
−5
1
2
4
−2
−2
−1
−1
0
This means:
2x3 − 5x2 + x + 2 = (x − 2)(2x2 − x − 1)
Now:
2x2 − x − 1 = 2x2 − 2x + x − 1
= 2x(x − 1) + 1(x − 1)
= (2x + 1)(x − 1)
1
So the other roots are − and 1.
2
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35
2.
2
−2
2
−3
− 11
6
−4
14
−6
−7
3
0
This means:
2x3 − 3x2 − 11x + 6 = (x + 2)(2x2 − 7x + 3)
Now:
2x2 − 7x + 3 = 2x2 − 6x − x + 3
= 2x(x − 3) − 1(x − 3)
= (2x − 1)(x − 3)
So the other roots are
1
and 3.
2
3.
12
−
3
2
12
This means:
16
−5
−3
− 18
3
3
−2
−2
0
3
12x3 + 16x2 − 5x − 3 = (x + )(12x2 − 2x − 2)
2
Now:
12x2 − 2x − 2 = 12x2 − 6x + 4x − 2
= 6x(2x − 1) + 2(2x − 1)
= (6x + 2)(2x − 1)
1
1
So the other roots are − and .
3
3
4.
3
−
1
3
3
This means:
7
− 22
−8
−1
−2
8
6
− 24
0
1
3x3 + 7x2 − 22x − 8 = (x + )(3x2 + 6x − 24)
3
Now:
3x2 + 6x − 24 = 3(x2 + 2x − 8)
= 3(x + 4)(x − 2)
So the other roots are −4 and 2.
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4.5.5
36
Long division
Tips: for polynomial long division, we write the polynomial being divided under the bar, and the divisor
outside. At every stage, we look only at what needs to be multiplied by the leading coefficient of the
divisor to make the leading coefficient of the dividend. Pitfalls: remember to put blanks where there are
powers of x missing (see number 3). Also be careful when subtracting in intermediate steps – you’re
subtracting all the terms.
1.
2x2 + 3x + 5
3x − 1
6x3 + 7x2 + 12x − 5
− 6x3 + 2x2
9x2 + 12x
− 9x2 + 3x
15x − 5
− 15x + 5
0
This one came out even, so the answer is 2x2 + 3x + 5.
2.
3x + 7
x−3
3x2 − 2x
+5
− 3x2 + 9x
7x + 5
− 7x + 21
26
So
3x2 − 2x + 5
26
= 3x + 7 +
x−3
x−3
3.
4x3 + 16x2 + 60x + 246
x−4
4x4
− 4x2
− 4x4 + 16x3
+ 6x
16x3 − 4x2
− 16x3 + 64x2
60x2 + 6x
− 60x2 + 240x
246x
− 246x + 984
984
That was a long one! Remember the blanks!
4.5.6
4x4 − 4x2 + 6x
984
= 4x3 + 16x2 + 60x + 246 +
x−4
x−4
Graphs of polynomial functions
1. This is just “multiplying out”:
y = 2(x − 1)2 + 4 = 2(x2 − 2x + 1) + 4 = 2x2 − 4x + 2 + 4 = 2x2 − 4x + 6
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37
2. We can ignore the last term as a start. The first two terms are 2x2 + 4x = 2(x2 + 2x). The perfect
square closest to the expression in parentheses is x2 + 2x + 1 = (x − 1)2 . If we were to replace that,
we would have: 2(x − 1)2 = 2x2 + 4x + 2. But what we have is 2x2 + 4x + 5, which is three more. So:
y = 2x2 + 4x + 5 = 2(x − 1)2 + 3
3. The vertex of the standard parabola y = x2 is (0, 0). This parabola is shifted right by 1 and up by 4.
So its vertex is at (1, 4).
4. We wrote this parabola in vertex form above. So the vertex is (−1, 3).
5. The vertex form will look like y = a(x + 1)2 + 3 if the vertex is at (−1, 3). If the y-intercept is
(0, −2), then the equation at that point reads:
−2 = a(0 + 1)2 + 3 ⇒ −2 = a + 3
So a = −5, and the equation is y = −5(x + 1)2 + 3.
6. If a polynomial has a root of x = a, it has a factor of (x − a). So a parabola with these roots could
have equation y = (x + 2)(x − 3). Multiplying this by a constant is not going to change where it
equals zero, so any multiple of this would be fine as well.
7. See the answers for the graph. General strategy for graphing in vertex form is to plot the vertex.
Then plug in x = 0 to get the y-intercept and plot that. Reflect that point over the axis of symmetry
(the vertical line through the vertex) to get a third point on the graph. Then draw the best parabola
you can through those three points.
8. We have already done the work to put this in vertex form, so using that as in the previous example is
the best bet.
4.5.7
Graphs of higher-degree polynomial functions
Graphing higher-degree polynomial functions is something of an art. But it’s not too bad to get a grasp on
some basics.
1. Polynomial functions of even degree have their ends pointing the same direction (both up or both
down); polynomial functions of odd degree have their ends pointing in opposite directions (one up,
one down).
2. The up and down is decided by the leading coefficient. Positive leading coefficients mean the graph
goes up to the right (and on both sides for a polynomial function of even degree). Negative
coefficients in polynomials of odd degree mean the left end goes up and the right end goes down.
Those are the absolute basics. What else do you have to go on?
1. The constant term gives the y-intercept.
2. If you can factor a little, the roots give you some x-intercepts.
Go on to calculus – you’ll learn even more techniques. For now, if you need more, check the online
lectures, section 2.2.
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4.6
38
Rational expressions
1.
2x2 + 5x − 3
x2 + 2x − 3
2x2 + 6x − x − 3
(x + 3)(x − 1)
(2x − 1)(x + 3)
(x + 3)(x − 1)
2x − 1
, x 6= −3
x−1
=
=
=
2.
−4x3 y2 −4xy3
·
2x2 y3 10xy2
=
=
−2x −2y
·
y
5
4xy 4x
= , x, y 6= 0
5y
5
3.
3
3x
+
x − 2 2x + 3
=
=
=
=
3(2x + 3)
3x(x − 2)
+
(x − 2)(2x + 3) (2x + 3)(x − 2)
3(2x + 3) + 3x(x − 2)
(2x + 3)(x − 2)
6x + 9 + 3x2 − 6x
(2x + 3)(x − 2)
3(x2 + 3)
(2x + 3)(x − 2)
4.
4x − 48
x2 − 144
=
=
4(x − 12)
(x + 12)(x − 12)
4
, x 6= 12
x + 12
5.
x2 − 9 x2 + 2x + 1
·
2x + 2 (x − 3)(x + 1)
=
=
(x + 3)(x − 3) (x + 1)(x + 1)
·
2(x + 1)
(x − 3)(x + 1)
x+3
, x 6= −1, 1, 3
2
6.
x2 + x − 2 x2 + 4x − 5 x2 + 2x − 15
÷
· 2
3x2 − 9x
6x − 18
x −x−6
=
=
(x + 2)(x − 1)
6(x − 3)
(x − 3)(x + 5)
·
·
3x(x − 3)
(x + 5)(x − 1) (x − 3)(x + 2)
2
, x 6= −5, −2, 1, 3
x
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4.6.1
39
Complex fractions
1.
x+5
3x2
x2 −25
6x3
=
=
=
x+5
6x3
·
3x2 x2 − 25
(x + 5) · 6x3
3x2 (x + 5)(x − 5)
2x
, x 6= 0, −5
x−5
9
1
2 = 2 =9
2.
1 1
1
+
3 6
2
4+
3.
5
x−5
5
x+3
3
+ x+3
3
+ x−5
=
=
=
4.6.2
5(x+3)
3(x−5)
(x−5)(x+3) + (x+3)(x−5)
5(x−5)
3(x+3)
(x+3)(x−5) + (x−5)(x+3)
5x+15+3x−15
(x+3)(x−5)
5x−25+3x+9
(x+3)(x−5)
8x
x
8x
=
=
, x 6= −3, 5
8x − 16 8(x − 2) x − 2
Graphs of rational functions
General principles: to graph a rational function:
1. Factor top and bottom completely.
2. If possible, cancel any common factors. If any factors cancel, take note.
3. If, after the previous step, the degree of the numerator is one more than the degree of the
denominator, there is a slant asymptote. To find the slant asymptote, perform long division of the
polynomials, and ignore the remainder. The result will be linear, the equation of your slant
asymptote. Plot it as a dotted line.
4. If the degree of the denominator is greater than the degree of the numerator, the denominator is
“more powerful”, so the graph has a horizontal asymptote of y = 0. This means that the left and right
ends of the graph will head towards zero.
5. If the degree of the numerator and denominator are the same, the graph will approach (horizontally)
the quotient of their leading coefficients. Plot this as a dotted (horizontal) line.
6. Any roots of the numerator (after cancelling) are zeros, or x-intercepts, of the function. Plot these on
the graph.
7. Any roots of the denominator (after cancelling) are vertical asymptotes. Plot vertical dashed lines for
these. Your final graph will not cross these lines.
8. Plug in x = 0 to find the y-intercept, and plot it.
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9. At this point, dotted lines will divide your graph into segments. You now need to sketch in the graph,
not crossing the vertical asymptotes, and at the end approaching the horizontal (or slant) asymptote.
You need a point in every vertical region to decide whether the graph is above or below the
horizontal (or slant) asymptote. Your x-intercepts and y-intercept will be a good starting point. Plot a
few more points, if necessary, and make sure that your graph does not have new zeros that you didn’t
find above.
10. If a factor disappeared when simplifying, you need to plot an empty circle in the graph over that
x-value, whose y-value corresponds to the value that x produces when plugged in to the simplified
function.
Tips for the specific problems:
1. This function has a zero at x = −2, a vertical asymptote at x = 3, a horizontal asymptote of y = 1. At
x = 4, y = 6. These points allow you to tell where the graph lies. Then just follow the asymptotes.
2.
x2 − 4x + 3 (x − 3)(x − 1) x − 3
=
=
, x 6= 1
xx − 2
(x + 2)(x − 1) x + 2
The simplified function has x-intercept (3, 0), vertical asymptote x = −2, horizontal asymptote y = 1,
and a “hole” at (1, − 23 ).
3.
x−1
x−1
=
x2 − x − 2 (x + 1)(x − 2)
This function has an x-intercept of (1, 0) and two vertical asymptotes: x = −1 and x = 2. Its
horizontal asymptote is y = 0, since the degree of the denominator is larger. A little more point
plotting fills out the graph.
4.
x2 + 3x + 2 (x + 1)(x + 2)
=
x−1
x−1
The degree of the numerator is one greater than the degree of the denominator. The result of long
division is x + 4 plus a remainder. So y = x + 4 is a slant asymptote. The zeros are x = −1 and
x = −2, and there is a vertical asmptote of x + 1. Plugging in x = 2 gives enough information to
finish the plot.
4.7
Exponential and Logarithmic functions
To start: the rules of exponents with the corresponding rules of logarithms:
1. am · an = am+n
1. loga mn = loga m + loga n
2. am /an = am−n
2. loga (m/n) = loga m − loga n
3. (am )n = amn
3. loga xm = m loga x
4. a0 = 1, if a 6= 0.
4. loga 1 = 0
1.
√
3
82/3 = (81/3 )2 = ( 8)2 = 22 = 4
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2. Negative exponents indicate taking a reciprocal. They do not make the answer negative!
8−2/3 =
1
1
=
82/3 4
3.
(−8)2/3 = [(−8)1/3 ]2 = (−2)2 = 4
4.
(−8)−2/3 =
1
1
=
2/3
4
(−8)
5. Order of operations here: exponents take precedence over the negative sign.
−82/3 = −4
6. Powers are applied to numerator and denominator:
√ 3
3/2
43/2
4
8
4
23
= 3/2 = √ 3 = 3 =
9
3
27
9
9
7. This logarithm asks the question, “what power of 2 is 12 ?” Since
is -1.
1
2
is the reciprocal of 2, the answer
8. This one asks, “what power of 3 is 27?” The answer is 3.
9. What power of 9 is 3? Since 3 is the square root of 9, the answer is 21 .
1. Rewrite this one in exponential form. When log has no shown base, the base is ten:
x = 10−3 = 0.001
2.
x=2
−1/2
√
1
2
=√ =
2
2
3.
e2x − 2 = 0 ⇒ e2x = 2
⇒ 2x = ln 2
1
⇒ x = ln 2
2
4.
8 − 3 · 20.5x = −40 ⇒ −3 · 20.5x = −48
⇒ 20.5x = 16
⇒ 0.5x = 4 ⇒ x = 8
5.
log3 9x = 3 ⇒ 9x = 33 ⇒ 9x = 27 ⇒ x = 3
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6. x occurs twice in this problem in places that cannot be combined. So we must find a way to factor:
e2x − 3ex = 28 ⇒ (ex )2 − 3(ex ) − 28 = 0
The left side looks like y2 − 3y − 28, which factors as (y − 7)(y + 4), but with y replaced by ex . So:
(ex − 7)(ex + 4) = 0,
which means that ex = 7 or ex = −4. Since ex is never negative, the only possiblity is ex = 7, which
means x = ln 7.
7. We want to use rules of logarithms to combine the terms on the left:
x+1
= log 3
x−1
Now we have an equality of two logarithms. Since the logarithm function is one-to-one, the
arguments of the logarithms must be equal:
log(x + 1) − log(x − 1) = log 3 ⇒ log
x+1
= 3 ⇒ x + 1 = 3(x − 1) ⇒ x + 1 = 3x − 3 ⇒ 4 = 2x ⇒ x = 2
x−1
Note that any time we solve a logarithm equation, we must check that the solution can be plugged in
to the original equation. In this case, x = 2 is fine. If plugging it in made the argument of a logarithm
negative, we have a false solution, which must be discarded.
8.
x2 − 4
= ln(3 − x)
x+2
x2 − 4
= 3−x
⇒
x+2
(x + 2)(x − 2)
⇒
= 3−x
x+2
5
⇒ x−2 = 3−x ⇒ x =
2
ln(x2 − 4) − ln(x + 2) = ln(3 − x) ⇒ ln
And x = 25 does not make any logarithm arguments in the original equation negative (nor any
denominators zero!). So it is a valid solution.
Question: why combine logs? It makes equation solving easier.
1. Note here that the log terms cannot be combined until the coefficient 12 has been brought inside the
first as a power. In a sense, the rules of logarithms should be applied in the same order as order of
operations (the 21 becomes an exponent, so it’s first).
√
√
1
log2 x + log2 y = log2 x1/2 + log2 y = log2 x + log2 y = log2 y x
2
2. Same thing here:
3 log x − 2 log y − log z = log x3 − log y2 − log z
= log x3 − (log y2 + log z)
= log x3 − log y2 z
x3
= log 2
y z
Why did I factor out the negative sign in the second line? Remember that order of operations for
addition and subtraction goes right to left. So I would have to do two divisions if I didn’t factor. This
makes more room for error.
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Question: why do we even learn about separating out logs? Answer: You need to do it in calculus.
1.
√ x y
1
√
√
log
= log x y − log z = log x + log y − log z = log x + log y − log z
z
2
2.
log4 4x2 y3 = log4 4 + log4 x2 + log4 y3 = 1 + 2 log4 x + 3 log4 y
4.8
4.8.1
Matrices and Linear Algebra
Matrix Multiplication
The size of a matrix is given by its dimensions, in the form (rows) × (columns). For example, the matrix


1
7 3
 −2 0 3 


 4
3 1 
−2 −2 1
is 4 × 3.
Two matrices may be multiplied left-to-right if the number of columns of the left matrix is equal to the
number of rows of the right matrix.
Matrix multiplication is done like this: the result has, in row m and column n, the number gotten by
multiplying each entry in row m of the left hand matrix by the corresponding entry in column n of the right
hand matrix and adding the results. The resulting matrix has the same number of rows as the left matrix
and the same number of columns as the right matrix. In brief, a m × n matrix may be multiplied by an n × p
matrix and the result is m × p.
See the work below to learn how this works in practice. Also note that the order in which matrices are
multiplied is important. Some pairs of matrices can only be multiplied in one direction.
1.
4.9
Measuring and describing angles
1. A circle is divided into 360 equal pieces of arc called degrees. Another way is to measure the outside
in terms of the length of the radius. One radius measured out on the outside of a circle is called a
radian. Since the circumference of a circle is equal to 2πr, there are 2π radians in a circle. Thus the
2π
. So to convert radians to degrees, we multiply by 180
ratio of radians to degrees is 360
π . To go the
π
other way, multiply by 180 .
2. See previous item.
3. Coterminal angles look the same when drawn. In other words, they differ by multiples of 2π (in
radians) or 360◦ . So 390◦ is coterminal to 390◦ − 360◦ = 30◦ . Subtracting another 360 gives a
negative angle of −330◦ . To do part b) in radians, you should write 2π = 12π
6 , so it can be easily
17π
5π
added or subtracted with 6 . So a positive coterminal angle is 6 , and a negative would be − 7π
6 .
4. The complement of an angle is what needs to be added to make 90◦ or
what needs to be added to make a straight line (180◦ or π radians).
a) 90 ◦ −72◦ = 18◦ ;
b) 180◦ − 72◦ = 108◦ ;
π
2
radians. The supplement is
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c) Write
π
2
=
d) Write π =
4.10
6π
12 .
6π
6 .
Then
Then
6π
12
6π
6
π
− 12
=
− 5π
5 =
5π
12 .
π
6.
Trigonometric functions
Here’s a good way to remember the first quadrant trig function values:
θ (rad) θ (deg) sin θ√
0
0◦
0 = √20
1
1
π/6
30◦
2 = √2
2
π/4
45◦
0=
√ 2
3
π/3
60◦
2√
π/2
90◦
1 = 24
cos θ
1 = sqrt4
2
sqrt3
2
sqrt2
2
sqrt1
1
=
2
2
0 = sqrt0
2
π
For the angles in the other quadrants, we need to find a reference angle. For example, 5π
6 is 6 above the
π
negative x-axis. So its sine and cosine will be the same as that of 6 , but with different signs. In the second
quadrant (where 5π
6 lives), sine is positive and cosine is negative.
1. Standard first quadrant.
2. Standard first quadrant.
3. 135◦ is 45◦ above the negative x-axis. sec x =
√
1
√ 1 = 2.
sec(135◦ ) = cos(135
◦) =
2/2
1
cos x ,
and cos x is negative in the second quadrant. So
4. 300◦ is 60◦ below the positive x-axis (in the fourth quadrant). There, sin x < 0 and cos x > 0, so
tan x < 0.
√
sin 60◦
3/2 √
◦
tan(300 ) = −
=−
= 3.
◦
cos 60
1/2
5. cot x =
cos x
sin x .
Since sin 0 = 0. cot 0 is not defined.
6. Standard first quadrant.
7. Write 2π =
8.
6π
3 .
π
Adding this to − 5π
3 to get a coterminal angle gives 3 . This is a first quadrant angle.
3π
4
is between π2 and π, so in the second quadrant. It is
quadrant, cosine is negative and sine is positive, so
tan(
9. Since 4 > 3,
sin x < 0, so
4π
3
π
4
above the negative x-axis. In the second
3π
sin(π/4)
)=−
= −1.
4
cos(π/4)
is a bit more than π, so in the third quadrant. The angle is
10. cos π = −1, so sec π =
π
3
below the x-axis. There,
√
4π
1
2
2 3
csc( ) = −
=√ =
3
sin(π/3)
3
3
1
cos π
= −1.
Once you know the sine of an angle, you know almost all the trig functions. This is because
sin2 x + cos2 x = 1. This means that
p
cos x = ± 1 − sin2 x
To decide plus or minus, we need geometric information about the angle.
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1. θ is acute,
where
If sin θ = 0.6,
√ so in first quadrant,
√ all trig functions are positive.
√
0.6
cos θ = 1 − 0.62 = 1 − 0.36 = 0.64 = 0.8. Then tan θ = 0.8
= 43 .
2. Here, tangent and sine are both negative. That means cosine is positive. Since tangent is opposite
over adjacent, we can work from a right triangle with legs 8 and 15. 82 + 152 = 172 , so the
8
hypotenuse is 17. Taking signs into account, sin θ = − 15
17 and cos θ = 17 .
3. θ is in the second quadrant, so sin θ is positive and cos θ is negative. Since sine is opposite over
hypotenuse, we can work with a right triangle with hypotenuse 5 and one leg 4. Since 32 + 42 = 52 ,
the other (adjacent) leg is 3. Thus cos θ = − 35 and tan θ = − 43 .
4.11
Inverse trigonometric functions
1
Don’t be confused, sin−1 1 doesn’t mean
. It is asking the question, what angle has a sine of 1? There
sin 1
are multiple answers, since at least we can add a full rotation to the angle without changing the trig
functions. So to have a clearly defined inverse function, we choose the angle with the given value that is
closest to 0, and positive when possible. This makes the inverse trig functions work like:
function returns values in
sin−1
[−π/2, π/2]
cos−1
[0, π]
−1
tan
[−π/2, π/2]
π
5π
1. Sine gives the y-coordinate of a point on the
is π/6 above the x-axis, so 6π
6 − 6 = 6 .
unit circle. There’s only one point on the unit
4. Sine is negative in the fourth quadrant, for the
circle with y-coordinate 1, π/2.
purposes of computing sin−1 . A sine of 1/2 is
an angle of π/6, so the answer here is −π/6.
2. If an angle has tangent 1, it means
sin x
sin x = cos x, since tan x = cos
x . This happens
5. For inverse tangents of negative angles, we
at π/4.
look
√ in the fourth quadrant (negative angles).
3 is steeper than 1, so the answer is −π/3.
3. Negative cosines are in quadrant two or three.
Inverse cosine doesn’t take values in √quadrant
6. See number 3. Here the answer will be π/4
3
three. We need an x-coordinate of − 2 . That
above the negative x-axis, or 3π
4 .
4.12
Trigonometric Equations
1. sin x = cos x. This one is probably easiest to see by superimposing the graphs or looking at the unit
circle. Which points on the unit circle have the same x− √
and y−coordinates? Where the line y = x
meets the circle. So we are solving x2 + x2 = 1, or x = ± 2/2. These happen at the angles π/4 and
5π/4.
2. 3 sin x − 2 = 5 sin x − 1. Strategy here is to begin by isolating the trig function:
−1 2 sin x ⇒ sin x = −1/2
This happens at x = π/6 or 5π/6.
3. 2 cos2 x + cos x − 1 = 0. This equation is like a quadratic equation in cos x. Factor it as:
(2 cos x − 1)(cos x + 1) = 0
So we have either 2 cos x − 1 = 0 or cos x + 1 = 0. The first equation means that cos x = 1/2, which
happens when x = π/3, 5π/3. The second equation means cos x = −1, which happens at x = π.
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46
4. 2 cos2 x + 3 sin x = 0. This one has a mix of trig functions. To make it an equation of a single
function, we use an identity: cos2 x + sin2 x = 1, so cos2 x = 1 − sin2 x. Plugging this in to the original
equation gives:
2(1 − sin2 x) + 3 sin x = 0 ⇒ 2 − 2 sin2 x + 3 sin x = 0 ⇒ 2 sin2 x − 3 sin x − 2 = 0
Factor:
(2 sin x + 1)(sin x − 2) = 0
which yields sin x = −1/2 or sin x = 2. The first equation gives x = 7π/6, 11π/6. The second
equation cannot be solved, since sin x is never larger than 1.
4.13
Graphs of trigonometric functions
The rules for modifying the graphs of trig functions are the same as for other functions. However, some
special terminology is involved.
For example, for the function A sin(Bx +C) + D,
1. A is called the amplitude (this term is not used for the tangent function). Since sine and cosine take
values only between -1 and 1, when multiplied by A, they take values between −A and A.
2. D is a shift up (D positive) or down (D negative). It doesn’t have a special name.
3. Remember that inside of a function, horizontal shifts are performed before stretching. So the
function shifts left (C positive) or right (C negative). After this, the graph is stretched (or
compressed) by a factor of 1/B. The former value of the function at 0 is now shifted to −C, then
stretched to −C/B. This number is called the phase shift. The sine and cosine functions have a
period of 2π (meaning they repeat every 2π). The period of the tangent function is π. Because of the
π
stretch factor, for sine and cosine functions, the period is 2π
B , and for tangent, it is B .