1

A BRIEF REVIEW OF ALGEBRA AND

TRIGONOMETRY

Some Key Concepts:

1. The slope and the equation of a straight line

2. Functions and functional notation

3. The average rate of change of a function and the DIFFERENCE-

QUOTIENT OF A FUNCTION

4. Domains of some common types of functions

5. Graphs of functions and piecewise functions

6. Expanding, factoring, and simplifying expressions

7. Solving algebraic equations

8. Basic trigonometry

9. Solving trigonometric equations

2

1. The slope and the equation of a straight line:

(a) Slope of a straight line

Four different kinds of lines and their slopes:

(b) Equation of a straight line

The equation of a straight line can be written in any one of the following three ways:

(i) The Point-Slope Form

The equation of a straight line that passes through the point ),( 11 yx and having slope m is given by

x

y

xx

yy

ofchange

yofchange

run

risemPQlinetheofSlope

12

12

X

Y

Px1,y1

Qx2,y2

x1 x2

y1

y2

Dx=x2-x1

Dy=y2-y1

X

Y

Positive

Negative

Zero

Undefined

3

)( 11 xxmyy

(ii) The Slope-Intercept Form

The equation of a line with slope m and having the y-intercept ),0( b is given by

bmxy

(iii) The Standard Form

In this form, you move all x and y terms of the equation of the given line, to one side, and the constant

term to the other side. It is normally written as

CByAx

Example:

1. Find the equation of the line that passes through the points )2,1( and )9,2( . Provide the exact

answer in the standard form with integral coefficients.

Solution: Make sure to draw your own sketch (not included here). Let ),( 11 yxP be )2,1( and

),( 22 yxQ be )9,2( . Then the slope m of the line PQ is given by

The equation of this line PQ can be found either by using the point-slope form or by the slope-intercept

form. For example, using the point-slope form )( 11 xxmyy , its equation is given by,

)1(3

72 xy

Multiply both sides by 3 to get: )1(7)2(3 xy

7763 xy

In the standard form, the equation is: 1337 yx

(c) Parallel and perpendicular lines:

If two non-vertical lines are parallel, then they have equal slopes. In other words, if 1m and 2m are the

slopes of the parallel lines, then 21 mm . On the other hand, if two-non vertical lines are perpendicular,

then the slope of one line is equal to the negative reciprocal of the slope of the other. In other words, if

1m and 2m are the slopes of the perpendicular lines, then 21 /1 mm .

3/7)12/()29()/()( 1212 xxyym

4

Example:

1. Find the equation of the line that is perpendicular to the line 432 yx and which passes through

the point )3,1( . Provide the exact answer in the slope-intercept form.

Solution: Make sure to make a brief sketch – coordinate axes are not necessary (not included).

First, find the slope of the given line 432 yx . To do this, arrange this equation in the form

bmxy (i.e. solve for y) and look for the coefficient of x:

3

4

3

2;423;342 xyxyyx

Therefore, the slope of the given line is 2/3.

So, the slope of a line perpendicular to the original line is the negative reciprocal of 2/3, i.e.

2/3)3/2/(1 or .

Now, to find the equation of the required perpendicular line, use the point-slope form with 2/3m

and )3,1(),( 11 yx :

)1(2

33 xy

To put in the slope-intercept form, solve the above equation for y:

2

3

2

3 xy

2. Functions and the functional notation:

Quite often, a function can be given by an equation, such as 432 2 xxy . The same function can be

re-written as 432)( 2 xxxf , which is called the functional notation. Thus, )(xf is just another

name for “y”. Thus, as a specific example, )3(f means the y-value of the function when x is equal to 3.

So, this can be found by substituting each x-variable in the expression 432 2 xx by “3”. Therefore,

we have

1349184)3(3)3(2)3( 2 f

13)3( f

Note that 13)3( f just means that, when 3x , the y-value of the function f is equal to 13.

5

Examples:

1. Given )(2)( xSinxxf , find the exact value of )3/(f .

Solution:

6

334

2

3

3

2

332

3

Sinf

6

334

3

f

2. Given 143)( 2 xxxf , find x if 21)( xf .

Solution:

DO NOT PLUG IN 21x in the given equation, as that would be a major mistake! Rather set

21)( xf , and solve the resulting equation for x, as follows:

21143 2 xx

02043 2 xx

The above is a quadratic equation. BE VERY FAMILIAR with the methods of solving quadratic

equations and other equations. For example, one can use the Factoring Method or the Quadratic Formula

)2/()4( 2 aacbbx . Let us use the Factoring Method, since the right-hand side of the last

equation can be found to be factorable:

0)2)(103( xx

Now use the zero-product property: 020103 xorx

23/10 xorx

Thus, there are two x-values for which 21)( xf .

3. VERY IMPORTANT EXAMPLE:

Given 143)( 2 xxxf , find )( hxf . Expand and simplify your answer.

Solution:

This problem illustrates one of the subtle points of functional notation: DO NOT simply add “h” to the

right-hand side of the equation to obtain hxx 143 2 as the answer for )( hxf . In fact,

hxx 143 2 is equal to hxf )( , which is quite different from the desired )( hxf .

In order to find )( hxf , replace all the x-terms on the right hand-side of the given equation by hx :

1)(4)(3)( 2 hxhxhxf

6

Now use rules of algebra such as the Foil Method or special squaring formulas to expand the right-hand

side:

144363

144)2(3)(22

22

hxhxhx

hxhxhxhxf

144363)( 22 hxhxhxhxf

Just to summarize, given a function )(xf , the two quantities )( hxf and hxf )( are totally different!

For example, for the function xxf 1)( , we can find these two quantities as follows:

hxfhxfgeneralIn

hxhxfBut

hxhxhxf

xxfGiven

)()(,

1)(

1)(1)(

1)(:

3. Average rate of change of a function and the difference-quotient:

(a) The average rate of change of a function

In the above diagram, when the x-value is changed from 1x to 2x , the y-value of the function )(xf

changes from )( 11 xfy to )( 22 xfy . Then, the change of x, or the increment of x, is given by

12 xx , and is denoted by x . Similarly, the change of y, or the increment of y, is given by

)()( 1212 xfxfyy , and is denoted by y .

X

Y

P

Q

x1 x2

f x1

f x2y=fx

Dx=x2-x1

Dy=f x2-f x1

7

We define the average rate of change of )(xf from 1xx to 2xx as the ratio xy / . It is very

important to observe that, in the above diagram xy / is equal to the slope of the secant line PQ. A

secant line of a curve is a line joining any two points of the curve. Thus we have:

PQlineanttheofslopexx

xfxf

xx

yy

x

y

ofanytoeqaulisxxtoxxfromxfofchangeofrateaverageThe

sec)()(

)(

12

12

12

12

21

Examples:

1. Find the average rate of change of the function )2()( 2 xCosxf from 0x to 3/x .

Solution

First make sure to draw your own diagram to see the corresponding secant line (not included). The

graphical view point is very essential for a successful study of especially calculus.

Let 01 x and 3/2 x . Then the average rate of change of )(xf from 1xx to 2xx is given

by:

4

9

)3/(

)4/3(

)3/(

)1()2/1(

)3/(

)0()3/2(

0)3/(

)0()3/()()( 2222

12

12

CosCosff

xx

xfxf

(b) The difference-quotient of a function

In part (a), we just defined the average rate of change of a function )(xf from 1xx to 2xx .

The difference quotient of a function is just a special case of the average rate of change – in other

words, the difference-quotient is just the average rate of change of )(xf from x to hx . In order to

find a formula for the difference-quotient, let xx 1 and hxx 2 , and rewrite the previous average

rate of change of formula )(/)]()([ 1212 xxxfxf as follows:

h

xfhxf

xhx

xfhxfQuotientDifference

)()(

)(

)()(

Thus, one can look at the difference-quotient of a function in three different ways. As the algebraic

meaning, or the definition, the difference-quotient of a function is hxfhxf /)]()([ . As the

geometric meaning of the difference-quotient, it is equal to the slope of the secant line PQ where

))(,( xfxP and ))(,( hxfhxQ are two points on the graph of )(xfy . As the physical

meaning of the difference-quotient, it is equal to the average rate of change of the function )(xf from x

to hx . All these ideas are illustrated in the writings below:

8

9

Here are some worked examples on difference-quotients of several types of functions:

10

11

12

13

4. Domains of some common types of functions

Given a function )(xf , its domain is the set of all values of x, for which )(xf is defined. As a very

simple example, the domain of the function 2)( xxf is the entire real line, or the interval ),( ,

since the expression 2x is defined for any real value. The function 2)( xxf is just one example of a

polynomial. In fact, the domain of any polynomial is the entire real line.

However, on the other hand, the domain of the function )2/(1)( xxf is not the entire real line.

Note that the expression )2/(1 x is not defined at 2x , since 0/1 is not defined. However, for any

other x-value, the expression )2/(1 x is always defined. In other words, the domain of the function

)2/(1)( xxf is all real numbers except 2x . In the interval notation, we can write this domain as

),2()2,( .

Here are two very useful principles:

For T/1 to be defined, T must be a nonzero quantity, i.e. 0T

For T to be defined as a real number, T must be a nonnegative quantity, i.e.

0T

Examples:

1. Find the domain of 162

)(2

x

xxf . Express the exact answer in the interval notation.

Solution:

The numerator “x” is always defined. By the previous remarks, the domain of )(xf is all x-values such

that the denominator 162 2 x is nonzero. In order to find when this does happen, ask the opposite

question, i.e. set 162 2 x equal to zero, and solve for x.

22

8

162

0162

2

2

2

x

x

x

x

However, DO NOT be confused in saying that domain is equal to 22 . The domain is in fact, all real

values except 22 . Thus, in the interval notation, the domain of )(xf is equal to

),22()22,22()22,(

14

2. Find the domain of 73)( xxf . Express the exact answer in the interval notation.

Solution:

Recall that For T to be defined as a real number, T must be a nonnegative quantity, i.e. 0T .

Therefore, the domain of )(xf is all x-values such that 073 x . We can solve this linear inequality

as follows:

073 x

73 x

Now divide both sides of the inequality by 3 . However, note that when you multiply or divide both

sides of an equality by an negative quantity, the inequality sign must be reversed!!

3

7

3

3

x

3

7 x

Therefore, the domain of )(xf is all real values less than or equal to 7/3. In the interval notation, the

answer is ]3/7,( .

5. Graphs of functions and piecewise functions

(a) Some basic graphs

Your study of calculus will become more enjoyable, if you can memorize the shapes of some important

graphs:

(a) xy (b) 2xy (c) 3xy

(d) || xy (e) xy (f) 3 xy

-4 -2 2 4X

-4

-2

2

4

Y

-4 -2 2 4X

5

10

15

20

25

Y

-3 -2 -1 1 2 3X

-20

-10

10

20

Y

-3 -2 -1 0 1 2 3X

1

2

3Y

0 1 2 3X

0.5

1

1.5

Y

1 2 3X

0.5

1

1.5Y

15

(g) x

y1

(h) xy 2 (i) x

y

2

1

(j) xey (k) )ln(xy (l) 122 yx

Now here is the main advantage: Once you memorize the above shapes, you can build the graphs of

more complicated functions using transformation techniques.

Examples:

1. Draw a quick graph of 3|2|)( xxf .

Solution:

Note that the parent function for the given equation is || xy , and we already know the shape of its graph

by part (d) above. Now start with this graph (d), translate it 2 units to the left, and then 3 units down, to

obtain the required graph, The vertex of the final graph is )3,2( .

-3 -2 -1 1 2 3X

-3

-2

-1

1

2

3

Y

-3-2-1 1 2 3X

2

4

6

8

Y

-3-2-1 0 1 2 3X

2

4

6

8

Y

-2 -1 1 2X

1

2

3

4

5

6

7

Y

2 4 6 8X

-1

1

2

Y

-1.0 -0.5 0.5 1.0X

-1.0

-0.5

0.5

1.0

Y

16

(b) Piecewise functions and their graphs

One of the biggest concepts of calculus is the notion of the limit of a function. In order to understand this

concept successfully, it is better to be familiar with the idea of piecewise functions. These functions

normally have two or more pieces in its defining equation. Here is an example:

Examples:

1. Consider the function given by

418

422

22

)( 2

xif

xifx

xifx

xf

(a) Find )3(f (b) Find )3(f (c) Find )4(f (d) Draw a clear graph of )(xf

Solution:

(a) Note that 3x satisfies the inequality 2x , Therefore, to find )3(f , plug in 3x in the

expression 2x , the first “piece” of the piecewise function. Thus, 123)3( f .

(b) Here 3x satisfies the inequality 42 x . Therefore, to find )3(f , we use the second “piece” of

the given function. Thus, 112)3()3( 2 f .

(c) This part is VERY INTERSTING! In this case. 4x does not satisfy either inequality 42 x or

4x . Therefore, )4(f is undefined, or simply does not exist. What this means is that, the graph of

)(xf will have a hole corresponding to 4x . See the graph in part (d) below.

(d) First the graph the three functions ,2,2 2 xyxy and 18y by completing disregarding

the conditions given by the inequalities. This is where knowing the basic graphs will really pay

dividends!

-6 -4 -2 2X

-3

-2

-1

1

2

Y

17

2 xy 22 xy 18y

Now we are ready to cut each of the above graphs using appropriate vertical lines: As shown above, cut

the first graph using the vertical line 2x , and take the part to the left of this vertical line. The second

graph must be cut using two vertical lines, 2x and 4x , and take the part of the graph between these

two vertical lines. The third graph is cut by the vertical line 4x , and take the part to the right of this

vertical line.

Finally, we must glue together the remaining parts of these three graphs. However, one must take

care of the end points of the pieces which are glued together. The final assembled graph is given below

– note that a solid dot means the corresponding point is included, while a small circle means the point is

excluded.

-4 -3 -2 -1 1 2 3 4

-4 -2 2 4

5

10

15

20

25

-4 -2 2 4

5

10

15

20

25

30

35

-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 X

2

4

6

8

10

12

14

16

18

Y

18

6. Expanding, factoring, and simplifying expressions

Here are some crucial algebra formulas:

(A) Expansion Formulas:

1. 222 2)( bababa

2. 222 2)( bababa

3. 32233 33)( babbaaba

4. 32233 33)( babbaaba

(B) Factoring Formulas:

1. ))((,.22 biabiaarefactorscomplextheHoweverfactorsrealanyhavenotdoesba

2. ))((22 bababa

3. ))(( 2233 babababa

4. ))(( 2233 babababa

Examples:

1. Factor: 5416 3 x

Solution:

)964(322])3()2([2)278(25416 23333 xxxxxx

2. Factor: 44 16yx

Solution:

)4()2)(2()4()4(416 222222222244 yxyxyxyxyxyxyx

3. Factor: 3223 yxyyxx

Solution:

Use factoring by grouping method.

)()())()(()()()()( 222223223 yxyxyxyxyxyxyxyxyyxxyxyyxx

19

4. VERY IMPORTANT (LEARN THIS!!)

Simplify the following algebraic expression, and leave the answer in the factored form:

42322 )32.(2).4(2)2()32(4.)4( xxxxx

Solution:

)834()32)(4(4

]3282[)32)(4(4

])32()4(2[)32)(4(4

)32)(4(4)32()4(8

)32.(2).4(2)2()32(4.)4(

232

2232

232

42322

42322

xxxx

xxxxx

xxxxx

xxxxx

xxxxx

5. VERY IMPORTANT (LEARN THIS!!)

Simplify the following expression, and leave the answer in the factored form: 22

92

x

Solution:

22

2

22 2

)32)(32(

2

94

2

9

1

2

2

92

x

xx

x

x

xx

6. EXTREMELY IMPORTANT (LEARN THIS - YOU WILL NEED THIS SEVERAL

TIMES!!)

Simplify the following expression completely: 2/122/12 )4()2.()4(2

1. xxxx

Solution:

4

)2(2

4

42

)4(

)4(

1

)4(

)4(

)4()4(

)4()2.()4(2

1.

2

2

2

2

2/12

22

2/12

2/12

2

2/122/12

2

2/122/12

x

x

x

x

x

xx

x

x

x

xx

x

xxxx

20

7. EXTREMELY IMPORTANT (LEARN THIS - YOU WILL NEED THIS SEVERAL

TIMES!!)

Simplify the following expression completely: )9(

)2()9(2

1.2.)9(

2

2/1222/12

x

xxxxx

Solution:

2/32

2

2/32

22

2/32

32

2/122

2/122/12

32/12

2

2/12

32/12

2

2/1222/12

)9(

)18(

)9(

]218[

)9(

)9(2

)9(.)9(

)9(.)9(

)9(2

)9(

)9()9(2

)9(

)2()9(2

1.2.)9(

x

xx

x

xxx

x

xxx

xx

xx

xxx

x

x

xxx

x

xxxxx

REMINDER: Now try several problems from the transparency on “Some

Factoring Problems”

7. Solving algebraic equations

Solving various types of equations is an essential skill for anyone studying calculus. Here are some

examples.

Examples:

1. Solve 024 24 xx

Solution:

The above is not a quadratic equation. However, with a simple substitution, it can be transformed into a

quadratic equation. Let 2xu . Then the given equation becomes:

0242 uu

The right-hand side of the above equation does not factor. So use the quadratic formula with

,4,1 ba and 2c .

21

222

224

2

84

)1(2

)2)(1(4)4(4

2

4 22

a

acbbu

222 x

22 x

Therefore, the four solutions are 22,22,22,22 andx

2. Solve 0)12()34()12()34( 2332 xxxx

Solution:

DO NOT expand the right-hand side of the given equation – instead factor the right-hand side

0)34()12()12()34( 22 xxxx

0)26()12()34( 22 xxx

0)13()12()34(2 22 xxx

Use the zero product property to get: 034 x or 012 x or 013 x

3/1,2/1,4/3 x

REMINDER: Now try several problems from the transparency on “Some

Equation Solving Problems”

22

8. Basic trigonometry

Make every attempt to be familiar with the following facts from trigonometry:

(a) Trig functions of special acute angles (b) Trig functions of all special angles in “color

wheel”

(c) Signs of trigonometric functions in various quadrants

0 30 45 60 90

Sin 0

2

1

2

1

2

3

1

Cos 1

2

3

2

1

2

1

0

Tan 0

3

1

1 3 undefined

23

(d) Basic Trigonometric Identities (e) The Double Angle Formulas:

Examples:

1. Find the exact value of the expression )3(46

33

22 323

CosTanCos

Solution:

4

19)1(4

3

13

8

12)1(4

3

13

2

12

)3(46

33

22

3

23

323

CosTanCos

2. SOMETHING VERY USEFUL TO KNOW!

Find all the values for which Sine, Cosine, and Tangent functions are zero.

Solution:

Just use the information gained by above (a) and (b).

(a) Sin is zero for any angle that falls on the x-axis. In other words, this happens for

,....3,2,,0,,2....., . All these values are integer multiples of . Thus we can conclude the

following:

nifonlyandifSin 0 where n is any integer

24

(b) 0Cos is zero for any angle that falls on the y-axis. In other words, this happens for

,....2/5,2/3,2/,2/,2/3....., . All these values are odd integer multiples of 2/ . Thus

we can conclude the following:

20

nifonlyandifCos where n is any odd integer.

(c) Just like Sin , Tan is zero for any angle that falls on the x-axis. This we can write:

nifonlyandifTan 0 where n is any integer

9. Solving trigonometric equations

Solving trigonometric equations quite often uses the facts (a) and (b) of the previous section 8.

Examples:

Try your BEST to understand the first two examples. The others are built from these two.

1. Solve 2 1 0 where 40 .

Solution:

First, isolate the trigonometric function by solving for :

2 1

12

Now draw the quadrant diagram, and use the “ASTC” principle to find out which quadrant the angle

belongs to. We know that is a negative number. So the angle must belong to either quadrant II

or quadrant III. We will locate these two quadrants by circling them as given in the diagram below:

I II

III IV

The next step is to find the reference angle. It is the

acute angle between the X-axis and the terminal side

of the angle .  So, by dropping the negative side of

the right-hand-side of the equation 1/2,

we ask that the cosine of what acute angle is equal to

1/2.  It is 60° or /3 radians.  Thus, the reference

angle is equal to /3. We will now draw this

reference angle in the circled quadrants:

25

Thus, there are 4 solutions to the given equation. They are 2 /3, 4 /3, 8 /3, 10 /3.

2. Solve 034 2 Cos where 20 .

Solution:

34 2 Cos

4

32 Cos

2

3Cos

We can treat the above like two separate equations, 2/3Cos and 2/3Cos and solve them.

For both of these equations, the reference angle is 30 or 6/ radians.

(i) To solve 2/3Cos , observe that Cos is negative only in quadrants II and III. Draw the

reference angle of 6/ radians in these two quadrants as follows:

I II

III IV

/

/

/

/ All you have to do now is to read off the angles: The angles are always read from the positive X-axis. This is where we use the given domain 0 4 for the first time. So here no negative angles are allowed. Thus, we will turn counterclockwise up to two complete turns, because 4 radians represents two complete turns – see the “red turnings” in the given diagram. The first answer is 2 /3. The second answer is 4 /3. These are the only two answers up to one complete turn of 2 radians. However, recall that we have to go up to two complete turns. They are not shown in the diagram, but the final two angles are 8 /3 and 10 /3.

Therefore, the two solutions for this part are 5 /6 and 7 /6.

Never attempt Without a diagram!

26

(ii) On the other hand, to solve 2/3Cos , observe that Cos is positive in quadrants I and IV.

Draw the reference angle of 6/ radians in these two quadrants as follows:

The final solutions are obtained by combining the answers for the above parts (i) and (ii). Therefore, the solutions are 6/11,6/7,6/5,6/ and

3. Solve 04)2( SinSin where 42 Solution: Use the Double Angle Formula CosSinSin 2)2( to convert the entire right-hand side of the equation to the single angle , as follows:

042 SinCosSin 0)2(2 CosSin 020 CosorSin

20 CosorSin

Now solve like two different problems. Make sure to supply your own diagrams (omitted here). For any , Cos is always between -1 and 1, so the equation 2Cos doesn’t have any solutions. Thus, only solutions for the original equation can come from 0Sin . However, in example 2 of section 8 we learned that 0Sin only happens when the angle lies along the x-axis. Therefore, all the solutions strictly between 2 and 4 are 3,2,,0, and .

Therefore, the two solutions for this part are /6 and 11 /6.

27

4. Solve 2)()2( xSinxCos where 20 x .

Solution: Use the Double Angle Formula xSinxCos 221)2( to convert the entire right-hand side of the equation to the single angle x . Also, in the process, the entire equation will be in terms of one type of trigonometric function, i.e. in terms of )(xSin only. 2)())(21( 2 xSinxSin

03)()(2 2 xSinxSin 0)1()32( xSinxSin 2/3)( xSin or 1)( xSin Now solve like two different problems. Make sure to supply your own diagrams (omitted here). First, the equation 2/3)( xSin does not have any solutions, since for any angle x, )(xSin always lies between -1 and 1. On the other hand, the only solution for 1)( xSin is 2/x in the given domain

20 x . Therefore, the only solution to the given equation is 2/x .