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

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

)( 11 xxmyy −=−

xy

xxyy

ofchangeyofchange

runrisemPQlinetheofSlope

∆∆

=−−

====12

12

X

Y

PHx1,y1LQHx2,y2L

x1 x2

y1

y2

Dx=x2-x1

Dy=y2-y1

X

Y

PositiveNegative

Zero

Undefined

(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(372 +=− 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

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:

34

32;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 2/3−=m and

)3,1(),( 11 −=yx :

)1(233 +−=− xy

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

23

23

+−= 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 3=x , the y-value of the function f is equal to 13.

Examples:

1. Given )(2)( xSinxxf += , find the exact value of )3/(πf .

Solution:

6334

23

32

332

3+

=+=

+

=

πππππ Sinf

6

3343

+=

∴

ππf

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

Solution:

DO NOT PLUG IN 21=x 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

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

144363144)2(3)(

22

22

+−−++=

+−−++=+

hxhxhxhxhxhxhxf

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:

hxfhxfgeneralInhxhxfBut

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

X

Y

P

Q

x1 x2

f Hx1Lf Hx2L y=fHxL

Dx=x2-x1

Dy=f Hx2L-f Hx1L

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∆ .

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

xfxfxxyy

xy

ofanytoeqaulisxxtoxxfromxfofchangeofrateaverageThe

sec)()(

)(

12

12

12

12

21

=−−

=−−

=∆∆

==

Examples:

1. Find the average rate of change of the function )2()( 2 xCosxf = from 0=x to 3/π=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 π=x . Then the average rate of change of )(xf from 1xx = to 2xx = is given by:

πππππ

ππ

49

)3/()4/3(

)3/()1()2/1(

)3/()0()3/2(

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

12

12 −=−

=−−

=−

=−−

=−− CosCosff

xxxfxf

(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:

hxfhxf

xhxxfhxfQuotientDifference )()(

)()()( −+=

−+−+

=−

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:

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

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 2=x , 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 2=x . In the interval notation, we can write this domain as ),2()2,( ∞−∞ U .

Here are two very useful principles:

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

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

Examples:

1. Find the domain of 162

)( 2 −=

xxxf . 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

81620162

2

2

2

±=∴

=

=

=−

x

xx

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,( ∞−−−∞ UU

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. 0≥T . 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!!

37

33

−−

≤−− x

37

≤∴ 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 4 X

-4

-2

2

4

Y

-4 -2 2 4X

5

10

15

20

25Y

-3 -2 -1 1 2 3X

-20

-10

10

20

Y

-3 -2 -1 0 1 2 3 X

1

2

3Y

0 1 2 3 X

0.5

1

1.5

Y

1 2 3 X0.5

1

1.5Y

(g) x

y 1= (h) xy 2= (i)

x

y

=

21

(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

8Y

-3-2-1 0 1 2 3X

2

4

6

8Y

-2 -1 1 2X

1

2

3

4

5

6

7

Y

2 4 6 8X

-1

12Y

-1.0 -0.5 0.5 1.0X

-1.0

-0.5

0.5

1.0Y

-6 -4 -2 2X

-3

-2

-1

1

2Y

(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

><≤+

<+=

418422

22)( 2

xifxifx

xifxxf

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

Solution:

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

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

(b) Here 3=x 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. 4=x does not satisfy either inequality 42 <≤ x or

4>x . 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 4=x . See the graph in part (d) below.

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

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

2+= xy 22 += xy 18=y

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

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

lines. The third graph is cut by the vertical line 4=x , 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

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 +−=+−−=−−=−−−=+−−

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

++++=

+++++=

+++++=

+++++=

+++++

xxxxxxxxx

xxxxxxxxxx

xxxxx

5. VERY IMPORTANT (LEARN THIS!!)

Simplify the following expression, and leave the answer in the factored form: 2292x

−

Solution:

22

2

22 2)32)(32(

294

29

12

292

xxx

xx

xx+−

=−

=−=−

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

Simplify the following expression completely: 2/122/12 )4()2.()4(21. −+− − xxxx

Solution:

4)2(2

442

)4()4(

1)4(

)4(

)4()4(

)4()2.()4(21.

2

2

2

2

2/12

22

2/12

2/12

2

2/122/12

2

2/122/12

−

−=

−

−=

−−+

=

−+

−=

−+−

=

−+− −

xx

xxx

xx

xx

x

xx

x

xxxx

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

Simplify the following expression completely: )9(

)2()9(21.2.)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(21.2.)9(

xxx

xxxx

xxxx

xx

xxxxx

xxxxx

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

2=c .

222

2242

84)1(2

)2)(1(4)4(42

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”

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 o30 o45 o60 o90

θSin 0 21

21

23

1

θCos 1

23 2

1 21

0

θTan 0 3

1 1 3 undefined

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

Examples:

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

33

22 323 πππ CosTanCos −

+

Solution:

419)1(4

313

812)1(4

313

212

)3(46

33

22

323

323

=−−

+

−=−−

+

−=

−

+

πππ 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) θSin is zero for any angle θ that falls on the x-axis. In other words, this happens for

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

following:

πθθ nifonlyandifSin == 0 where n is any integer

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

,....2/5,2/3,2/,2/,2/3....., πππππθ −−= . All these values are odd integer multiples of 2/π . Thus we

can conclude the following:

20 πθθ nifonlyandifCos == where n is any odd integer.

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

πθθ 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:

1. Solve 034 2 =−θCos where πθ 20 <≤ .

Solution:

34 2 =∴ θCos

432 =θCos

23

±=θCos

We can treat the above like two separate equations, 2/3−=θCos and 2/3=θCos and solve them. For both of these equations, the reference angle is o30 or 6/π radians. (i) To solve 2/3−=θCos , observe that θCos is negative in quadrants II and III. Therefore, the required solutions between 0 and π2 are 6/5π and 6/7π . (ii) On the other hand, to solve 2/3=θCos , observe that θCos is positive in quadrants I and IV. Therefore, the required solutions between 0 and π2 are 6/π and 6/11π . 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/ ππππθ and= .

2. Solve 04)2( =+ θθ SinSin where πθπ 42 <<− Solution: Use the Double Angle Formula θθθ CosSinSin 2)2( = to convert the entire right-hand side of the equation to the single angle θ , as follows:

042 =+ θθθ SinCosSin 0)2(2 =+∴ θθ CosSin 020 =+= θθ CosorSin

20 −== θθ CosorSin

, for anyθ , θCos is always between -1 and 1, so the equation 2−=θCos doesn’t have any solutions. Thus, only solutions for the original equation can come from 0=θSin . However, in example 2 of section 8 we learned that 0=θSin only happens when the angleθ lies along the x-axis. Therefore, all the solutions strictly between π2− and π4 are ππππ 3,2,,0, and− .