MA 109 College Algebra Chapter 1.pdf

7/28/2019 MA 109 College Algebra Chapter 1.pdf

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Chapter 1: Algebra and Geometry Review

1.1 Algebra

Algebra is a part of mathematics which solves problems by representing quantities by symbols (often called

variables), expressing the relationships between the quantities as equationsor inequalities, and manipulating

these expressions according to well defined rules in order to find additional properties of the quantities and

solve the problem.

This course course assumes that you have already had considerable experience in using algebra, and that thematerial in this chapter is, for the most part, simply review. The remainder of this section will deal with rules for

manipulating algebraic expressions, for solving linear and quadratic equations in one unknown, for solving

systems of equations in more than one unknown, and for solving inequalities.

1.1.1 Simplifying Expressions

We begin with a review of the basic rules for simplifying algebraic expressions. You probably already know

hundreds of these, and the purpose here is point out those that we think are the most important. This will

reduce the number of such rules down to a few dozen; in subsequent chapters, we will see that even this short

list can be pared down to a few fundamentals from which all the rest follow.

Some really fundamental rules are:

Commutativity of Addition and Multiplication: , and . Note that subtraction and

division are NOT commutative.

1.

Associativity of Addition and Multiplication: , and . Because

of the Associative law, one often simply omits the parentheses: we write a + b + c because the value of

the expression is the same regardless of how it is parenthesized.

2.

Multiplication distributes over addition: , and . Does addition

distribute over multiplication?

3.

Additive and Multiplicative Identities: and .4.

Inverses: For every number a there is a number denoted -a such that . Furthermore, if

, then there is a number denoted such that .

5.

The following examples show how these three fundamental rules can be combined to simplify more complicated

expressions:

Example 1: You can distribute over longer sums by applying the distributive law multiple times:

Example 2: One can multiply out products of arbitrary sums using the Example 1:

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Operator Precedence: In the above expressions we have been omitting parentheses. For example, we write

ab + ac which mightbe interpreted as a(b + a)c, but we know that the intention is (ab) + (bd). We know this

because of the following precedence rules:

Parenthesized expressions are computed first.1.

Powers are computed next.2.

Unary negation is applied next.3.

Multiplication and Division are next in precedence.4.

Addition and subtraction have least precedence and are done last.5.

Within the same precedence level, binary operations are done left to right except for exponentiation.

Successive exponentiations should be completely parenthesized to avoid any possibility of ambiguity.

6.

Note: Depending on the context, multiple exponentiations without parentheses might be interpreted as being

grouped left to right, righjt to left, or as simply being syntactically incorrect. The best practice is to always use an

explicit parenthesization to avoid any possible misinterpretation.

Example 3: Be careful to evaluate operators of equal precedence from left to right. For example,

and not 3/8.

Negatives: These are the main properties of the negation operator (unary minus):

Property Example

Fractions: These are the main properties of fractions. In all the formulas, one assumes that the quantities in the

denominators are all non-zero.

Property Example

if and only if

Powers: One can raise arbitrary real numbers to integer powers. One defines and, by induction,

. For negative integers , one defines . For non-negative real numbers and positive

integers , one can define the root to be unique real number such that . This allows one to define


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rational powers of non-negative numbers by . The following properties are true for these

rational powers:

Property Example

1.1.2 Solving Equations

The last section was concerned with the problem:

Given the values of some quantities, how do you calculate expressions involving those quantities?

Most of the course will be concerned with the inverse problem:

Given the value of some expressions involving quantities, how do you find the values of the

quantities.

The two principal arithmetic operations are addition and multiplication. Here is a problem of the second type:

Problem 1: Find two numbers given the values of their sum and product.

The solution of this problem was already known to the Babylonians, and is one of the most important algebra

problems known to them. How can we solve it?

First let use represent the two numbers by the letters x and y and represent their sum and product by the letters

a and b. The problem can then be expressed as: Given a and b, find x and y such that

Now there are lots of pairs x and y such that x + y = a. One possibility is to take two equal numbers, this would

give x = y and our equation becomes 2x = 2y = a. So, x = y = a/2. Now, if , we would have

our solution.

On the other hand, if this weren't the case, we would not have the solution. This would be the case where the

two numbers x and y are not equal. We can think of this is x is different from a/2 and so x = a/2 + t for some

number t. Since we still want x + y = a, increasing x by t means that we have to decrease y by the same

amount, i.e. y = a/2 - t. Now, let's try this as the solution by putting it in the second equation:

Although it was not clear before how to choose the exact value for t, this last equality tells us that

and so . We know the square of the number t we need, and so

. But then and . Another solution is

obtained by letting t be its second possible value, this simply switches the values of x and y.

Problem 1 is the most important algebra problem solved in antiquity. The approach we have taken is that of

Diophantus of Alexandria. The presentation was very quick; so let us go back and comment on a number ofimportant points:

From the statement of the problem, one might have assumed that there was a single answer -- in fact,

we seem to have come up with multiple answers, both of which are (hopefully) equally valid.

i.

Logically, we started off by assuming that there was a solution given by x and y and then deduced whatii.


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the values of x and y must be. We know that the problem can't have any other solutiosn, but we do not

yet know that the values of x and y are solutions. So, check this:

The verification for the second solution is analogous -- or one could note that it follows by commutativity

of addition and multiplication.

But is this enough? We had been computing without being concerned about whether or not the

operations we were doing actually made any sense. In some cases, they may not: notably, we found the

square root of . When , there is a problem. There is no real number whose

square is negative. There are two ways of resolving this:

At that point in the argument, note that if , then there is no t of the required type

and so there can be no solution to the problem.

a.

Extend our number system to include numbers whose squares are negative. The extended

number system is called the complex numbers.

b.

Both of these approaches will be used. It is convenient to have a notion of number in which we canproceed to take square roots with abandon, i.e. without first checking to see if the number is

non-negative. A portion of the course will be devoted to showing that one can indeed extend the domain

of numbers in this manner. On the other hand, one needs to distinguish the types of numbers -- although

one computes with the complex numbers, the solution may be required to be a real number, or may have

even more stringent conditions, like it may be required to be a positive integer.

iii.

So, we typically have two values of t or none. There is also the possibility that there be exactly one value

of t -- this would occur if . So, even though the expressions for (x, y) given for the two

values of t appeared different, they might actually correspond to the same numbers.

iv.

Let's go back to our original problem: Given a and b, find x and y such that

One could have interpreted the first equation as x = a - y and substituted into the second equation to get

. This can be rewritten as . Once one has y, it is easy to obtain x as

x = a - y and so our problem is reduced to

Problem 2: Given a and b, find y such that

Our original problem is symmetric in x and y. So, we could just as well have solved for y = a - x andsubstituted into the second equation and seen that x satisfied exactly the same quadratic equation as we

have seen y satisfies.

On the other hand, suppose that y is a solution of Problem 2. Then we could set x = a - y. We would then

have x + y = a and . So, we see that solving Problem 2 is equivalent to

solving Problem 1. In other words, solving a system of simultaneous linear equations (Problem 1) is the

same as solving a quadratic equation in one variable (Problem 2). You may have already recognized this

as our solution is exactly what one would expect from the quadratic formula applied to Problem 2.

v.

Again, consider the quadratic equation . What we have seen is that the coefficients a

and b have an interpretation -- a is the sum of the roots and b is the product of the roots.

vi.

1.1.3 Solving Equations in one Variable

Solving systems of equations was the topic of the last section. Let's concentrate in this section on the special


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case in which there is only one equation and only one variable. The next section will return to the more general

case.

There are two basic ideas in working with equations:

If two quantities are equal, then applying the same procedure to each of the quantities yields equal

results.

1.

If a product is zero, then so is at least one of the factors.2.

In other words:

Proposition 1: For all real numbers , , and :

If , then and .i.

If , then either or . Conversely, one has .ii.

Example 4: To solve the general linear equation ax + b = c, we apply the first principle. Assuming that both

sides are equal, we can add -b to both sides to get (ax + b) + (-b) = c + (-b) or ax = c - b. If a is non-zero, then

one can then multiply both sides by to obtain , which simplifies to .

Doing the same operations with numbers, one can start with 2x + 3 = 5. Assuming that both sides are equal, wecan add -3 to both sides to get (2x + 3) + (-3) = 5 + (-3) or 2x = 5 - 3. Since the coefficient 2 is non-zero, one

can multiply both sides by to obtain , or . Now, substitute 1 for x in

the original equation to check to see that it is indeed a solution.

If you want to solve: for x, then, assuming that both sides of the equations are equal, one can

multiply them by to get

So, the solution is x = 1. Again, one should substitute this back into the original equation to check that it is

indeed a solution.

In each example, we started out by assuming that we had a solution, solved to find out what the solution must

have been, and then checked to see that the value actually worked. This last step is NOT just a check for errors

in algebra, but is a NECESSARY step in the procedure. For example, suppose you want to solve:

. Assuming that you have a solution, we can proceed exactly as in the last example:

So, the solution can only be x = 2. But, when you try to substitute this value back into the original equation, you

see that the denominator is zero. So, x = 2 is NOT a solution. This means that there are NO solutions to the


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original equation.

Example 5: Suppose you want to solve . Assume that x is a solution to the equation. The left

side can be factored to get . Using the second part of Proposition 1, we know that either x -

2 = 0 or x - 4 = 0. So, x = 2 or x = 4. Substituting each of these values back into the original equation verifies

that each of two possibilities is indeed a solution of the original equation.

Quadratic Formula: Suppose we want to solve the general quadratic equation: .

If a = 0, then the equation is bx + c = 0. This is a simple linear equation. If b is not zero, then it has a single

solution x = -c/b. If b = 0 and c is not zero, then there are no solutions. Finally, if both b and c are zero, then

every real number x satisfies the equation.

If a is not zero, then we can divide through by a to put the equation in the form . This is

of the form of Problem 2 of the last section. The solutions obtained there were:

(provided that all the operations make sense, i.e. a is non-zero and ).

Completing the square: This is another approach to solving a quadratic equation, and it is a method we will

use in many other problems as well.

Let us assume that we wish to solve the equation . If we could rewrite the equation in the form

, then it would be easy to solve for x. This can be accomplished by the operation of

completing the square. To see how to do it, just expand out this last expression to get .

If this is to be the same as , then corresponding coefficients in the two equations must be

equal, i.e. we must have:

But this system of equation is easy to solve for d and e. We have and .

Let's rework our last example using this method:

Example 5: Solve .

We want to express this in the form . To do this, we take d = a/2 = -6/2 = -3. So, we would get

. Rather than trying to remember the formula for e, we can just multiply out the square to get

. Comparing the constant terms, we need 9 + e = 8 and so e = -1, i.e. .

We can now solve for x, we get and so . So and so the solutions

are x = 4 or x = 2. It is easy to verify that these are both solutions of the original equation.

1.1.4 Solving Systems of Equations

When there is more than one variable, one often has more than one equation which the values of the variables

are required to simultaneously satisfy. For example, in completing the square, we needed to find all solutions of


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the system of equations: and where the variables were d and e, and a, b, and c were

constants. We were not interested in pairs of numbers (d, e) which satisfied just one of the two equations; they

needed to satisfy both equations.

Another example is the general system of 2 linear equations in two unknowns x and y: ax + by = e, cx + dy = f.

Now, in the special case of the system in the last paragraph, it was easy to solve for the variables because one

of the equations involved only one of the variables. We could use it to solve for that variable. Having its value,

we could substitute its value into the other equation and obtain an equation involving only the second variable;

this equation was then solved and we had all the possible solutions of the system.

What makes the general system of 2 linear equations look more difficult is that both equations involve both

variables. There are two approaches:

One could solve the first equation for one of the variables in terms of the other. Then substitute this into

the second equation giving an equation in only the second variable. Then proceed as in the easy case.

1.

One could subtract an appropriately chosen multiple of one equation from the other in order to obtain an

equation involving only one variable.

2.

As before, one needs to check that the possible solutions one obtains do indeed satisfy the original equations.

Example 6: Consider the system: 2x + 3y = 5, 4x - 7y = -3. Assume that x and y are satisfy both equations. One

can proceed using either method:

Solving for x using the first equation, gives x = (5/2) - (3/2)y. Substituting this into the second equation

gives 4((5/2) - (3/2)y) - 7y = -3, which can be used to find y = 1. Substituting this back into our expression

for x yields x = 1. One then checks that the values x = 1, y = 1 do indeed satisfy the original equations.

1.

If one subtracts twice the first equation from the second equation, one gets (4x - 7y) - 2(2x + 3y) = -3 -

2(5) or -13y = -13. This gives y = 1 and substituting this value back into the first equation gives 2x + 3 = 5

or x = 1. As before, we need to check that x = 1, y = 1 does indeed satisfy the original two equations.

2.

Example 7: Find all the solutions of the system of equations: , . Assume that one has a

solution x, y. Solving the second equation for y, one gets a value y = 1 -x which when substituted into the firs

equation gives: or . Collecting terms, we get a quadratic

. Factoring and using the second part of Proposition 1, gives x = 0 or x = 1. Substituting these

values into our expression for y, gives two possible solutions (x,y) = (0, 1) and (x,y) = (1, 0). Substituting each of

these into the original equations, verifies that both of these pairs are solutions of the original system of

equations.

If we have more than two variables and more equations, we can apply the same basic strategies. For example,

if you have three linear equations in three unknowns, you can use one of them to solve for one variable in terms

of the other two. Substituting this expression into the two remaining equations gives two equations in two

unknowns. This system can be solved by the method we just described. Then the solutions can be substituted

back into the expression for the first variable to find all possible solutions. When you have these, substitute

each triple of numbers into the original equations to see which of the possibilities are really solutions. One can

also use the second approach as is illustrated by the next example.

Example 8: Solve the system of equations: x + y + z = 0, x + 2y + 2z = 2, x - 2y + 2z = 4. Assume that (x, y, z)

is a solution. Subtracting the first equation from each of the other two equations gives y + z = 2, -3y + z = 4.

Now subtracting the first of these from the second gives -4y = 2 or y = -1/2. Substituting this into the y + z = 2

gives z = 5/2. Finally, substituting these into the first of the original equations gives x = -2. So the only possible

solution is (x, y, z) = (-2, -1/2, 5/2). Substituting these values into the original equations shows that this possible

solution is, in fact, a solution of the original system.

1.1.5 Inequalities

In addition to the four basic algebraic operations on real numbers, there is also an order relation. The basic

properties are:

Proposition 2: Let a, b, and c be real numbers.

(Trichotomy) For any pair a, b of real numbers, exactly one of these conditions holds true: a < b, a = b, or

a > b.

1.


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(Transitivity) If a < b and b < c, then a < c.2.

If a < b, then a + c < b + c.3.

If a < b and c > 0, then .4.

If a < b and c < 0, then .5.

Just as Proposition 1 allowed one to operate with equations, Proposition 2 allows one to work with inequalities.

The main difference is that multiplication tends to complicate things as there are two cases depending on

whether the multiplier is positive or negative.

Example 9: Solve the general linear inequality ax + b < 0. Using Proposition 2, this implies ax < - b. If a > 0,

Proposition 2 gives x < -b/a. On the other hand, if a < 0, then x > -b/a. Finally, if a = 0, then there are no possible

solutions if or every real number is a solution if and .

We still need to check that our possible solutions are indeed solutions. This is somewhat more difficult because

we do not simply have a small number of values of substitute:

Case 1: If a > 0 and x < -b/a, then Proposition 2 shows that ax < -b and so ax + b < 0. So all the x

smaller than -b/a are solutions of the inequality.

Case 2: If a -b/a, then Proposition 2 shows that ax < -b and so ax + b < 0. So all the x larger

than -b/a are solutions of the inequality.

Case 3: If a = 0 and , there are no possible solutions. On the other hand, if , then any

number x clearly satisfies .

The absolute value |r| of a real number r is defined to be r or -r depending on whether or not r is non-negative

or not. For example, |2| = 2, but |-2| = -(-2) = 2.

Example 10: Find all solutions of |2x - 3| > 5. There are two cases:

Case 1: If , then |2x - 3| = 2x - 3 and so the inequality becomes 2x - 3 > 5. Solving the

inequality gives x > 4. Such x satisfy both and |2x - 3| > 5.

Case 2: If , then |2x - 3| = -(x - 3) and so the inequality is -(2x - 3) > 5. Solving this inequality

gives x < -1. Such x satisfy both 2x - 3 < 0 and |2x - 3| > 5.

So, the solutions are all real numbers x that are either less than -1 or bigger than 4.

1.2 Geometry

Geometry is the mathematical study of the relationships between collections of points, curves, angles, surfaces

and solid objects including the measurement of these objects and the distances between them. This review will

discuss the correspondence between the real numbers and the points of a line, the general idea of analytic

geometry, as well as the equations of lines, circles, and parabolas as well as the measurement of distance

between points.

1.2.1 The Pythagorean Theorem

The most important result of classical synthetic geometry is:

Theorem 1: If is a right triangle with hypotenuse of length c and legs of length a and b, then

.


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To understand its proof, we need to know

Proposition 3: The sum of the angles of any triangle is 180 degrees.

Proof: Let be an arbitrary triangle. Construct a line DE parallel to the base BC and through the third

vertex A as shown in the diagram below.

Since BA is transversal to the pair of parallel lines BC and DE, one has . Similarly, CA is

transversal to the pair of parallel lines BC and DE. So, . Since DAE is a straight line, we have

. From the diagram, we see that

as was to be shown.

The proof of the Pythagorean Theorem will also require some simple facts about the areas of squares and

triangles. For future reference, let's state these and other results as

Proposition 4:(Areas and Perimeters)

The area of a rectangle (or even a parallelogram) is the product of the lengths of its base and its height.i.

The area of a triangle is half the product of the lengths of its base and its height.ii.

The area of a circle is where r is its radius. The circumference (perimeter) of of a circle isiii.


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.

Now it is easy to see why the Pythagorean Theorem is true. Start with an arbitrary right triangle where C

is the right angle and the opposite side (the hypotenuse) is of length c. Let a and b be the lengths of the legs

which are opposite to the angles A and B respectively.

Construct the figure below where the square in the middle has sides of length c. On each of its sides,

place a triangle congruent to using the side of the square as the hypotenuse.

a.

The interior angles of the square are, of course, right angles and the sum of the angles opposite the legs

of each of the triangles is by Proposition 3. It follows that the legs of the triangles at each of the

vertices of the square actually form a straight line segment (because the sum of the three angles there is

precisely 180 degrees. So our figure is actually a large square with sides of length a + b.

b.

Now the large square can also be considered to be the union of the smaller square with side c and four

congruent right triangles with base a and height b. So we can calculate the area of the figure in two ways

(using Proposition 4) and equate the answers: . When you multiply out and

simplify this equation, one finds , as was to be proved.

c.

There are at least hundreds of extant proofs of the Pythagorean Theorem. The proof given here is quite

different from the proof appearing in Euclid's Elements.

Corollary 1: i. The length of the diagonal of a rectangle with sides of lengths a and b is precisely .

ii. Let T be a triangle with sides of length a, b, and c opposite angles A, B, and C respectively. Then T is

a right triangle with right angle C if .

Proof: i. The first assertion is an obvious consequence of Theorem 1.

ii. Let T' be a triangle where angle C' is a right angle and the sides opposite angles A' and B' are of

length a and b respectively. By Theorem 1, the length of the side opposite angle C' is of length .

Then corresponding sides of the triangles T and T' are of equal length; so the two triangles are congruent. But

then angle C must be equal to angle C' and so angle C is a right angle.

We will need to know about similar triangles. Two triangles and are defined to be similar if

, , and . By Proposition 3, if two of these equations hold, then so does the

third.

Proposition 5: If the two triangles and are similar, then the ratio of the lengths of a pair of sides


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of one triangle is equal to the ratio of the lengths of the corresponding pair of sides of the other triangle, e.g.

.

1.2.2 Analytic Geometry

The rough idea of analytic geometry is to model the plane as the set of all pairs of real numbers. A curve is then

the set of solutions of some algebraic equation. In order to solve a geometric problem, one first translates it into

an algebra problem about the sets of algebraic equations. Then one uses algebra to solve this problem, and

finally translates the answer back into geometric terms.

In the case of a line, one can model it as the set of real numbers.

Given the line L, choose two points called 0 and 1. Make the distance between 0 and 1 our unit length,

so the distance between 0 and 1 is one unit. From 1, mark off another unit distance to determine a new

point which we will call 2. Repeat the process to get 3, 4, etc. Then mark off -1, -2, etc. in the opposite

direction. Since we know the distance between 0 and 1, one can construct line segments of length 1/n

for

1.

For every positive integer n, one can construct a segment of length 1/n. To do this, construct a right

triangle with one leg AC of length 1, with right angle at C, and with hypotenuse of length n. Mark

off a point D on the hypotenuse such that the distance from A to D is 1 unit. Drop a perpendicular from D

to the leg AC. The point of intersection of the perpendicular and AC is denoted E. Then is similar

to . Using Proposition 5, one sees that AE is of length 1/n.

2.

Again starting from 0, one can mark off points 1/n, 2/n, etc. as well as points -1/n, -2/n, etc. just as we

labeled the integer points. Since we can do this for all positive integers n, we now have all the rational

points labeled.

3.

Any real number is determined by its (infinite) decimal expansion. By taking more and more digits of the

decimal expansion we get rational numbers which approach the real number. By taking the limiting point

of the corresponding points on the line, we get the point on the line which corresponds to the given real

number. This completes our mapping of the real numbers to the points of the line L.

4.

To handle the plane, start with two perpendicular lines in the plane. Their point of intersection is denoted 0. The

first line is called the x-axis and the second line is called the y-axis. Using the same unit distance as before,one can map the real numbers onto each of the two lines. With any pair (a, b) of real numbers, we can

associate a point in the plane: From the point marked a on the x-axis, erect a line perpendicular to the x-axis.

Similarly, from the point marked b on the y-axis, erect a line perpendicular to the y-axis. The point of intersection

of these two lines is the point associated with (a, b).

Assumption 1: The above mapping is a 1-1 and onto mapping between the set of all pairs (a,b) of real

numbers and the points of the plane.

Because of this assumption, we will usually not distinguish between the ordered pair (a, b) and its

corresponding point in the plane. In particular, we will refer to the point as being the ordered pair (a,b).


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Remark: By using three pairwise perpendicular lines intersecting at a point 0, one can name points in three

dimensional space as simply triples (a, b, c) of real numbers. Of course, one could then define n-dimensional

space, as simply being the set of ordered n-tuples of numbers.

Definition 1: The distance between the points P = (a,b) and Q = (c, d) is defined to be .

This is called the distance formula.

Although the formula is a bit complicated, this is the obviousdefinition as you can see by examining thediagram below in which PQ is simply the hypotenuse of a right triangle whose legs are of length |c - a| and |d -

b| respectively. In light of the Pythagorean Theorem the formula for the length of this hypotenuse is the value

given in the definition.

Example 11: The points on either axis labeled a and b are of distance |b - a| from each other -- where distance

is calculated by Definition 1.

Example 12: The distance between the points (3, 5) and (2,7) is .

Translation of Graphs If the point (x,y) satisfies the equation f(x, y) = 0, and if A and B are real numbers, then

the point (x + A, y + B) satisfies the equation f(x - A, y - B) = 0 (because f((x+A) - A, (y+B) - B) = f(x, y) = 0).

Another way of saying this is: if the graph of a function y = f(x) is moved A units to the right and B units to the

right, then one obtains the graph of y - B = f(x - A). This amounts to the simple rules:

Motion Change in Equation

Move Graph A units to right Replace x with x - A

Move Graph A units to left Replace x with x + AMove Graph B units upwards Replace y with y - B

Move Graph B units down Replace y with y + B

Graph Compression and Expansion If the point (x,y) satisfies the equation f(x, y) = 0, and if A and B are

non-zero real numbers, then the point (Ax, By) satisifes the equation f(x/A, y/B) = 0 (because f((Ax)/A, (By)/B) =

f(x, y) = 0). Another way of saying this is: if the graph of a function y = f(x) is expanded by a factor of A

horizontally and by a factor of B vertically, then the result is the graph of y = Bf(x/A). Again, you can give simple

rules:

Expansion or Compression Change in Equation

Expand Graph horizontally by a factor A Replace x with x/A

Compress Graph horizontally by a factor A Replace x with Ax

Expand Graph vertically by a factor B Replace y with y/B

Compress Graph vertically by a factor B Replace y with By


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Reflecting a graph across an axis If the point (x, y) satisfies the equation f(x, y) = 0, then the point (-x, y)

obtained by reflecting the point (x, y) across the y-axis satisfies f(-x, y) = 0. Similarly, the point (x, -y) obtained

by reflecting the point (x, y) across the x-axis satsifes the equation f(x, -y) = 0.

1.2.3 Lines, Circles, and Parabolas

Lines, circles, and other curves like parabolas should simply be the sets of solutions of certain algebraic

equations. This appears to be the case. For example, we can define a vertical line to be the set of solutions of

some equation x = a, where a is a constant. Similarly, a horizontal line is the set of solutions of an equation ofthe form for some constant b.

Definition 2: A line is either a horizontal or vertical line or else it is the set of solutions of an equation of the

form y = mx + b where m and b are constants.

The equation y = mx + b (or x = a) is called the equation of the line which is the set of solutions of the

equation. Note that this is a 1-1 correspondence between the set of lines and the set of equations of lines. In

particular, different equations correspond to different lines; in fact, you can easily convince yourself that through

any two points there is precisely one line. The constant m is called the slope of the line and the constant b is

called the y-intercept of the line. Note that the y-intercept is the y-coordinate of the point where the line

intersects the y-axis. The slope and y-intercept of a vertical line is not defined. There are many algebraically

equivalent forms of the equation of a line; the particular one y = mx + b is called the slope-intercept form ofthe equation of the line.

If for i = 1, 2 are two distinct points on the line y = mx + b, then one has for i = 1, 2

and, taking differences, one gets

Solving for m gives:

Proposition 6: The slope of the non-vertical line through and is given by .

Example 13: The equation of the line through the two points and is given by

This is the so-called 2 point form of the equation of the line. It is proved by substituting in the two points.

Example 14: The equation of the line with slope m which passes through the point is

. This is called the point-slope form of the equation of the line.

Example 15: Consider the line through the two points (2, 3) and (5,7). Its slope is . Its

equation in point-slope form is (or equivalently ). To find the

y-intercept, just substitute x = 0 into either of these equations and solve for y; the y-intercept is 1/3. So the

slope-intercept form of the equation of the line is y = (4/3)x + 1/3. To find points on this line, just substitute in

arbitrarily chosen values of x and solve for the y-coordinate.

Two lines which have no point of intersection are said to be parallel.

Example 16: Two distinct lines are parallel if and only if they are either both vertical or else they both have the

same slope.

Suppose the two lines are not vertical. Let and be their equations and let's assume

(x, y) is a point of intersection. Then it satisfies both equations. Equating them, we get

and so . If the slopes were unequal, the coefficient of x would be non-zero and we could


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solve for and substitute this back into either of the original equations to obtain the value for y. If the

slopes are not equal, it is easy to then verify that this pair is indeed a point of intersection, i.e. satisfies both

equations. On the other hand, if the slopes were equal, then we would have which

would mean that the lines were not distinct (because they have the same equation).

In case the two lines are vertical, the equations are of the form and where . Clearly, no

pair (x, y) could satisfy both of these equations; so distinct vertical lines are parallel.

Finally, if one line has the equation and the other the equation , then the point (a, ma + b) is a

point of intersection. This completes the proof of the assertion.

We say that two lines are perpendicular if they intersect at right angles to each other.

Example 17: Two lines are pendendicular if and only if one of the following is true:

One line has slope 0 and the other is vertical.1.

Neither line is vertical and the product of the slopes of the lines is -1.2.

It is easy to verify the result in case either of the two lines is vertical, either of the two lines is horizontal, or if the

lines are are parallel or coincident. So, assume that none of these conditions are true. Then the two lines have

slopes and their equations can be written for i = 1, 2 where is the point of

intersection of the two lines. The lines have points A and B with x-coordinates, say, a + 1. Using the equations

of the lines, we see that the points are and . By the Pythagorean

Theorem, the lines are perpendicular if and only if . Using the distance formula, this

amounts to

If you multiply out and simplify this expression, you will see that it is equivalent to .

Remark: The last example shows both the power and danger of analytic geometry. There was no

understanding involved, the result came from brute algebraic computation.

A circle with center (a, b) and radius r is the set of points (x,y) whose distance from (a, b) is precisely r. By the

distance formula, this circle is the set of solutions of

The set of solutions of an equation of the form where , b, and c are constants is called

a parabola. The following properties are easy to verify:

If a > 0, the parabola opens upward; if a < 0, it opens downward.1.

If a > 0, the lowest point on the parabola (i.e. the one with smallest y-coordinate) is at (b, c). This point is

called the vertex of the parabola.

2.

The parabola is symmetric about its axis x = b (i.e. For every real number z, the points on the parabola

with x-coordinates have the same y-coordinates).

3.

If you interchange the x and y variables, one obtains the equations of parabolas which open to the right and left

instead of up and down.

Example 18: Find the equation of the line which passes through the two points of intersection of the circle

centered at the origin of radius 1 and the circle centered at (1,1) with radius 1.

The equations of the two circles are and . If (x, y) is a point of

intersection, it must satisfy both equations. Expanding out the second equation, one gets


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. Subtracting the first equation and simplifying gives the equation x + y = 1. If we knew

that there were two points of intersection, then they would both satisfy this equation and it is the equation of a

line; so it must be the desired line. (In order to verify that there are indeed two points of intersection, finish

solving the system of equations and verify that the two solutions satisfy the original equations.)

1.2.4 Conic Sections

Let L be a non-vertical line through the origin. By rotating the line L around the y-axis, one obtains a pair of

infinite cones with vertex at the origin. By intersecting this pair of cones with various planes, one obtains curvescalled ellipses, parabolas, and hyperbolas as well as some degenerate cases such as a single point or a single

line. These intersections were called sections and so ellipses, parabolas, and hyperbolas were referred to as

conic sections, i.e. sections of a cone.

Nowadays the importance of conic sections is not that they arise by intersecting a cone with a plane, but rather

that they can be used to categorize all curves whose equations are polynomials of degree 2. So, lines are the

curves represented by equations which are polynomials of degree 1, and conic sections are the curves

represented by equations which are polynomials of degree 2.

In this section, we will define ellipses, parabolas, and hyperbolas without reference to sections of cones, and

obtain equations for each in some special cases.

1.2.4.1 Parabolas

We have already defined parabolas to be the graphs of functions of the form . Let us give a

more traditional definition:

Definition 3: Let F be a point and D be a line in the plane which does not contain F. Then the parabola with

focus F and directrix D is the set of points P = (x, y) such that the distance from P to F is the same as the

distance from P to D. (By the distance from P to the line D, we mean the shortest distance from P to any point

of D, i.e. the length of the line segment obtained by dropping a line through P perpendicular to D.)

Now, let's look at a special case in which the vertex F = (0, d) and the line D is y = -d. If P = (x,y) is on the

parabola with focus F and directrix D, then the distance formula tells us that

When you multiply this out and collect terms, one sees that this is just or .

Given the parabola, , we see that a = 1/(4d) and so the focus is at (0, d) = (0, 1/(4a)) and the directrix is

y = -d = -1/(4a). Where is the focus and directrix of ?

1.2.4.2 Ellipses

One could define ellipses and hyperbolas in a manner exactly analogous to Definition 3, viz.

Definition 3a: Let F be a point and D be a line in the plane which does not contain F and e be a positive real

number. Then the conic section with focus F, directrix D, and eccentricity e is the set of points P = (x, y)

such that the distance from P to F is equal to e times the distance from P to D. If e = 1, the conic section is

called a parabola. The conic section is called an ellipse if e < 1 and a hyperbola if e > 1.

Such a definition leads to rather complicated formulas, and so, instead, we will define

Definition 4: Let and be two not necessarily distinct points in the plane and let a be a positive real

number. Then the ellipse with foci and and semimajor axis a is the set of points P = (x, y) such that

the sum of the distances from P to the foci is exactly 2a.

Clearly, if a is less than half the distance between the foci, then the ellipse with semimajor axis a is the empty

set.


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Again, let us consider a special case: Let and where is a real number. Let a

be a real number greater than c. Then if P = (x, y) is a point of the ellipse with these points as foci and with

semimajor axis a, then we have by the distance formula:

This is the equation of the ellipse, but it is better to simplify it a bit. To do so, subtract the second square root

from both sides and then square both sides to get:

Moving all the terms except for the square root from the right side to the left side and simplifying gives

Dividing both sides by 4 and squaring again gives

which can be re-arranged to get

Finally, since a is larger than c and both are positive, we can let b be a positive number such that .

Substituting this into our last formula and dividing both sides by gives us the standard form of the

equation of the ellipse:

The quantity e = c/a is called the eccentricity of the ellipse. The quantity b is called the semiminor axis of the

ellipse. In particular, if the eccentricity of the ellipse is 0, then the two foci coincide, the semimajor and

semiminor axes are equal, and the ellipse is a circle whose radius is precisely the semimajor axis.

Of course, one can also have ellipses with a vertical semimajor axis. If we simply interchange the roles of x and

y in the above calculation, the final formula is the same, except that quantities a and b are interchanged. You

distinguish the two cases by looking to see which of the two is the larger. You can also translate the graph of the

equation to obtain ellipses centered at points (a, b) instead of at the origin (0, 0). As usual, the formulas look the

same except that x is replaced with x - a and y is replace by y - b. Expansion and compression either

horizontally or vertically just expands or contracts the values of a and b.

1.2.4.3 Hyperbolas

In analogy with Definition 4, we have

Definition 5: Let and be two not necessarily distinct points in the plane and let a be a positive real

number. Then the ellipse with foci and and semimajor axis a is the set of points P = (x, y) such that

the difference of the distances from P to the foci is exactly 2a. (Note that we have to subtract the smaller of two

distances from the larger one in order to get 2a.)

Taking the foci (0, -c) and (0, c) as before, it is easy to use the distance formula to write down the equation ofthe hyperbola. You should go through steps just as with the ellipse. When you are done, you will see that the

equation of the hyperbola reduces down to


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where b is a positive number such that . As before, we call the eccentricity of the

hyperbola. Note that e > 1 for hyperbolas and less than 1 for ellipses.

1.2.5 Trigonometry

Angle Measurement: Angles are measured in radians. Start with the unit circle centered at the

origin (0, 0). Make one side of the angle the positive x-axis. The other side of the angle is another ray from the

origin, say 0A in the diagram below. The magnitude of the radian measure of the angle is equal to the length of

the arc of the circle swept out as the positive x-axis is rotated through the angle to the ray 0A. The radian

measure is positive if the arc is swept out in a counterclockwise direction and is negative otherwise. If we take

instead of the unit circle, the circle centered at the origin of radius r, then the arc swept out by the same action is

of length where is the radian measure of the angle.

Again referring to the diagram above, if A is the point on the unit circle corresponding to the second ray of the

angle, then its coordinates are by definition the cosine and sine of the angle , i.e. .

Since A lies on the unit circle, we have the identity:

. Further, it is clear by the symmetry of the circle that we have:

If B is the point on the ray 0A where the ray intersects the circle of radius r: , then is equal

to the ratio of the x-coordinate of B and the the hypotenuse r (because of similar triangles). This explains the

common definition of the cosine as being the ratio of the adjacent side to the hypotenuse; when using this, be

careful, that that the adjacent side may be either a positive or negative number. Similarly, the sine of the same

angle is the quotient of the opposite side over the hypotenuse (where the opposite side might be negative).

These definitions allow one to create tables of trigonometric functions for practical use in solving right triangles.

For example, Ptolemy computed tables in half degree increments.


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The remaining trigonometric functions are:

Example 19: Dividing the identity , by and using the above definitions, one

obtains the identity:

Similarly, by dividing by , one obtains the identity:

Example 20: Let be the angle opposite the side of length 3 in a right triangle whose sides are of length 3, 4,


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and 5. Then the trigonometric functions of are:

The values of the trigonometric functions at a number of commonly occurring angles are given in the table

below:

Degrees Radians sine cosine tangent cotangent secant cosecant

0 0 0 1 0 undefined 1 undefined

30 1/2 2

45 1 1

60 1/2 2

90 1 0 undefined 0 undefined 1

Rather than memorize the table, it is easier to keep in mind the figure above used to define the trig functionsand reconstruct the two standard triangles shown below.


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There are many more important results from trigonometry. The most important are:

Proposition 7:(Addition Formulas) For arbitrary angles and , one has:

.i.

.ii.

and

Proposition 8: Let be any triangle and a, b, c be the lengths of the sides opposite angles A, B, and C

respectively. Then

( Law of Cosines) .i.

(Law of Sines) .ii.

You will have an opportunity to prove these results in the exercises.

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Revised: July 12, 2001

All contents copyright 2001 K. K. Kubota. All rights reserved


MA 109 College Algebra Chapter 1.pdf

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