Arithmetic Operations

The simplification of algebraic expressions with radicals will involve the simplification and combination of the quantities within the radical sign. To add and subtract radicals, one must ensure that the index and the radicand have the same value.

In multiplying and dividing radicals, one must ensure that the indices of the radicals have the same value, in order to use the algebraic laws governing radicals.

The distributive property is used when addition and multiplication are combined in expressions involving radicals.

Exponents and Radicals

The concept of exponentiation, which represents "repeated multiplication" for integer exponents, can be extended to the real numbers. The following laws for exponents apply to all real numbers (and thus to all integers):

(1) x^mxⁿ = x^m+n

(2)
x^m = x^m-n
xⁿ

(3) (x^m)^m = x^mn

(4) (xy)^m = x^my^m

(5) (a/b)^m = a^m/b^m

(6) x⁰ = 1

(7) b^-n = 1/bⁿ

 

EX. x² · x⁴ = x²⁺⁴ = x⁶

 

[(a^-2)(b³)]² = [(a^-2)²(b³)²] = (a^-4)(b⁶)

(a/b)³ = a³ / b³

b^-2 = 1/b²

Exponentiation involving fractional exponents gives rise to the concept of roots and radicals. The relationship between exponents and roots is expressed as follows:

In the above equation, b is known as the radicand, n is the index, and the expression on the right-hand side is the nth root of b. The 2nd root of a number is known as its square root, while its 3rd root is known as its cube root.

EX. 25^1/2 = ±5

If we set a to be the nth root of a real number b, then

a = b^1/n

aⁿ = b^(1/n)n = b¹ = b

The roots of real numbers may be either real or complex numbers. In particular, the nth root of a negative radicand, where n is even, has to be a complex number, since the nth power of any real number, where n is even, has to be a positive number. We will restrict our discussion of exponents and roots to real number solutions.

Given that n is an even integer,

(-b)ⁿ = (-1·b)ⁿ = (-1)ⁿ bⁿ = bⁿ

Therefore, if the nth root of a real number, where n is even, can assume a value a, it can also assume the value -a. To avoid confusion, we define the principal nth root of a real number, where the nth root is a real number, to be the positive nth root of the number. When an algebraic expression addresses the nth root of a number, and the root is a real number, it is often customary to refer to the principal (positive) nth root of that number.

EX. Since 4 ² = (-4)² = 16, the square root of 16 can have the value 4 or -4. Its positive square root is 4. That is,

16^1/2 = 4

By applying the laws of exponentiation, we get the following equality:

By letting m = n

b^m/m b¹ = b

The nth root of a product of two numbers is evaluated as

EX.

Similarly, the nth root of a quotient of two numbers is evaluated as

EX.

The process of factoring a real number involves expressing the number as a product of prime factors. Because a prime number has only two factors, the number 1 and the prime number itself, they are considered to be irreducible factors. Similarly, one factors a polynomial expression by representing it as a product of irreducible polynomials, i.e., polynomials that cannot be further reduced into other factors aside from the number 1 and itself.

EX.
12 = 2 × 2 × 3

x² + x = x (x + 1)

15x²y + 18xy² = 3xy (5x + 6y)

x³ + 5x² = x² (x + 5)

xy + 2y + 7x + 14
= y(x + 2) + 7(x + 2)
= (y + 7) (x + 2)

The quadratic expressions considered herein are of the form

ax² + bx + c

where x is the (only) variable, and a, b and c are nonzero real numbers.

When a = 1, the quadratic expression simplifies to

x² + bx + c

If we set x² + bx + c equal to the product of x + c₁ and x + c₂, then

x² + bx + c = (x + c₁) (x + c₂)
= x² + (c₁ + c₂)x + c₁c₂

and thus b = c₁ + c₂ and c = c₁c₂.

Using this line of thinking, we can formulate a procedure for (possibly) factoring the quadratic expression x² + bx + c :

(1) Find two real numbers whose sum is equal to b and whose product is equal to c.
(2) If two real numbers satisfying the above criteria are found, and we denote them as c₁ and c₂ , then (x + c₁) and (x + c₂) are factors of x² + bx + c.

In most cases, b and c are integers, so c₁ and c₂ will also be integers.
Obviously, there are quadratic expressions of the type x²+ bx + c for which no two real numbers can be found, whose sum is equal to b and whose product is equal to c. Either they don't exist, or they are not easily determined (especially when they are not integers). In this case, a different procedure for factoring is called for.

EX.
Factor x² + 7x + 12

First, list the combinations of integers whose product is equal to 12.

They are

12 and 1 ( 12 · 1 = 12 )
6 and 2 ( 6 · 2 = 12 )
4 and 3 ( 4 · 3 = 12 )

Then find the sum of these combinations:

12 + 1 = 13
6 + 2 = 8
4 + 3 = 7

The integers 4 and 3 have a sum of 7 and a product of 12.
\ The factors of x²+ 7x + 12 are ( x + 3 ) and ( x + 4 ).

EX.
Factor y²- 3y - 10

The combinations of integers whose product is -10, and their sums are

10 · -1 = -10 10 + ( -1 ) = 9
5 · -2 = -10 5 + ( - 2) = 3
-10 · 1 = -10 -10 + 1 = -9
-5 · 2 = -10 -5 + 2 = -3

The integers -5 and 2 have a sum of -3 and a product of -10.
\ The factors of y²- 3y - 10 are ( y -5 ) and ( y + 2 ).

The quadratic expression ax²+ bx + c can be rewritten as

If b/a and c/a are integers, then the quadratic expression

can be factored using the above procedure. Oftentimes, though, this is not the case. Thus, a different procedure is utilized to factor this quadratic expression.

The quadratic expression ax²+ bx + c can be algebraically manipulated as follows:

ax²+ bx + c = ax²+ (b₁ + b₂)x + c
= ax²+ b₁ x + b₂x + c

If b₁ b₂ = ac, then (b₁ b₂ / a) = c, and thus

Therefore, the following procedure can be used to factor the expression
ax²+ bx + c:

(1) Find two real numbers b₁ and b₂ whose sum is b and whose product is equal to the product of a and c.
(2) If two real numbers b₁ and b₂ are found satisfying the above criteria, then

are factors of the expression ax²+ bx + c and may be further simplified.

This procedure is most useful when a, b and c are integers. Of course, this method will not always work, since there is no guarantee that the numbers b₁₂ can be found. Other factoring techniques will have to be employed in these cases. and b

EX.
Factor 6x²+ 19x + 10.

As a = 6, b = 19 and c = 10, ac = (6)(10) = 60. Some combinations of integers whose product is 60, along with their sums, are

60 · 1 = 60, 60 + 1 = 61
30 · 2 = 60, 30 + 2 = 32
20 · 3 = 60, 20 + 3 = 23
15 · 4 = 60, 15 + 4 = 19

The two numbers 15 and 4 have a sum of 19 and a product of 60.
Letting b₁ = 15 and b₂ = 4, we can express 6x² + 19x + 10 as the product of

Therefore, 6x² + 19x + 10
= (1.5x + 1) (4x + 10)
= (1.5x + 1) 2 (2x + 5)
= (3x + 2) (2x + 5)

EX.
Factor 10x+ 7x - 12.

As a = 10, b = 7 and c = -12, ac = (10)(-12) = -120.
Some combinations of integers whose product is 120, along with their sums, are

-10 · 12 = -120, -10 + 12 = 2
10 · -12 = -120, 10 + ( -12) = -2
-15 · 8 = -120, -15 + 8 = -7
15 · -8 = -120, 15 + ( -8) = 7

The two numbers 15 and -8 have a sum of 7 and a product of 120. Letting b₁ = 15 and b₂ = -8, we can express 10x²+ 7x - 12 as the product of

Therefore, 10x+ 7x - 12 = (-1.25x + 1) (-8x - 12)
= (-1.25x + 1) -4 (2x + 3)
= (5x - 4) (2x + 3)

Rational Expressions Rational expressions are represented as the quotient of two algebraic expressions. Thus, they can be manipulated like fractions. The following properties of fractions can be used to deal with rational expressions: (1) For any real numbers a and b, where b ¹ 0, -a/b = a/-b = -(a/b) (2) For any real numbers a and b, where b ¹ 0, -a/-b = a/b (3) For any real numbers a, b and k, where b ¹ 0 and k ¹ 0, (a·k)/(b·k) = a/b (4) For any real numbers a and b, where a ¹ b, (a-b)/(b-a) = -1 We can use the above properties for fractions to simplify rational expressions: EX.

Since a rational expression is the quotient of two algebraic expressions, it can be represented in fractional form. In doing so, it is often desirable to eliminate all terms involving radicals from the denominator of the fraction. This is known as rationalizing the denominator.

EX.

 

When the denominator is a binomial radical expression, the equality

(a + b) (a - b) = a² - b²

can often be used to rationalize it.

EX.

Two common factoring situations involve (1) the difference between two squares, and (2) the sum or difference of two cubes.

a² - b² = (a + b) (a - b)
a³ + b³ = (a + b) (a² - ab + b²)
a³ - b³ = (a - b) (a² + ab + b²)

EX.
y² - 9 = y² - 3² = (y + 3) (y - 3)
9t² - 4 = (3t)² - 2² = (3t + 2) (3t - 2)
(x + 1)² - y² = [(x + 1) + y][(x + 1) - y]

y³ + 8 = y³ + 2³ = (y + 2) (y²- 2y + 4)
x³ - 1 = x³ - 1³ = (x - 1) (x²+ x + 1)

8s³+ 27 = (2s)³ + 3³ = (2s + 3) (4s²- 6s + 9)
64t³ - 125 = (4t)³ - 5³ = (4t - 5) (16t²+ 20t + 25)

An arithmetic progression is a sequence in which each term (after the first) is determined by adding a constant to the preceding term. This constant is called the common difference of the arithmetic progression. An arithmetic progression can be defined as follows:

The arithmetic progression { a_n } = a₁, a₂, a₃, ...., a_n ,
where n = 1, 2, 3, . . .
Its terms are determined by the equation:

a_n = a₁ + (n - 1)d, where

a₁ is the first term of the arithmetic progression
a_n is the nth term of the arithmetic progression
n is the term number
d is the common difference of the arithmetic progression

The sum of the first n terms of an arithmetic progression is
calculated as

S_n = n ( a₁ + a_n ) / 2

or

S_n = n ( 2a₁ + (n - 1)d ) / 2 where a_n = a₁ + (n - 1)d

EX. For the sequence { a_n } = 1, 3, 5, 7, 9, ..... where a_n = 2n - 1

a_n = 2n - 1 = 1 + 2n - 2 = 1 + 2(n-1)

The sequence { a_n } = 1, 3, 5, 7, 9, ..... is an arithmetic sequence with a₁ = 1 and d = 2. The 6th to 10th terms of this arithmetic progression are

a₆ = 1 + 2(6-1) = 1 + 10 = 11
a₇ = 1 + 2(7-1) = 1 + 12 = 13
a₈ = 1 + 2(8-1) = 1 + 14 = 15
a₉ = 1 + 2(9-1) = 1 + 16 = 17
a₁₀ = 1 + 2(10-1) = 1 + 18 = 19

The sum of the first n terms of the sequence { a_n } = 1, 3, 5, 7, 9,. . . is
S_n = n (2(1) + (n - 1)2) / 2 = n (2 + 2n - 2) / 2 = 2n² / 2 = n²

We can verify this for the first 5 terms:

S₁ = 1² = 1
S₂ = 2² = 1 + 3 = 4
S₃ = 3² = 1 + 3 + 5 = 9
S₄ = 4² = 1 + 3 + 5 + 7 = 16
S₅ = 5² = 1 + 3 + 5 + 7 + 9 = 25

The binomial theorem is a useful formula for determining the algebraic expression that results from raising a binomial to an integral power. It provides one with a quick method for finding the coefficients and literal factors of the resulting expression.

The binomial theorem is stated as follows:

where n! is the factorial function of n, defined as

n! = n (n-1) (n-2) ..... 1

and 0! = 1 by convention.

EX.

The binomial theorem can also be used to find the rth term of the expansion of (a + b)ⁿ.

The first term of this expansion is aⁿ, and the (n+1)th term is bⁿ. By looking at the statement of the binomial theorem, we can see that the literal factors of the rth term are a^n-r+1 · b^r-1, where the sum of the exponents of a and b must be equal to n. Since all terms in the expansion of (a + b)ⁿ can be written as

we can set p = r - 1 so that the literal factors of the above term will be a^n-r+1 · b^r-1 . Thus, the rth term of the expansion of (a + b)ⁿ is

EX. The 7th term of ( x + y )¹³ is computed as

A geometric progression is a sequence in which each term (after the first) is determined by multiplying the preceding term by a constant. This constant is called the common ratio of the arithmetic progression. A geometric progression can be defined as follows:

The geometric progression { a_n } = a₁, a₂, a₃, ...., a_n , ....
where n = 1, 2, 3, ....

Its terms are determined by the equation a_n = a , where

a₁ is the first term of the geometric progression
a_n is the nth term of the geometric progression
n is the term number
r is the common ratio of the geometric progression

The sum of the first n terms of an geometric progression is calculated as

S_n = ( a₁ - a₁ rⁿ ) / ( 1 - r)

or

S_n = ( a₁ - a_n r ) / ( 1 - r) where a_n r = a1 · r^n-1 = a₁ rⁿ

EX. In the sequence { a_n } = 1, 2, 4, 8, 16 , ...., where n = 1, 2, 3, ....

a_n = (1) 2^n-1= 2^n-1

Thus, the sequence { a_n } = 1, 2, 4, 8, 16 , .... is a geometric progression with a₁ = 1 and r = 2. The 6th to 10th terms of this geometric sequence is

a₆ = 2^6-1 = 2⁵ = 32
a₇ = 2^7-1 = 2⁶ = 64
a₈ = 2^8-1 = 2⁷ = 128
a₉ = 2^9-1 = 2⁸ = 256
a₁₀ = 2^10-1 = 2⁹ = 512

The sum of the first n terms of the sequence { a_n } = 1, 2, 4, 8, 16 , . . is

S_n = (1 - (1) 2_n ) / (1 - 2) = (1 - 2ⁿ) / (-1) = 2ⁿ - 1

We can verify this for the first 5 terms:

S₁ = 2¹- 1 = 1
S₂ = 2²- 1 = 1 + 2 = 3
S₃ = 2³- 1 = 1 + 2 + 4 = 7
S₄ = 2⁴- 1 = 1 + 2 + 4 + 8 = 15
S₅ = 2⁵- 1 = 1 + 2 + 4 + 8 + 16 = 31

Given a geometric progression { a_n } = a₁, a₂, a₃, ...., a_n , ...., if the absolute value of the common ratio, | r | , is less than 1, the corresponding geometric series S_n = a₁ + a₂ + a₃ + .... + a_n + ... will converge to a finite value as n approaches infinity. This value can be calculated as follows:

S_¥ = a₁ / (1 - r) where

a₁ is the first term of the geometric progression
r is the common ratio of the geometric progression, | r | < 1
S_¥ indicates that the series sums up all the infinite terms in the
sequence.

Note that when | r | > 1, the terms in the geometric sequence will get progressively larger, approaching infinity as n goes to infinity. Therefore, the corresponding geometric series will not converge. So, as a warning, the above formula must not be used when | r | > 1.

EX. The sequence { a_n } = 1, 1/2, 1/4, 1/8, .... is a geometric progression with a₁ = 1 and r = 0.5. Since | r | = | 0.5 | < 1, the sum of all the infinite terms in this sequence will converge to a finite number. That number is

S_¥ = 1 / (1 - 0.5) = 1 / 0.5 = 2

The principle of mathematical induction is stated as follows:

If a given statement S_n concerning a positive integer n is true for n = 1, and if the truth of S_n for n = k, where k is a positive integer, implies that S_n is true for n = k + 1, then S_n is true for every positive integer n.

The premise behind this principle is self-intuitive. If a particular statement S_n (normally a mathematical equation or inequality involving the variable n) is true for n = 1, and it can also be proven that S_n being true implies that S_{n + 1} is also true, then the statement S_n is true for n = 1 + 1 = 2. Similarly, S_n is true for n = 2 + 1 = 3, n = 3 + 1 = 4, and so on for all positive integers.

It should be noted that the principle of mathematical induction can be extended to include whole numbers by simply proving S_n to be true for
n = 0. Then, if S_n being true implies that S_{n + 1} is true, S_n will be true for n = 0 + 1 = 1, n = 1 + 1 = 2 and so on for all positive integers. Thus, S_n will be true for all whole numbers (the set of positive numbers plus zero).

The principle of mathematical induction is very helpful in proving many statements about positive integers. According to this principle, a mathematical statement involving the variable n can be shown to be true for any positive integer n by proving the following two statements:

(1) The statement is true for n = 1.
(2) If the statement is true for any positive integer k, then it is also
true for k + 1.

The following examples should clarify the use of this principle:

Example 1: Use mathematical induction to prove that

1 + 3 + 5 + .... + (2n - 1) = n² is true for every positive integer n.

Solution: Utilize the two steps of mathematical induction outlined above.

Step 1. Replacing n by 1 in the above equation gives

2 (1) - 1 = 1 = 1² so n = 1 satisfies the equation.

Step 2 Assume that the equation is true for n = k. Then

1 + 3 + 5 + .... + (2k - 1) = k²
1 + 3 + 5 + .... + (2k - 1) + (2(k+1) - 1)= k² + ( 2(k+1) - 1 )
= k² + ( 2k + 2 - 1 )
= k² + 2k + 1
= (k + 1)²

The final equality proves that the equation is true for n = k + 1, given it is true for n = k.

The proof of the above equality by mathematical induction is complete.

Example 2: Use mathematical induction to prove that

0 + 1 + 2 + .... + n = n (n+1) / 2 is true for every whole number n.

Solution: Apply the two steps of mathematical induction, but prove the equality true for n = 0, since the statement is to be shown valid for all whole numbers (including zero).

Step 1 By plugging n = 0, we have

0 = 0 (0+1) / 2 = 0 so n = 0 satisfies the equation.

Step 2 Assume that the equation is true for n = k. Then

0 + 1 + 2 + .... + k = k (k+1) / 2
0 + 1 + 2 + .... + k + (k + 1) = k (k + 1) / 2 + (k + 1)
= (k + 1) ((k/2) + 1)
= (k + 1) (k + 2) / 2

The last equality proves that n = k satisfying the above equation implies that n = k + 1 also satisfies it.

Thus, the above equality is satisfied for n = 0, as well as n= 0 + 1 = 1,

n = 1 + 1 = 2, and so on for all positive integers; it is true for all whole numbers.

The following example illustrates a helpful method in formulating the required proofs in Step 2. In this example, the mathematical statement in question is explicitly written out for n = k and n = k + 1. Then, the attempt is made to derive the latter statement from the former, using axioms and previously-proven theorems. This procedure may facilitate the thinking process required in establishing the appropriate proofs, as is the case here.

Example 3: Use mathematical induction to show that

n² < 2ⁿ for all positive integers n > 4

Solution: While mathematical induction can still be applied to prove the above statement, a few changes must be made to the solution process. Since the equation is to be shown true for all integers greater than 4, the starting point should be n = 5 instead of n = 1. Furthermore, the fact that n > 4 should be used in Step 2.

Step 1. Using n = 5, we have

5² = 25 < 2⁵ = 32 so n=5 satisfies the inequality

Step 2 Assume that the inequality is true for n = k, where k > 4.

Then

k² < 2^k
2k² < 2·(2^k) = 2^k+1

To show that n = k + 1 satisfies the inequality, we must show that (k+1)² < 2 ^{k + 1}, where k > 4. If it can be proven that (k+1)² < 2k², then (k+1)² < 2^k+1 by transitivity.

Since k > 4, k - 1 > 3 > 2. As k + 1 > k,
(k - 1) (k + 1) > 2k
k₂ - 1 > 2k
k₂ > 2k + 1
2k² > k² + 2k + 1 = (k + 1 )²

Since and (k + 1)² < 2k² and 2k² < 2^k+1 < 2^k+2, so n = k + 1 satisfies the above inequality whenever n = k does.

This completes the proof.

A sequence is a function whose domain is the set of positive integers
{ 1, 2, 3, ... }. The functional values or range elements are called the terms of the sequence. A sequence can be defined as follows:

{ a_n } = a₁, a₂, a₃, ...., a_n , .... where n = 1, 2, 3, ....
a₁, a₂, a₃, ...., a_n , .... are terms of the sequence { a_n }
a_n = f(n), where f(n) is some function of n, n = 1, 2, 3, ....

A sequence with a first and last term is called a finite sequence, while a sequence with an infinite number of terms is called an infinite sequence.

Associated with any sequence is a series, which is defined as the sum of all the terms of the sequence. Therefore, for the sequence { a_n } = a₁, a₂, a₃, ...., a_n , .... , its corresponding series S_n is calculated as

S_n = a₁ + a₂ + a₃ + .... + a_n + ...

EX. { a_n } = 1, 3, 5, 7, 9, ..... where a_n = 2n - 1
S_n = 1 + 3 + 5 + .... + 2n - 1

There are two special kinds of sequences that will be discussed in this section. These are the arithmetic and the geometric sequences (or progressions). An arithmetic sequence is a sequence in which there is a constant difference between successive terms, while a geometric sequence is a sequence wherein each term after the first can be obtained by multiplying the preceding term by a common multiplier. Each of these sequences, along with its associated series, have special properties that will studied here.

Application of e and Exponential Functions

In the calculation of interest, the following formula is often used:

where
A is the money accumulated.
P is the principal (beginning) amount
r is the annual interest rate
n is the number of compounding periods per year
t is the number of years
EX. An initial investment of $1,000 is invested at a 12.5% annual interest rate, compounded annually, for 10 years. Using the above formula, we calculate the amount accumulated after 10 years as follows:

P = $1000
r = 12.5% = .125
n = 1
t = 10 years

A = $1000(1 + .125)
A = $1000(1.125)
A = $1000(3.247)
A = $3,247

EX. An initial investment of $2,000 yields $10,000 after 15 years. The interest is compounded quarterly. Using the interest formula, we can calculate the interest rate:

In calculations involving depreciation, the interest formula can also be used, provided that the depreciation rate is taken to be a negative interest rate.

EX. The principal cost of a car is $20,000. It depreciates at a rate of 6% per year. After 5 years, its book value is given by

P = $20,000
r = -6% or -.06
n = 1
t = 5

A = $20,000 (.94)
A = $14,680

In the interest formula

the value of A, given fixed values for P, r (where r> 0) and t, will increase as n increases, i.e., the more often the interest is compounded, the greater the amount of money accumulated. As n approaches infinity, the value of A converges to a finite value. It can be shown that

Therefore

This is known as continuous compounding of interest. The formula used herein is:

A = Pe^rt

whereA is the money accumulated.

P is the principal amount
r is the interest rate compounded continuously
t is the number of years

EX. An investment of $200 is invested at an interest rate of 5% compounded continuously. In 5 years, it will yield an amount computed as

A = Pe^rt

P = $200
r = 5% or .05
t = 5

A = $200e^(.05)(5)
A = $200e^.25
A = $200(1.285)
A = $257

The number e is also used in applications involving exponential growth and decay. Entities that are subject to exponential growth increase at a rate that is proportional to the amount of substance in the entity; entities subject to exponential decay decrease at a rate proportional to the amount of substance in the entity. The formula often used herein is

Q(t) = Q₀e^kt

where Q(t) is the amount of the substance at time t, Q0 is the amount of the substance at t = 0, and k is a constant that depends on the characteristics of the entity. If k > 0, then Q(t) increases as t increases; this represents exponential growth. If k < 0, then Q(t) decreases as t increases; this is known as exponential decay.

EX. A certain bacteria culture has a growth rate of k = .03 and an initial count of 10000.

After 3 hours, the population of the bacteria culture is calculated as

Q(t) = Q₀e^kt
Q₀ = 10000
k = .03
t = 3

Q(t) = 10000 e^0.03(3)
Q(t) = 10,940

EX. A certain radioactive particle decays at a rate of k = .06. After 5 hours, a 100-gram sample will decay to an amount calculated as .

Q(t) = Q₀e^kt
Q_o = 100 g
k = -0.06
t = 5

Q(t) = 100 g e^-0.06(5)
Q(t) = 74.08 g

Exponential Functions

The exponential function with base b, where b > 0, has the following form:

f(x) = b^x
-¥ < x < ¥

where b is a constant. In an exponential function, the base of f(x) is a constant real number, while the exponent is the dependent variable of the function.

The graphs of all exponential functions (with a positive base) share similar properties:

(1) Since the value of b^x is positive for b > 0, the graph of the exponential function

f(x) = b^x

lies above the x-axis, but never touches it, as the value of b^x can never be zero.

(2) When b > 1, the value of b^x increases towards infinity as x approaches infinity, while it decreases to (but does not become) zero as x approaches negative infinity. Therefore, the graph of f(x) = b^x curves to the right and moves upward infinitely as x ® ¥, while it approaches the x-axis asymptotically as x ® -¥.

When 0 < b < 1, the value of b^x decreases to (without reaching) zero as x approaches infinity, while it increases towards infinity as x approaches negative infinity. Therefore, the graph of f(x) = b^x approaches the x-axis asymptotically as x ® ¥, while it curves to the right and moves upward infinitely as x ® -¥.

When b = 1, the value of b^x is always equal to 1, since 1^x = 1 for any value of x. Thus, the graph of f(x) = b^x is the ordinate y = 1.

(3) Since b⁰ = 1 for any real number b, the graph of the exponential function

f(x) = b^x

for any real number b, includes the point (0,1).

(4) For b > 0,

(1/b)^x = (b^-1)^x = b^-x

Thus, the graphs of f(x) = (1/b)^x and f(x) = b^x are reflections of each other through the y-axis.

EX. The exponential function

f(x) = 3^x

contains the following points:

 

x	3x	(x, y)
-2	1/9	(-2, 1/9)
-1	1/3	(-1, 1/3)
0	1	(0, 1)
1	3	(1, 3)
2	9	(2, 9)

EX. The exponential function

contains the following

x	(1/3)x	(x, y)
-2	9	(-2, 9)
-1	3	(-1, 3)
0	1	(0, 1)
1	1/3	(1, 1/3)
2	1/9	(2, 1/9)

Note that the graphs of f(x) = 3^x and f(x) = (1/3)^x are reflections of each other across the y-axis.

The quantity b^x, where b > 0 and b ¹ 1, obey the following axioms:

(1) b^x = b^y if and only if x = y.
(2) If b > 1 and r < s < t, then b^r < b^s < b^t
If 0 < b < 1 and r < s < t, then b^r > b^s > b^t

EX.
3^x = 27
3^x = 3³
x = 3

4^x = 128
(2²)^x = 2⁷
(2)^2x = 2⁷
2x = 7
x = 3.5

2^x < 2^y
x < y

(0.25)^m > (0.25)ⁿ
m < n

Logarithmic Functions

Given the exponential function

f(x) = y = b^x

it is often desirable to solve for x in terms of y. In order to accomplish this, we define a new function called the logarithm (log for short) to the base b, b > 0 , defined as follows:

g(x) = y = log _b x whenever x = b^y

That is, the logarithm to the base b of a real number x is the real number y such that b raised to the yth power is equal to x. The quantity b^y, where b > 0 , will always have a positive value for any real number y.

If x = b^y, then y = log _b x and x must be positive in either equality. Thus, the logarithm function has a domain that consists only of positive real numbers.

Note that if b = 1, then b^y = 1 for any real number y. Thus, the logarithm function having base 1 has only one element in its domain, the number 1, and so it has no real significance.

The quantity log _b x, where b = 10, is often referred to as simply log x.

The functions f(x) = b^x and g(x) = log b x are inverse functions of each other. To verify this, note that for f(x) = b^x and g(x) = log _b x ,

f(g(x)) = b^{log3 x}
z = log _b x implies that b^z = x

b^{log3 x} = x

g(f(x)) = log _bb^x = x since b^x = b^x.

EX. log ₅ 25 = 2 since 5² = 25
log ₁₀ 0.00001 = -5 since 10^-5 = 0.00001
log _0.5 0.0625 = 4 since (0.5)⁴ = 0.0625

Since the functions f(x) = b^x and g(x) = log _b x are inverse functions of each other, their graphs are mirror images of each other across the line y = x. The graph of g(x) = log _b x, where b > 0 and b ¹ 1, will also curve to the right and move upwards without bound as x goes to infinity, although more slowly than the graph f(x) = b^x. It will approach the y-axis asymptotically as x approaches 0, since b^x® 0 as x ® -¥ and thus

log _b x ® -¥ as x ® 0

EX. The logarithmic function

f(x) = log ₃ x

contains the following points:

x	log 3 x	(x,y)
0.004	-5	(0.004, -5)
0.012	-4	(0.012, -4)
0.037	-3	(0.037, -3)
0.111	-2	(0.111, -2)
0.333	-1	(0.333, -1)
1	0	(1, 0)
3	1	(3, 1)
9	2	(9, 2)
27	3	(27, 3)
81	4	(81, 4)
243	5	(243, 5)

Because the exponential function and the logarithmic function having the same base b are inverse functions of each other, it is true that

blog _b x = x and log _b b^x = x for x > 0, b > 0 and b ¹ 1

From the second equality, we get two more equalities (for b > 0 and b ¹ 1):

(1) log _b b¹ = log _b b = 1
(2) log _b b⁰ = log _b 1 = 0

The logarithmic function also possesses the following properties:

(1) log _b xy = log _b x + log _b y
(2) log _b (x / y) = log _b x - log _b y
(3) log _b x^r = r log _b x

where x, y and b are positive real numbers, with b ¹ 1.

EX. log _x 16 = 4
x⁴= 16
x⁴= 2
x = 2

EX. log ₅ x = 3
5³= x
x = 125

EX. log ₁₀ 10000 = x
10^x= 10000
10^x= 10⁴
x = 4

EX. 5^x= 17
log 5^x= log 17
x log 5 = log 17

EX. log 5x - log (x - 5) = 1
log 5x - log (x - 5) = log 10

5x = 10(x - 5)
5x = 10x -50
5x = 50
x = 10

When dealing with a change in the base of a logarithmic function, the following equality can be used to facilitate the conversion:

where a, b and c are positive real numbers, where b ¹ 1 and c ¹ 1.

EX.

e and the Natural Logarithm

One of the most important numbers used as the base for exponential and logarithmic functions is denoted as e. It is an irrational number, with its value approximated as e » 2.7182818. Exponential and logarithmic functions with base e occur in many practical applications, including those involving growth and decay, continuous compounding of interest, alternating currents and learning curves.

The natural logarithm of a positive real number is defined as the logarithm to the base e of the number. The natural logarithm of x, x > 0, is denoted as ln x. Symbolically,

ln x = log_e x where x>0

By definition, ln x = y implies that e^y = x.

When converting from base 10 to base e, we can use the following formula:

log x = .4343 ln x

where log ₁₀ e = .4343.

Since the function f(x) = e^x and f(x) = ln x are inverse functions of each other,

ln e^x = x and e^{ln x} = x

The natural logarithm possesses the same properties as common logarithms.

EX.
e^{x + 3} = 17
ln e^{x + 3} = ln 17
x + 3 = ln 17
x + 3 = 2.833
x = -0.167

EX.
ln x + ln (x - 2) = ln 15
ln [x(x - 2)] = ln 15
x²- 2x = 15
x²- 2x - 15 = 0
x²+ 3x -5x - 15 = 0
x(x + 3) -5(x + 3) = 0
(x - 5)(x + 3) = 0

x - 5 = 0 or x + 3 = 0
x = 5 or x = -3

Since x cannot be negative, the solution set is {5}.

( f + g )(x) = f(x) + g(x)

( f - g )(x) = f(x) - g(x)

( f · g )(x) = f(x) · g(x)

( f / g )(x) = f(x) / g(x)

(f o g) (x) = ( f (g(x) ),

The domain of f _o g such that x is in the domain of g and that g(x) is the domain of f.

ex.

f(x) = x and g(x) = x + 2 find (f _o g) (x).
(f _o g) (x) = f (g(x))

= f ( x + 2 )
= ( x + 2 )²

The domain of g(x) = x + 2 is the set of all real numbers, and so is the domain of
f(x) = x² thus the domain of (f _o g) (x) is the set of real numbers.

The inverse of (a,b) is (b,a).
Functions are said to be inverse of each other if f _o g = g _o f.

Finding Inverse Functions

ex.

f(x) = 3x - 4
y = 3x - 4 replace f(x) with y
x = 3y - 4 replace x with y and y with x.
3y = x + 4 solve for y

y = (x+4)/3 replace y with f^-1(x)

f^-1(x) = (x+4)/3

The inverse function of 3x - 4 is (x+4)/3.

To test if the example above are inverse of each other, do the inverse function test.

Functions are said to be inverse of each other if f _o g = g _o f.

f(x) = 3x - 4

f _o g = f (g(x))

g(x) = f^-1(x) = (x+4)/3

f(x) = 3x - 4
g _o f = g (f(x))

= g ( 3x - 4 )

They are inverse of each other.

Graph of Inverse Functions:
If the graph of the function f is known then the graph of f^-1 is reflected across the line

y = x. ( or f(x) = x)

ex.

A set of ordered pairs is called a relation.
ex.
(1,2), (3,5), (7,10)...are relations.

The first components in the ordered pairs (x-coordinate) is called the domain, and the second components (y-coordinate or f(x) coordinate) is called the range.

A function is a relation were each member of the domain is reserved one and only one member of the range.

Functions are named by using single letters.
ex.

f = {(x,y) | y = x + 4 }

or

f (x) = x + 4

The equations above are representations of functions. The second example is the most commonly used representation; it is called the function notation.
The ordered pair for functions is (x, f (x)).

consider the example;

f (x) = x + 4 at x = 2.
f (2) = 2 + 4
f (2) = 6

the ordered pair is (2,6)

ex.
find the range; given the domain {2,4,6} of f(x) = 3x - 1

f(2) = 3(2) - 1 = 6 -1 = 5
f(4) = 3(4) - 1 = 12 - 1 = 11
f(6) = 3(6) - 1 = 18 - 1 = 17

the range is {5,11,17} and the ordered pairs are (2,5); (4,11); (6,17)

Find the domain and range of:

f(x) = 2 / (x-3)

Since x cannot be zero; the domain is D = { x| x ¹ 3} and f(x) can have any value; thus the range is R = { f(x) | any real number}

Graph of Functions: [ to graph functions, just change y into f(x) from graphing equations]

Graphing functions is the same as graphing equations with the exception of naming the domain and the range.

Linear Functions:
ex.
f(x) = 3x + 2:
the function is in the slope-intercept form (y = mx + b)
m(slope) = 3
y-intercept = 2

Quadratic Functions:[ f(x) = ax²+ bx + c ]
The same as graphing quadratic equations:
ex.
graph:
f(x) = 2x²+ 8x + 9
y = 2x²+ 8x + 9 replace f(x) with y
y - 9 = 2x+ 8x complete the square
y - 9 + 4 = 2( x²+ 4x + 4)
y - 1 = 2(x + 2)² parabola with vertex at (-2,1)
y = 2(x + 2)²+ 1 solve for y and replace y with f(x)
f(x) = 2(x + 2)²+ 1

Other functions:

Rules for graphing other functions:
1. Determine the domain of the graph.
2. Check for symmetry.

if f(-x) = f(x), then symmetric about the y-axis.
if f(-x) = -f(x), then symmetric about the origin.
( functions rule out the possibility that the graph has an x-axis
symmetry)

3. Find the x-intercept and y-intercept of the graph

Evaluate f(0) to find the y-intercept.
To find the x-intercept, find the value or values that will make f(x) = 0.

4. Plot some points for the graph and reflect, according to the symmetry
test.

ex.

graph:

f(x) = | x |

the domain is D = {x | x is any real number}
f(-x) = | -x | = | x | = f(x) symmetric about the y-axis.
f(0) = | 0 | = 0
f(x) = 0 at x = 0
the intercepts are at the origin.

 

x	f(x)
1	1
-2	2
-1	1
2	2

Vertical Line Test:
Given a graph, it can be determined if it is a function or a relation.
In order for the graph to be a function, the vertical line must only intersect the graph at one and only one point.

Horizontal Line Test:
The horizontal line test implies a one to one function, meaning that there is only one value of x associated with each value of f(x).

Consider two points C(h,k) and P(x,y) ;and the distance between them is r. From the distance formula:

square both sides
r²= ( x - h )²+ ( y - k )²

r²= ( x - h )² + ( y - k )² is the standard equation for circles.

Ax²+ Ay²+ Dx + Ey + F = 0
is the general form of the equation for circles.

ex.
Write the general equation of the circle with center at C(4,-5) and a
radius of 5.

r²= ( x - h )²+ ( y - k )²
5²= ( x - 4 )²+ ( y -(-5))²
25 = x² -8x + 16 + y² + 10y + 25

x²+ y² -8x + 10y + 16 = 0 is the general equation.

Find the center and the radius of the circle with equation x²+ y²-10x - 4y + 16 = 0.

x²+ y²-10x - 4y +16 = 0

(x-10x ) + (y- 4y ) = -16 complete the square.
(x²-10x + 25) + (y²- 4y + 4) = -16 + 25 + 4
( x -5)² + ( y - 2 )² = 13

The center is at (5, 2) and the radius is

If the constant term in the standard equation is positive, then a graph of a circle exist. If it is zero, a single point exist, If it is negative, there is no graph.

standard form of the equation of an ellipse:

where b²= a² - c²

center (0,0) (0,0)major axis on x-axis, length 2a on y-axis, length 2aminor axis on y-axis, length 2b on x-axis, length 2bfoci (±c,0) (0,±c)vertices (±a,0) (0,±a)covertices (0,±b) (±b,0)</span>

a is always larger than b; and a,b, and c are related by c²= a² - b²

ex.
Graph 9x²+ 4y²= 36

which is an ellipse.
a² = 9 ; b² = 4
center (0,0)
major axis: y-axis
vertices: (0, ± 3)
covertices: (± 2,0)

Graph 9x² + 25y² = 225

a²= 25 ; b²= 9
center (0,0)
major axis: x-axis
vertices: (± 5,0)
covertices (0,± 3)

c²= a² - b²
c²= 25 - 9
c² = 16

foci: (± 4,0)

Hyperbola:

The standard equation for hyperbolas is:

where b² = c² - a²

vertices (± a,0) (0, ± a)foci (± c,0) (0, ± c)transverseaxis on x-axis, on y-axis,length 2a length 2aconjugateaxis on y-axis, on x-axis,length 2b length 2b

a is always larger than b; and a,b, and c are related by c² = a² + b²

ex.
Graph 9x² - 16y²= 144

a² = 16 ; b² = 9
major axis: x-axis
vertices: (± 4,0)

c² = a+ b
c² = 16 + 9
c² = 25

foci: (± 5,0)

 

Graph 36x² - 4y² + 144 = 0
36x² - 4y² = -144 factor -1 out
4y² - 36x² = 144

a² = 36 ; b² = 4
major axis: y-axis
vertices: (0,± 6)

c²= a² + b²
c²= 36 + 4
c² = 40

(to find the asymptotes, let the x term equal the y term and solve for y)

Standard Equation: x²= 4py y²= 4px
vertical parabola horizontal parabola
axis y-axis x-axis
vertex (0,0) (0,0)
p is the focus of p>0 opens up p>0 opens right
the parabola
p<0 opens down p<0 opens left
focus (0,p) (p,0)
directrix y = -p x = -p
length of latus rectum 4|p| 4|p|

ex.
Graph x²= 6y

The standard equation is x² = 4py,
which implies a vertical parabola with the vertex at the origin and the axis is the y-axis.

4p = 6

P > 0, the parabola opens up and the focus is at (0, 3/2)
The directrix is the line y = -3/2

4|p| = 6, the end points of the latus rectum is (3, 3/2) and (-3, 3/2)

Graph y²= -16y

The equation is in the form: y² = 4px,
which implies a horizontal parabola with the vertex at the origin and the axis is the x-axis.

4p = -16
p = -4
p < 0, the parabola opens left and the focus is at ( -4,0).
The directrix is the line x = 4.
± 2p = ± 8; \ the end points of the latus rectum are points (-4, 8) and (-4, -8).

Coordinate Systems and Graphing

Consider two number lines, one horizontal and one vertical. The two lines are perpendicular to each other at there zero points. These number lines make up the axes of a coordinate system. The horizontal line is called the x-axis, which is positive to the right and negative to the left. The vertical line is called the y-axis, which is positive going up and negative going down. The intersection of both x-axis and y-axis at there zero points is called the origin. The intersection of the axes also divides the plane into four regions called quadrants. The quadrants are numbered counterclockwise form one to four.

Ordered Pairs are pairs of real numbers that describe a point in the coordinate system. The ordered pair describes a point in the plane by its x and y coordinates. (x,y) is a point in the plane. The first number describes the distance in the x-axis; also called the abscissa, and the second number describes the distance in y-axis, also called the ordinate. The point (0,0) is the origin of the graph. Points are named by using capital letters, and point O often refers to the origin (0,0). (x,y) is the coordinate of the point.

Each quadrant have a distinct pattern;

• In the first quadrant, both the x and y-coordinate are positive.

• In the second quadrant, the x-coordinate is negative and the y-coordinate is positive.

• In the third quadrant, both the x and y-coordinate are negative.

• In the fourth quadrant, the x-coordinate is positive and the y-coordinate is negative.

Distance Between Points:

Distance Formula:

ex.

Find the distance between a(3,6) and B(7,9).

Graphing Linear Equations and Inequalities:

A solution for an equation with two variables is an ordered pair of real numbers that satisfy the equation. Consider the equation y = x + 3; for every number that is substituted into x a number y is produced and an ordered pair is formed.

Notice that there is an unlimited number of solutions to the equation.

This is the basis for plotting points or graphing an equation or inequality. Linear Equations have the basic form of Ax + By = C.

Graphing Linear Equations:

x and y-intercepts:

x-intercepts of a graph are the x-coordinates that are common with the x-axis.
To solve for the x-intercept, let y = 0, then solve for x.
y-intercepts of a graph are the y-coordinates that are common with the y-axis.
To solve for the y-intercept, let x = 0, then solve for y.

ex.

graph 2x -3y = 6

First, find the intercepts.

x = 0; -3y = 6

 y=2; the point (0,-2) is on the line.y = 0; 2x = 6x = 3; the point (3,0) is on the line.

Now, check for a third point in the equation,

Let x = 6; 2(6) -3y = 612 -3y = 6-3y = -6y = 2

Graph y = 3x

x = 0; y = 3(0) = 0

The origin is part of the graph, and another two points must be found.
A table of values would be useful for this.

Vertical and Horizontal Lines:

x = 4, y = -5 are examples of vertical and horizontal equations of lines.

Graphing Linear Inequalities:

Graphing linear inequalities is almost the same as graphing linear equations, but with a slight difference.

ex.
graph 2x + 3y < 6

First graph the line 2x + 3y = 6, and use a broken line to graph the line since the line is not really part of the solution.

Choose a test point that is above and below the line to see if that point is a valid solution to the inequality. In this case, choose the origin (0,0) to see if it is a valid point in the solution.

2(0) + 3(0) < 60 < 6  the origin is part of thesolution.

Now, try a point above the line. Try point ( 5,4 ).

2(5) + 3(4) < 610 + 12 < 6   the point (5,4) is not part of the solution.

\ the solution graph is:

Any point in the blue area is a solution to the inequality.

Finding the Equation of a Line in the form Ax + B = C:
ex.

Find the equation of a line with a slope of -2 and a point at (4,6).

It is also known that a point (x,y) is on the line then,
From the formula:

Find the equation of the line that has the points (2,3) and (-5, 8)

First find the slope of the line.

Now that the slope is defined, the equation of the line can be found.

From the slope of the line, one of the given points and the variable point on the line (x,y); the equation of the line is from the formula;

Find the equation of the line with a slope of 3/2 and a y-intercept at 4.

The y-intercept implies that point (0,4) is in the line.
Now just solve as before:

3( x - 0 ) = 2( y - 4 )
3x = 2y - 8
3x - 2y = -8

To Translate the Graph:
ex.
(y - k) = | ( x - h) |
Replace (x - h) with x' and (y - k) with y':
Graph the equation with x' and y' with (h,k) as the center or vertex:
y' = | x' | at (h,k)

Slope-Intercept Form

Now, given a slope m and y-intercept at b.
What is the equation of the line?

y-intercept at b implies that the point (0,b) is on the line.
From the point slope form;

y - b = m ( x - 0 )
y - b = mx
y = mx + b This is the Slope-Intercept Form of a line.

ex.
From the previous example.

Find the equation of the line with a slope of 3/2
and a y-intercept at 4.

y-intercept at implies that the point (0,4) is on the line.

The slope-intercept form:

y = mx + b

2y = 3x + 8 multiplied 2 to both sides
3x - 2y = -8

Point-Slope Form:

The point-slope form is derived from the formula;

The equation of a line with point (x,y) and slope m is

THis is the point-slope form of a line.

ex.
From the previous example:
The equation of a line with slope -2 and the point (4,6)

y - 6 = -2( x - 4)
y - 6 = -2x + 8
2x + y = 14

If P₁(x₁, y₁)and P₂(x₂, y₂)are two different points, then the slope of the line between them is;

 

If the slope of a line is zero (0) then the line is a horizontal line.
If the slope of a line is undefined, then the line is a vertical line.

ex.
Find the slope of the line, m, between A(-3,1) and B(4,6)

Slope-Intercept Form

Now, given a slope m and y-intercept at b.
What is the equation of the line?

y-intercept at b implies that the point (0,b) is on the line.
From the point slope form;

y - b = m ( x - 0 )
y - b = mx
y = mx + b This is the Slope-Intercept Form of a line.

ex.
From the previous example.

Find the equation of the line with a slope of 3/2
and a y-intercept at 4.

y-intercept at implies that the point (0,4) is on the line.

The slope-intercept form:

y = mx + b

2y = 3x + 8 multiplied 2 to both sides
3x - 2y = -8

Y-axis symmetry:
The graph of an equation is symmetric with respect to the y-axis if an equivalent equation is obtained when x is replaced by -x.

ex.
x² = 2y
(-x)² = 2y = x² = 2y

this equation is symmetric about the y-axis.

X-axis symmetry:
The graph of an equation is symmetric with respect to the x-axis if an equivalent equation is obtained when y is replaced by -y.

ex.
y⁶= 4x
(-y)⁶ = 4x = y⁶ = 4x

this equation is symmetric about the x-axis.

Origin Symmetry:
The graph of an equation is symmetric with respect to the origin if an equivalent equation is obtained when x is replaced by -x and y replaced by -y.

ex.
x³ = y
(-x)³ = -y
-x³ = -y = x³ = y

this equation is symmetric about the origin.

Conic Sections Translation of Axes:

Ellipse:

standard equations for ellipse:

where b² = a² - c²

center: (h,k) (h,k)major axis: x = h, length 2a y = k, length 2aminor axis: y = k, length 2b x = h, length 2bfoci: (h ± c,k) (h, k ± c)vertices: (h ± a,k) (h, k ± a)covertices: (h, k ± b) (h ± b,k)

a is always larger than b; and a,b, and c are related by c² = a² - b²

ex.
graph 16x² + 25y² - 64x-200y + 64 = 0
convert to standard form
16x² - 64x + 25y² -200y = - 64
complete the square for the x and y terms

(16x² - 64x ) + (25y² -200y ) = - 64
16(x² - 4x ) + 25(y² - 8y ) = -64 complete the square
16(x²- 4x + 4) + 25(y²- 8y + 16) = -64 + 64 + 400
16(x - 2)² + 25(y - 4)² = 400

a = 5; b = 4
center: (2,4)
major axis: x = 2, length 10
minor axis: y = 4, length 8
c² = a² - b²
c² = 25 - 16
c² = 9
c = 3

foci: (2 ± 3,4) (5,4) and (-1,4)vertices: (2 ± 5,4) (7,4) and (-3,4)covertices: (2, 4 ± 4) (2,8) and (2,0)

ex.
graph 25x² + 9y²+ 200x + 54y + 256 = 0
convert to standard form
25x² +200x + 9y²+ 4y = -256
complete the square for the x and y terms

(25x²+200x )+ (9y²+ 54y ) = - 256
25(x² + 8x ) + 9(y² + 6y ) = -256
25(x² + 8x + 16) + 9(y² + 6y + 9) = -256 + 400 + 81
25(x + 4)² + 9(y + 3)² = 225

a = 5; b = 3
center: (-4,-3)
major axis: y = -3, length 10
minor axis: x = -4, length 6
c² = a² - b²
c² = 25 - 9
c² = 16
c = 4

foci: (-4, -3 ± 4) (-4,1) and (-4,-7)vertices: (-4, -3 ± 5) (-4,2) and (-4,-8)covertices: (-4 ± 3,-3) (-1,-3) and (-7,-3)

Standard Equation for Hyperbolas:

where b² = c² - a²

vertices (h ± a,0) (0, k ± a)foci (h ± c,0) (0, k ± c)</span>

a is always larger than b; and a,b, and c are related by c² = a² + b²

ex.

Graph 9x² - 25y²-54x + 250y -769 = 09x²- 54x - 25y²+ 250y = 769(9x² - 54x ) - ( 25y²- 250y ) = 7699(x²- 6x + 9) - 25(y²- 10y + 25) = 769 +81 - 6259(x - 3)² - 25(y -5)²= 225</span>

a = 5 ; b = 3

Center (3,5)

asymptotes

vertices (3 ± 5,5)

ex.

16x² - 9y²- 224x - 54y + 847 = 0
16x² - 224x -9y² - 54y = -847
(16x² - 224x ) - (9y² - 54y ) = -847
16( x² - 14x + 49) - 9( y² + 6y + 9) = -847 +784 -81
16( x - 7)² - 9(y + 3)² = -144
9(y + 3)² - 16( x - 7)² = 144 Factor -1 out of both sides

a = 4; b = 3

Center (7,-3)

vertices (7,-3 ± 4)

c² = a² + b²
c² = 16 + 9
c² = 25
c = 5

foci (7,-3 ± 5)

Conic Sections Translation of Axes:

Parabola:

The standard equation for translated parabolas is:

(x - h)² = 4p(y - k)² or (y - k)= 4p(x - h)vertex (h,k) (h,k)axis x = h y = kp is the focus of the parabola p > 0 opens up opens rightp < 0 opens down opens leftfocus (h, k + p) (h + p,k)directrix y = k - p x = h - plength of latus rectum 4|p| 4|p|endpoints of latus rectum (h ± 2p,k + p) (h + p, k ± 2p)

ex.

graph x²- 6x - 8y + 49 = 0 convert to standard formx²- 6x = 8y - 49 complete the square on the x termsx²- 6x + 9 = 8y - 49 +9(x - 3)² = 8y - 40(x - 3)² = 8(y - 5) vertical parabola

vertex: (3,5)
axis: x = 3
4p = 8
p = 2 > 0; opens up
focus: (3,7)
directrix: y = 3
length of latus rectum: 4|2| = 8 (or ± 2p)
endpoints of latus rectum: (7,7) and (-1,7)

ex.

graph y²+ 8y + 12x - 8 = 0 convert to standard formy²+ 8y = -12x + 8 complete the square on the x termsy² + 8y + 16 = -12x + 8 +16(y + 4)² = -12x + 24(y + 4)² = -12(x - 2) horizontal parabola

 

vertex: (2,-4)
axis: y = -4
4p = -12
p = -3 < 0; opens left
focus: (-1,-4)
directrix: x = 5

length of latus rectum: 4|-3| = 12 (or ± 2p)
endpoints of latus rectum: (-1,2) and (-1,-10)

The general equation for conic sections is :

Ax² + By² + Dx + Ey + F = 0

were A and B are not zero.
If AB = 0 then it is a parabola
If AB > 0 then it is an ellipse
If AB < 0 then it is a hyperbola

Given the system of equation;

Then

ex.
solve the system:
7x - 5y = -50
2x + y = -7

Third Order Determinants, are 3×3 determinants, which is written as,

The minor of an element in a determinant is a determinant that is obtained after the row and column in which the element appears are deleted.

ex.
Given the determinant; find the minor of a₁

To calculate third order determinants, a method called expansion of a determinant by minors can be used.

This process is called expansion of a determinant by minors about the first column.

It is possible to expand a determinant about any row or any column, provided that the right sign is applied to the terms of the expansion. The following sign array is very useful in determining the signs of the term of the expansion.

ex.
Expand the determinant by minors about the second column.
The second column in the sign array is - + -.

= -7( 5 - 2) -9(3 - 4) -6(-3 -10)
= -7(3) -9(-1) -6(-13)
= -21 + 9 + 78
= 66

Expanded Cramer's Rule:

Given the system of equation;

Then

ex.
5x -2y + 3z = -1
3x + y - 2z = 25
2x - 4y + 5z = 16

= 5(5-8) + 2(15-(-4)) + 3(-12 -2)
= 5(-3) + 2(19) + 3(-14)
= -15 + 38 - 42
= -19
D = -19

= -1(5-8) -25(-10 - (-12)) - 29(4 - 3)
= -1(-3) -25(2) -29
= 3 - 50 -29
D_x= -76

= 1(15 - (-4)) + 25 (25 -6) + 29 (-10 -9)
= 19 + 25(19) + 29(-19)
= 19( 1 + 25 - 29)
= 19(-3)
D_y = -57

= -1(-12 - 2) -25( -20 - (-4)) - 29(5-(-6))
= -1(-14) -25(-16) -29(11)
= 14 + 400 - 319
= 95
D_z = 95

NOTE: The expansion by minors method can be used with larger order of determinants, but it becomes more tedious as the determinant gets bigger.

A matrix with the same number of columns as rows is called a square matrix. The determinant is a real number which is associated with each square matrix having real number entries.

ex.

which is defined as;

The determinant of a square 2×2 matrix is defined as;

This is commonly called the determinant. A 2×2 determinant is also called a second order determinant.

ex.
Compute the determinant:

Determinants are used also to solve system of equations with the help of Cramer's Rule.

Solving Systems using Matrices:
A matrix is an array of numbers arranged in rows and columns.
ex.

This is a 2 by 3 matrix, meaning there are 2 rows and 3 columns

Generally, matrices of x rows and y columns are called dimension x × y or order x × y.

With every system of equation, there is a matrix of the coefficients and constant term that is associated with the system.

ex.
The system of equation:
3x + 5y = -4
x - 3y = 5

has a matrix that looks like;

This matrix is commonly called an augmented matrix, the dotted line separates the coefficients and the constant terms.

For any augmented matrices of linear system of equations, the following operations will produce an equivalent matrix:
1. Interchanging any two rows of the matrix.
2. Multiplying a row with any nonzero real number.
3. Any row of matrix can be changed by adding a nonzero multiple of
another row to that row.

With the operations above, reducing the matrix into a reduced echelon form will solve the system.

ex.
from the example;
2x + 3y + z = 1
5x + 2y - 3z = 8
x - 4y - z = 18

the matrix looks like:

To solve the matrix;

 

The solution implies that x = 5, y = -4 and z = 3, which is similar to the previous example.

The solution set is {(5,-4,3)}.

System of two linear equations in two variables:

Ax + By = C
Ex + Fy = G

The Substitution Method:

ex.
x + y = 2
2x - y = 7

1. Solve one of the equations for one variable in terms of the other,
try to avoid fractions.

x + y = 2
y = -x + 2

2. Substitute the expression above to the other equation.

2x - y = 7
2x - (-x + 2) = 7
2x + x - 2 = 7
3x - 2 = 7
3x = 9
x = 3

3. Substitute the solution above to one of the equations in the system
and solve.

x + y = 2
3 + y = 2
y = 2-3
y = -1

4. Check to see if the solutions obtained are right.

x + y = 2
3 - 1 = 2
2x - y = 7
2(3) - (-1) = 7
6 + 1 = 7

Elimination by Addition Method:

x + y = 2
2x - y = 7

1. Multiply one of the equation with an appropriate real number that will eliminate one of the variables when the equations are added together.
Multiply the first equation with 1 and add the first equation to the second.

2. Substitute x = 3 to one of the equation to find the value of y.

x = 3
x + y = 2
3 + y = 2
y = -1

3. Check the solutions with the equations.

x + y = 2
3 - 1 = 2
2x - y = 7
2(3) - (-1) = 7
6 + 1 = 7

System of Three Equations in Three Variables:

Ax + By +Cz = D
Ex + Fy +Gz = H
Ix + Jy +Kz = L

Solving system of three equations in three variables is similar to solving system with two variables. The solution set is a set of ordered triple of real numbers(x,y,z).

ex.

Now, add the first and third equations.

Add the two new solutions,

Substitute x = 5 into one of the two new solutions,

x + y = 1
5 + y = 1
y = -4

Substitute y = -4 and x = 5 into one of the original three equations,

x - 4y - z = 18
5 - 4(-4) - z = 18
5 + 16 - z = 18
5 + 16 - 18 = z
3 = z

The solution set is { 5,-4,3}

Solving systems of inequalities rely heavily on graphing of the equations to find the solutions.

ex.
x > 3
y < -2

first, graph the lines x = 3 and y = -2

the gray area represent the solution to the system of inequalities

x > 3
y < -2

ex.
y = x + 2
x² + y² = 9

Graphing the two equations reveal a circle with radius of 3 and the origin as its center and a line whose x and y intercepts are x = -2 and y = 2.

From the graph, the solutions are the points were the line and the circle intersect, and the intersections occur in two points, therefore there are two solutions to the system.

Solving the system analytically finds the exact intersection points, the substitution method works best,.

y = x + 3
x² + y² = 9

substitute

x² + (x + 3)² = 9
x² + x²+ 6x + 9 = 9
2x² + 6x = 0
2x ( x + 3) = 0

x = 0 or x = -3

substitute the values of x into one of the original equations to find the solution points.

y = x + 3 or y = x + 3
y = 0 + 3 or y = -3 + 3
y = 3 or y = 0

The solution points are {(0,3),(-3,0)}

Complex Numbers

Complex numbers are the numbers with the format a + b i, where a and b are real numbers and
i ² = - 1. If we denote the set of complex numbers by C, then

C = { a + b i , where a and b are real numbers, i² = -1 }

If in the number x = a + b i , b is set to zero, then x = a, where a is a real number. Thus, all real numbers are complex numbers, i.e., the set of complex numbers includes the set of real numbers.

Integers are the numbers that are in either (1) the set of whole numbers, or (2) the set of numbers that contain the negatives of the natural numbers. If the set of integers is denoted by I, then

I = {......, -3, -2, -1, 0, 1, 2, 3, ......}

Positive integers are the numbers in I greater than 0. Negative numbers are the numbers in I less than 0.
The number zero is neither positive nor negative, i.e., it is both nonpositive and nonnegative.

Given the above definitions, the following statements about integers can be made:

(1) N is the set of positive integers.
(2) W is the union of N and the number 0.
(3) The set of numbers that contain the negatives of the numbers in N is
the set of negative integers.
(4) I is the union of W and the set of negative integers.

Natural numbers, also known as counting numbers, are the numbers beginning with 1, with each successive number greater than its predecessor by 1. If the set of natural numbers is denoted by N , then

N = { 1, 2, 3, ......}

Rational numbers are the numbers that can be represented as the quotient of two integers p and q, where q is not equal to zero. If the set of rational numbers is denoted by Q , then

Q = { all x, where x = p / q , p and q are integers, q is not zero}

Rational numbers can be represented as:

(1) Integers: (4 / 2) = 2, (12 / 4) = 3
(2) Fractions: 3 / 4, 13 / 3
(3) Terminating Decimals: (3 / 4) = 0.75, (6 / 5) = 1.2
(4) Repeating Decimals: (13 / 3) = 4.333....., (4 / 11) = .363636......

Conversely, irrational numbers are the numbers that cannot be represented as the quotient of two integers, i.e., irrational numbers cannot be rational numbers and vice-versa. If the set of irrational numbers is denoted by H, then

H = { all x, where there exists no integers p and q such that x = p / q, q is
not zero }

Typical examples of irrational numbers are the numbers p and e, as well as the principal roots of rational numbers. They can be expressed as non-repeating decimals, i.e., the numbers after the decimal point do not repeat their pattern.

Real Numbers

Real numbers are the numbers that are either rational or irrational, i.e., the set of real numbers is the union of the sets Q and H. If the set of real numbers is denoted by R , then

R = Q È H

Since Q and H are mutually exclusive sets, any member of R is also a member of only one of the sets Q and H. Therefore, a real number is either rational or irrational (but not both). If a real number is rational, it can be expressed as an integer, as the quotient of two integers, and it can be represented by a terminating or repeating decimal; otherwise, it is irrational and cannot be represented in the above formats.

Whole numbers are the numbers beginning with 0, with each successive number greater than its predecessor by 1. It combines the set of natural numbers and the number 0. If the set of whole numbers is denoted by N, then

N= { 0, 1, 2, 3, .......}

A variable is a symbol that denotes a numerical quantity. Most variables are alphabetic characters. Algebra utilizes the concept of a variable to generalize mathematical concepts to a particular group of numbers, e.g. the set of real numbers. The number that the variable represents is called its value.

A constant is a symbol that represents a definite mathematical quantity. The most common examples of constants are the numerals used to represent real numbers, e.g., 3, 4.9, 2/3, etc. Symbols such as p and e are also constants, approximately denoting the values 3.141593 and 2.718282, respectively. Also, symbols that always represent the same quantity when used, such as g (acceleration due to gravity) or c (speed of light), are also called constants.

A term is a product with an unspecified number of factors, where the factors are variables or constants. The variables of a term are called literal factors, and the constants the numerical coefficient (or simply coefficient), of the term. Terms whose only factors are constants are called constant terms. Terms that have the same literal factors differ only in their numerical coefficients, and are called similar terms.

An algebraic expression is an additive combination of any number of terms.

By applying the distributive property, two or more similar terms can be combined into one term. The new term has the same literal factors as the similar terms, but its coefficient is the sum of the coefficients of the similar terms. This process is known as combining similar terms.

When each variable in an algebraic expression is replaced by a number, the algebraic expression takes on a numerical value. This process is known as evaluating algebraic expressions.

EX. In the algebraic expression 2x + 5ex²y²- p + 5.2x ; 2x, 5e x²y², -p and 5.2x are the terms. -p is a constant, so it is also a constant term. The term 5ex²y² has a coefficient of 5e and literal factors of x²y² . Since the terms 2x and 5.2x have the same literal factor of x, they are similar terms.

The similar terms 2x and 5.2x can be combined into one term with literal factor x and coefficient equal to (2 + 5.2) = 7.2. The simplified expression is now 7.2x + 5ex²y² - p. If we set x = 1 and y = 2, the expression is evaluated as

7.2(1) + 5e(1²)(2²) - p = 7.2 + 5e(4) - p = 58.4240

An algebraic equation is a mathematical statement equating two algebraic expressions, i.e., it states that one algebraic expression is equal to another algebraic expression.

The following additive and multiplicative properties apply to equations.

Addition Property of Equality:
For all real numbers a, b, and c,
a = b if and only if a + c = b + c.

Multiplication Property of Equality:
For all real numbers a, b, and c, where c ¹ 0,

a = b if and only if ac = bc.

An algebraic inequality is a mathematical statement comparing two unequal algebraic expressions. It states that one algebraic expression is greater than (>), less than (<), greater than or equal to (³), or less than or equal to (£), another algebraic expression.

The following additive and multiplicative properties apply to inequalities.

Addition Property of Inequality:
For all real numbers a, b, and c,
a > b if and only if a + c > b + c.

Multiplication Property of Inequality:
For all real numbers a, b, and c, where c ¹ 0,
(1) If c > 0, then a > b if and only if ac > bc.
(2) If c < 0, then a > b if and only if ac < bc.

A monomial is a term containing variables with only nonnegative integers as exponents. The degree of a monomial is the sum of the exponents of its variables. The finite sum (or difference) of a set of monomials is known as a polynomial. The polynomial is also an algebraic expression, being an additive combination of terms.

A monomial refers to a polynomial with just one term. Similarly, a polynomial with two terms is a binomial; a polynomial with three terms is a trinomial.

The degree of a polynomial is the degree of the monomial (in the polynomial) with the highest degree.

EX. The following are examples of polynomials

(1) 3x is a monomial of degree 1.
(2) 3x + 4y is a binomial of degree 1.
(3) -2x² + 3xy²z³+ w³y²z² is a trinomial. The term -2x² has

degree 2, the term 3xy²z³ has degree (1+2+3) = 6, and the term w³y²z² has degree (3+2+2) = 7. Thus, this polynomial has degree 7.

The solution of equations and inequalities involves finding the value (or values) of the variable(s) that make the mathematical statements true. The addition and multiplication properties of equations and inequalities, as well as the properties of real numbers, are used to simplify the equation/inequality as much as possible, prior to formulating the solution set for the variable in question.

Solving linear equations in one variable, i.e., equations dealing with algebraic expressions having one variable with an exponent of one, are the simplest to solve for.

EX.
2x + 3 = 7
2x + 3 + (-3) = 7 + (-3)
2x = 4
(0.5)2x = (0.5) 4
x = 2

EX.
4x + 5 = -15
4x + 5 + (-5) = -15 + (-5)
4x = -20
¼(4x) = ¼(-20)
x = -5

When the equation involves fractions, it is helpful to multiply both sides of the equation by the least common denominator of all fractions:

EX.
¾ x (40) + 4/5 (40) = 7/8 (40)

where the LCD (lowest common denominator) of 4, 5 and 8 is 40

30x + 32 = 35
30x + 32 - 32 = 35 - 32
30x = 3
x = 3 / 30 = 1/10

When the equation involves decimals, it is helpful to multiply both sides of the equation by the lowest power of 10 that converts all decimal numbers to integers:

EX.
2.5x - 3.2 = 12.5
2.5x - 3.2 + 3.2 = 12.5 + 3.2
2.5x = 15.7

By multiplying each side by 10, we get

25x = 157
x = 157/25 = 6.28

The process of solving inequalities is similar to that of solving equations, except that different additive and multiplicative properties apply.

EX.
2x - 3 > 5
2x - 3 + 3 > 5 + 3
2x > 8
½ (2x) > ½ (8)
x > 4
The solution set is { x | x > 4}

EX.
-3x - 4 < 8
-3x - 4 + 4 < 8 + 4
-3x < 12
-1/3 (-3x) > -1/3 (12)

(note the change in the inequality from < to >)
x > -4

The solution set is { x | x > -4}

EX.
1/3 x -3/4 < 1/2
1/3 x (12) -3/4 (12) < 1/2 (12)

where the LCD of 2, 3 and 4 is 12

4x - 9 < 6
4x - 9 + 9 < 6 + 9
4x < 15
x < 15/4

The solution set is { x | x < 15/4}.

If the equation or inequality involves absolute values, then its solution will involve solving two equations or inequalities.

(1) | ax + b | = k is equivalent to ax + b = k or ax + b = -k.

EX.
|2x + 1| = 5
2x + 1 = 5 or 2x + 1 = -5
2x = 4
2x = -6
x = 2 or x = -3
\ the solution set is {2 , -3}.

(2) | ax + b | < k is equal to -k < ax + b < k, where k > 0.

EX.
|3x + 2| < 7
-7 < 3x + 2 < 7
-9 < 3x < 5
-3 < x < 5/3

\ the solution set is the interval (-3, 5/3).

(3) | ax + b | > k is equal to ax + b > k or ax + b < -k, where k > 0.

EX. | 2x - 1 | > 3
2x - 1 > 3 or 2x - 1 < -3
2x > 4 or 2x < -2
x > 2 or x < -1

\ the solution set (-¥ , -1) È ( 2 , ¥ )

Descartes' Rule Of Signs:

Given a polynomial equation with real coefficients:

1. The number of positive real solutions to the given equation is either equal to the number of variations in the sign of the polynomial or two less than the number of the variations in the sign of the polynomial.

From the same polynomial:

with x replaced with -x; the number of negative real solutions to the polynomial equation is either equal to the number of variations in the sign of the polynomial or two less than the number of the variations in the sign of the polynomial.

number of variations is the number of sign changes from one term to the other.

ex.
x³- 3x²+ 4x - 5 = 0

There are three different sign changes in the equation.

(+ x³ to - 3x² is a sign change or variation)

therefore there are either three or one real positive solution to the equation.

Replace x with -x in the equation gives:

(-x)³ - 3(-x)²+ 4(-x) - 5 = 0
-x³- 3x²- 4x - 5 = 0

There are no variations in the sign, therefore there are no real negative solutions to the equation.

Graphing Polynomial Functions:

Standard Form for Linear Functions:

f(x) = ax + b

Standard Form for Quadratic Functions:
f(x) = ax²+ bx + c

The linear and quadratic functions above are special case polynomial functions.

Standard Form for Polynomial Functions:

f(x) =

This is a polynomial function with degree n.

n is a nonnegative integer.

a_n is a real number greater than zero.
a_n, a_n-1 ,. . . , a₁, a₀ are real numbers.

Consider the polynomial function:

f(x) = axⁿ

If n = 1, then it is an equation of a line that passes through the origin.
If n = 2, then it is a parabola with vertex at the origin and symmetric about the y-axis.
If a ¹ 1 and n = 3 ( f(x) = x³) the graph looks like:

Graphs of the function f(x) = axⁿ were n > 2 and a ¹ 1 are:

If n is even the curve is symmetrical about the y-axis.
If n is odd the curve is symmetrical about the origin.

see translation of axes:

Graphing Factored Polynomial Functions

ex.
f(x) = (x + 1)(x - 2)(x + 3)

find all the zero points of the graph:

f(x) = 0 = (x + 1)(x - 2)(x + 3)

x + 1 = 0 or x - 2 = 0 or x + 3 = 0
x = -1 or x = 2 or x = -3

Substitute values of x from each interval to the function to see what the graph will look like. For each value of x, f(x) is either always positive or always negative. Meaning that the graph is either above the x-axis if positive and below the x-axis if negative.

A function of the form:

were p(x) and q(x) are polynomial functions; this is a rational function.

examples of rational functions:

Some suggestions on how to go about graphing rational functions:

Check for symmetry; about the y-axis or the origin or both.
Find any vertical asymptotes.
Find any horizontal asymptotes.
Find out what the graph does near the asymptotes.
Plot points to see the shape of the graph.

Finding the Vertical and Horizontal Asymptotes:

Finding the Vertical Asymptotes:
Take the denominator, make it equal to zero and solve.

ex.

The denominator is x + 1, and the vertical asymptote is:

x + 1 = 0
x = -1

The vertical asymptote is the line x = -1

Finding the Horizontal Asymptote:

From the example above.

The vertical asymptote is x = -1.

To find the horizontal asymptote, divide the variable in the numerator with the numerator and the denominator.

 

As x gets very large from negative infinity ( - ¥ ), f(x) rises slightly from 2, and as x gets smaller from positive infinity, f(x) is almost 2, and is slowly falling.

ex.

the graph is symmetric about the y-axis.

vertical asymptotes:

x²- 25 = 0
x² = 25
x = ±5

Since every value of x is squared the horizontal axis is the x-axis.

A polynomial equation or function with degree n has n number of solutions; for example, a polynomial with a degree of three has three solutions.

ex.
(x - 4)³(x + 7)²(x + 2) = 0

This equation has a degree of six
This can also be written as:

(x - 4)(x - 4)(x - 4)(x + 7)(x + 7)(x + 2) = 0

which implies that there are six solutions to the equation:

 

x-4 = 0	or	x-4 = 0	or	x-4 = 0
x = 4	or	x = 4	or	x = 4 or
x+7 = 0	or	x+7 = 0
x = -7	or	x = -7	or
x+2 = 0
x = -2

The general solution for the equation is{4,-7,-2}, but it is also said that the equation has a solution 4 with multiplicity of three, a solution -7 with multiplicity of two.

Multiplicity is a repetitive solution to an equation, from the above example, the solution 4 has a multiplicity of three, meaning that the solution 4 is repeated three times.

Any solution of multiplicity p is counted p times.

Solving Polynomial Equations( with degree 3 or greater):

ex.
divide (2x⁴- 5x³+ 7x²+ 3x + 2) by (x - 3)

In the divisor, use 3 instead of -3, so that addition is used instead of subtraction.

The last row indicate a quotient of 2x³ + x² + 10x + 33 with remainder 101

ex.

divide 3x³ - 4x + 3 by (x + 1), using synthetic division.
3x- 4x + 3 is equal to 3x+ 0x- 4x + 3

The solution is 3x² - 3x -1 remainder 4.

If f(c) = 0 then x - c is a factor of the polynomial function f(x).

ex.

Is (x + 5) a factor of x³+ 7x²+ 7x- 15.

evaluate f(x) = x³+ 7x²+ 7x- 15 at f(-5).
f(-5) = (-5)³+ 7(-5)²+ 7(-5) - 15
= -125 + 175 - 35 -15
= 0
(x + 5) is a factor of x³+ 7x²+ 7x -15.

the remainder is 0, thus (x + 5) is a factor of x³+ 7x³+ 7x -15.

In the polynomial;

a_n xⁿ + a_n-1 x^n-1 + · · · + a₁ x + a₀ = 0

where the coefficients a_n ,a_n-1 , . . . ,a₁ ,a₀ are integers.

If a rational number c/d, is factored to its lowest terms, is a solution to the equation, then c is a factor of the constant term of the polynomial and d is a factor of the leading coefficient of the polynomial.

ex.
find solutions to the equation:
6x³- 13x²- 13x + 30 = 0

c = constant term = 30
factors of 30 are: ±1,±2,±3,±5,±6,±10,±15,±30
d = leading coefficient = 6
factors of 6 are: ±1,±2,±3,±6

use synthetic division to find a factor for the polynomial:
NOTE:
(x - c/d) is factor of the polynomial if the remainder is zero.
try the integers first;
1 is not a factor since the remainder is not 0.

-1 is not a factor since the remainder is not 0.

2 is a factor since the remainder is 0.

 

(x-2)(6x²- x - 15) = 0

 

6x²- x - 15 can be factored by using the conventional factoring techniques.

(x-2)(6x²-10x + 9x - 15) = 0
(x-2)[2x(3x - 5) + 3(3x - 5)] = 0

(x-2)(2x + 3)(3x - 5) = 0

x = 2 or x = -3/2 or x = 5/3

ex.
6x⁴- x³- 21x²- 6x + 8 = 0

c = 8; ±1,±2,±4,±8
d = 6; ±1,±2,±3,±6

try the integers;

1 is not a factor since the remainder is not 0.

factor out 6x³- 7x²- 14x + 8 by using the same process as above.

c = 8; ±1,±2,±4,±8
d = 6; ±1,±2,±3,±6

try the integers;

(x + 1)(x - 2)(6x²+ 5x -4) = 0

factor 6x²+ 5x -4.

(x + 1)(x - 2)(6x² + 5x - 4) = 0
(x + 1)(x - 2)(6x² - 3x + 8x - 4) = 0
(x + 1)(x - 2)[3x(2x - 1) + 4(2x - 1)] = 0
(x + 1)(x - 2)(2x - 1)(3x + 4) = 0

6x⁴- x³- 21x²- 6x + 8 = (x + 1)(x - 2)(2x - 1)(3x + 4) = 0

x = -1 or x = 2 or x = 1/2 or x = -4/3

If the polynomial function f(x) is divided by ( x - c ) then the remainder is equal to f(c).

ex.
evaluate f(x) = x³ - 3x² + 2x -1 at f(3) by synthetic division and the
remainder theorem.

ex.
Find the remainder of 3x³- x²+ 5x - 4 divided by (x + 2).
Use both synthetic division and the remainder theorem.

Using the remainder theorem:
Write x + 2 as x - (-2), thus f(-2), then evaluate f(x) = 3x³- x²+ 5x - 4
at f(-2).

f(-2) = 3(-2)³- (-2)²+ 5(-2) - 4
= 3(-8) - (4) + (-10) - 4
= -24 - 4 -10 - 4
= -42 the remainder is -42.

using synthetic division:

Completing the Square: ex. x2 + 10x + 5 = 0 x2 + 10x = -5 subtract 5 to both sides. x2 + 10x + 25 = -5 + 25 half of 10 squared, ( 10 ÷ 2 = 5; 52 = 25) added to both sides. (x + 5)2 = 20 factor x2 + 10x + 25. x2 + 14x - 3 = 0 x2 + 14x = 3 add 3 to both sides. x2 + 14x + 49 = 3 + 49 half of 14 squared, ( 14 ÷ 2 = 7; 72 = 49) added to both sides. ( x + 7 )2 = 52 factor x + 14x + 49. 3x + 6x + 7 = 0 3x + 6x = -7 3( x + 2x ) = -7 3( x + 2x + 1) = -7 + 1 3(x + 1)2 = -6 multiply 1/3 to both sides

Quadratic Equations (2nd degree equations in one variable. )
Standard form is ax²+ bx + c.

ex.
s² + 2s - 8 = 0
( s + 4 )( s - 2 ) = 0
s + 4 = 0 or s - 2 = 0
s = -4 or s = 2
\ the solution set is { -4,2 }

(3t - 9)^1/2 = t-3

[(3t-9)^1/2]² = (t-3)²
3t - 9 = t² -6t + 9
0 = t² -6t - 3t + 9 + 9
0 = t² - 9t + 18
0 = ( t - 3 )( t - 6 )

t - 3 = 0 or t - 6 = 0
t = 3 or t = 6

Then Check by SUBSTITUTING the values back in to see if the equation makes sense.

\ the solution set is { 3,6 }

For any number a,

x² = a if and only if x = or x = -.

ex.
y² = 64
y = ±8

\ the solution set is { ± 8 }

The general forms for quadratic inequalities are:
ax²+ bx + c > 0
ax²+ bx + c < 0
ax²+ bx + c >= 0
ax²+ bx + c =< 0

The number line plays an important role in solving factorable quadratic inequalities.

Consider the following :

x²+ x -12 >= 0
x²+ x -12 >= 0
( x + 4 )( x - 3 ) >= 0
( x + 4 )( x - 3 ) = 0

factor and solve for x, if the inequality is equal to zero.

x + 4 = 0 or x - 3 = 0x = -4 or x = 3

The numbers -4 and 3 are the critical numbers for the inequality, which is then plotted in the number line.

 

After the critical numbers are plotted, the number line is divided into three intervals;
-, -4; -4, 3 and 3, .

Now, find test numbers in each interval to find its affects on the signs of the factors, x + 4 and x - 3, also consequently, the product of these factors;
( x + 4 )( x - 3).

 

The diagram shows the sign of the factors for each interval and also the sign of the products of the factors for each interval.

From the number line it can be clearly seen the solution to the inequality.

x²+ x -12 >=0

\ the solution set is (-, -4] [ 3, ).

Consider the following:

x²-35 =< 2x
x²-2x -35 =< 0
( x + 5 )( x - 7 ) =< 0

( x + 5 )( x - 7 ) = 0

x + 5 = 0 or x - 7 = 0
x = -5 or x = 7

 

( x + 5 )( x - 7 ) = < 0

\ the solution set is [ -5, 7 ]

Quadratic Rational Inequalities: Consider the example: (x + 4) / (x - 2) < 2 x + 4 = 0 and x - 2 = 0 x = -4 x = 2 (x + 4) / (x - 2) < 0 The solution set is ( -4, 2 ) ex. x - 4 = 0 x + 2 = 0 x + 3 = 0 x = 4 x = -2 x = -3 the solution set is (-4, -2] [ 4, )

ax²+ bx + c = 0

The quadratic formula is used to solve unfactorable quadratic equations.

ex.
x² + 3x + 6 = 0
a =1; b = 3; c = 6.

To check to see if the roots are correct for the equation, use the sum and product rule of two roots. The sum of the roots produces the relationship of -b/a.

ex.
x² + 3x + 6 = 0

a = 1, b = 3, c = 6

-b/a = -3/1 = -3
( The sum of the roots)

The product of the roots produces the relationship c/a.

b² - 4ac is called the discriminant. It provides a way to find roots for the quadratic formula without solving the whole equation.

If b² - 4ac = 0 then, the equation has a double root or one real solution.
If b² - 4ac > 0 then, the equation has two real but unequal roots.
If b² - 4ac < 0 then, the equation has two nonreal, complex, unequal roots; which are complex conjugates of each other.

ex.
2x² - 4x + 2 = 0
a = 2, b = -4, c = 2

b² - 4ac
(-4)² - 4(2)(2)
16 - 16 = 0
b² - 4ac = 0

there are two real and equal roots.

3x² - 5x - 2 = 0
a = 3, b = -5, c = -2

b² - 4ac
(-5)2 - 4(3)(-2)
25 - (-24) = 25 + 24 = 49
b² - 4ac > 0

there are two real and unequal roots.

5x² + 4x + 1 = 0
a = 5, b = 4, c = 1

b² - 4ac
(4)2 - 4(5)(1)
16 - 20 = -4
b² - 4ac < 0

there are two nonreal and complex roots.

Absolute Values

The absolute value of a real number is the distance between its corresponding point on the number line and the number 0. The absolute value of the real number a is denoted by |a|.

From the diagram, it is clear that the absolute value of nonnegative numbers is the number itself, while the absolute value of negative integers is the negative of the number. Thus, the absolute value of a real number can be defined as follows:

For all real numbers a,
(1) If a >= 0, then |a| = a.
(2) If a < 0, then |a| = -a.

Examples:
| 2 | = 2
| -4.5 | = 4.5
| 0 | = 0

The arithmetic operations with real numbers are governed by the following axioms:

(1) Closure Axiom of Addition / Multiplication

For real numbers a and b,

a + b is a unique real number
ab is a unique real number

(2) Commutative Axiom of Addition / Multiplication

For real numbers a and b,

a + b = b + a
ab = ba

(3) Associative Axiom of Addition / Multiplication

For real numbers a, b and c,

( a + b ) + c = a + ( b + c )
(ab)c = a(bc)

(4) Identity Axiom of Addition

For any real number a,

a + 0 = 0 + a = a

(5) Identity Axiom of Multiplication

For any real number a,

a(1) = 1(a) = a

(6) Additive Inverse Axiom

For any real number a, there exists a unique real number -a such that

a + (-a) = -a + a = 0

The number -a is known as the additive inverse of a.

(7) Multiplicative Inverse Axiom

For any nonzero real number a, there exists a unique real number
( 1 / a ) such that

a ( 1 / a ) = ( 1 / a ) a = 1

The number ( 1 / a ) is known as the multiplicative inverse or reciprocal of a, where a ¹ 0.

(8) Distributive Axiom

For any real numbers a, b, and c,

a ( b + c ) = ab + ac
a ( b - c ) = ab - ac
( a + b) c = ac + bc
( a - b) c = ac - bc

As multiplication is related to the concept of "repeated addition", exponentiation involving integers is akin to "repeated multiplication". The use of positive integers as exponents is illustrated as follows:

bⁿ = b × b × .... × b where the number of b's to be multiplied together is n = x

where b is a real number and n is a positive integer. In the above equation, b is the base, n is the exponent, and x is the real number b raised to the nth power, or the nth power of b.

The following formulas are useful in algebraic manipulations involving exponents:

(1) a^m × aⁿ = a^m+n
(2) ( a^m )ⁿ = a^mn
(3) a⁰= 1, a ¹ 0
(4) a^-n = 1 / an , a ¹ 0
(5) a^m / aⁿ = a^m-n , a ¹ 0

where a is a real number, m and n are positive integers, and the above constraints are satisfied.

Formulas (1) and (2) can be derived from the definition of exponents using positive integers. Formula (3) defines exponentiation with zero; any real number raised to the zeroth power is 1. Formula (4) defines exponentiation using negative integers; a real number raised to the nth power, where n < 0, is the reciprocal of the same real number raised to the mth power, where m = -n > 0. Formula (5) is derived from Formulas (1) and (4).

The above formulas also apply when the exponents are positive real numbers. This will be shown in later sections.

The multiplication of two real numbers is similar to the concept of "repeated addition". This is best illustrated with the multiplication of two positive integers x and y:

x × y = xy = x + x + .... + x where the number of x's to be added together is y = y + y + .... + y where the number of y's to be added together is x = z

In the above equation, x is the multiplicand, y the multiplier, and the result z is the product of x and y. The similarity between multiplication and "repeated addition" is slightly less evident when non-integers and negative numbers are involved. With respect to multiplication of positive non-integers, the non-integer real number can be separated into its integer and fractional parts, and multiplication can then be defined as follows:

Given real numbers a, b, where a > 0, b > 0

b = c + d, c is an integer, 0 £ d < 1

ab = a × (c + d) = ac + ad
using the distributive axiom

ac = a + a + .... + a
where a is added together c times (a and c are positive integers)

ad is the real number whose ratio to a is equal to d,
i.e., is (100 × d) percent of a.

ab is the sum of ac and ad.

When negative numbers are involved in multiplication, the only difference in the result is its sign (positive or negative); the absolute value of the product remains unchanged. Specifically, the product of a positive real number and a negative real number is negative, and the product of two negative real numbers is positive.

The product of one and any real number is the real number itself; if a real number is added together once, the result is the same real number, of course. Similarly, the product of zero and any real number is zero; if a real number is added together zero times, nothing is added together, so the result is zero.

The division of two real numbers can be defined in terms of multiplication. Given any two real numbers a and b,

a / b = a × (1/ b) = c

where c is the result of the division of a by b. In the above equation, a is the dividend, b the divisor, and c the quotient of a and b.

"a / b" can be thought of as partitioning a into b equal parts, and then determining the value of the b equal parts. Since a real number cannot be divided up into zero equal parts, it is not plausible to consider the expression ( a / b ) when b = 0 (The equation a / 0 = c implies that 0 × c = a for some real number c, but 0 × c = 0 for all real numbers c.). Therefore, division by zero is undefined. However, division by one is acceptable; the quotient of a real number and one is the original real number.

The following are summary statements regarding the operations of multiplication and division:

(1) The product of zero and any real number is zero.
(2) The product of one and any real number is the real number.
(3) The quotient of zero and any real number is zero.
(4) The quotient of any real number and one is the real number.
(5) Division by zero is undefined, i.e., zero cannot be a divisor.
(6) The product of two positive integers or two negative integers is the product of their absolute values.

Ex. 2 × 2 = -2 × -2 = | -2 | × | -2 | = 4

(7) The product of a positive and a negative integer is the negative of the product of their absolute values.

Ex. 2 × -3 = -2 × 3 = - ( | -2 | × | -3 | ) = -6

(8) The quotient of two positive integers or two negative integers is the quotient of their absolute values.

Ex. ( 4 / 2 ) = ( -4 / -2 ) = | -4 | / | -2 | = 2

(9) The quotient of a positive and a negative integer is the negative of the quotient of their absolute values.

Ex. ( 6 / -3 ) = ( -6 / 3 ) = - ( | -6 | / | -3 | ) = -2

A real number can be represented in different ways. As mentioned before, a rational number (which is also a real number) can be expressed as an integer, as a quotient of two integers, and thus as a fraction (two integers separated by a "/"), or as a terminating or repeating decimal. Irrational numbers can have the form of fractions (where the numerator and/or denominator are not integers) or of non-repeating decimals. Real numbers can also be represented by expressions involving exponents and/or radicals, as well as by special symbols. The following are examples of real number representations:

4 = ( 8 / 2 ) = 4.00 = 2²
( 1 / 6 ) = 0.166666......
p = 3.14159265358979....
e = 2.718281828459....

Real numbers can also be expressed in scientific and/or set notation. Scientific notation is used for individual real numbers, while set notation is used to indicate groups of real numbers.

All real numbers have the following properties:

(1) Reflexive Property For any real number a, a = a. Example: 3 = 3, y = y, x + z = x + z (x, y and z are real numbers)

(2) Symmetric Property For any real numbers a and b, if a = b, then b = a. Example: If 3 = 1 + 2, then 1 + 2 = 3

(3) Transitive Property For any real numbers a, b and c, if a = b and b = c, then a = c. Example: If 2 + 3 = 5 and 5 = 1 + 4, then 2 + 3 = 1 + 4.

(4) Substitution Property For any real numbers a and b, if a = b, then a may be replaced by b, and b may be replaced by a, in any mathematical statement without changing the meaning of the statement. Example: If a = 3 and a + b = 5, then 3 + b = 5. (5) Trichotomy Property For any real numbers a and b, one and only one of the following conditions holds:

(1) a is greater than b ( a > b)

(2) a is equal to b ( a = b) (3) a is less than b ( a < b) Example: 3 < 4 , 4 + 2 = 6 , 7 > 5

A real number X expressed in scientific notation has the format

X = a × 10^k, where 1 £ |a| < 10 and k is an integer

Example:
1,201,000,000,000 = 1.201 × 10¹²
0.00000000000035 = 3.5 × 10^-13
-1345.89 = -1.34589 × 10^-3

To convert a real number from standard notation to scientific notation
(a × 10^k), the following procedure is useful:

(1) If the real number is not already in decimal form, convert it to a decimal number, terminating, repeating or otherwise. Note the location of the decimal point.
(2) Determine the first nonzero digit in the number. Place a marker immediately after this digit.
(3) Starting at the marker, count the number of places to the decimal point. If the movement is to the right, the count is considered to be positive; if the movement is to the left, the count is considered negative. If no movement is need, the count is zero.
(4) The marker indicates the location of the decimal point in the factor a. The count represents the exponent k to be used as the power of 10.

In the study of algebra, real numbers are often mentioned as a group, e.g. the set of real numbers greater than x, or the collection of real numbers satisfying a particular equation. Therefore, it is often convenient to indicate a particular group of real numbers using set notation. In this manner, the real numbers can be individually listed as part of a collection, or a continuum of real numbers can be represented concisely.

A set is a collection of objects. The objects are called the elements or members of the set. Conventionally, the pair of set braces, { } , are used to enclose the elements (or description thereof) of a set, using commas to separate the individual elements. Capital letters are normally used as names for sets.

If an object x is an element of set A, it is denoted by x Î A. Otherwise, the statement x Ï A, which means "x is not an element of A", is true.

The union of two sets A and B, denoted as A È B, is the set of all elements that are contained in sets A, B or both. The intersection of two sets A and B, denoted as A Ç B, is the set of all elements that belong to both sets A and B. An empty set has no elements; it is also called the null set and is denoted as Æ.

Two sets A and B are considered equal if they contain exactly the same elements. This is denoted by A = B. The statement A = B also implies that

A È B = A Ç B = A = B

Set-builder notation is generally used to represent a group of real numbers. It stipulates that sets be written in the format { x : x has property Y } , which is read as "the set of all elements x such that x has the property Y; the colon ":" means "such that". Using this notation, a set is often defined as the collection of real numbers that belong to either an open, closed, half-open, or infinite interval (of real numbers).

An open interval is a set of real numbers represented by a line segment of the real number line, whose endpoints are not included in the interval. This concept is made clear by the following definition:

( a, b ) = { x : a < x < b } where a < b

Since the endpoints of an open interval are not part of the interval,
a Ï ( a, b ) , b Ï ( a, b )

In contrast, a closed interval is a set of real numbers represented by a line segment of the real number line, whose endpoints are included in the interval. Its definition is as follows:

[ a, b ] = { x : a £ x £ b } where a < b

Since the endpoints of a closed interval are part of the interval,
a Î [ a, b], b Î [ a, b]

A half-open interval is also a set of real numbers represented by a line segment on the real number line, but with one endpoint included in the interval, and the other endpoint not included in the interval. They are defined as follows:

[ a, b ) = { x : a £ x < b } where a < b

( a, b ] = { x : a < x £ b } where a < b

An infinite interval is a set of real numbers, but it is represented by a ray or line on the real number line. As befits the name, an infinite interval does not have an endpoint in one or both directions. They are defined as follows:

(-¥, a ] = { x : x £ a }
[ a, -¥) = { x : x ³ a }
(-¥, ¥) = { x : x is a real number }

Similar definitions apply when the infinite interval is open at one end.