Christ the King College of Cavite Foundation, Inc.San Marino West, Salawag, Dasmariñas City, Cavite

MyMy Portfolio Portfolio

inin MathematiMathemati

cs IVcs IVKristel Anne D. Jazmin

Author

IntroductioIntroductionn

This Portfolio contains the lessons from Prerequisite IV to Quadratic Functions from the book E-Math (Advanced Algebra and Trigonometry). This is to show how far Seniors had learned in our first grading period.

Numbers are really hard to deal for, they seem irresolvable. So, I would like to impart some of the basic ideas in Mathematics, how to deal with these and how to make it easier to solve. Formulas have been given as well as some techniques and shortcuts in dealing with numbers. One must undergo in the long method first before going to the short one. So, follow the steps carefully. Besides, easy to understand ideas are laid out and examples are given for you to comprehend more. Just don’t miss a single word/number or you’ll miss one-third of your life!

This Portfolio is expecting that that the readers will be able to understand the basics in Trigonometry and refresh (for the Seniors) the tackled lessons. This time, readers must be familiar with it so that it would be easy for them in its application and there’s no need to go and read back the previous lessons once they need to apply it, especially for the next lessons in Trigonometry (Math – IV).

Just keep in mind that math is not a hard subject. It’s you who make it hard to deal with. Solving math is just CHALLENGING!

Table ofTable of ContentsContents

Introduction

Prerequisite4: Multiplying Polynomials

Prerequisite5: Factoring

Prerequisite6: Dividing One Polynomial by Another

Prerequisite7: Rational Expressions

Prerequisite8: Radicals

Prerequisite9: Complex Numbers

Prerequisite10: Ordered Pairs

Prerequisite11: Summation Notation

Lesson 1.1: Relations and Functions

Lesson 1.2: Evaluating Functions

Lesson 1.3: Operations and Functions

Lesson 1.4: Composite and Inverse Functions

Lesson 1.5: Linear Functions

Lesson 1.6: Quadratic Functions

Prerequisite 4:

Multiplying Polynomials

The Product Rule

For any number x and positive integers n and man • am = an+m , add the exponents of the same base.

Examples:1. x2 • x6 = x2+6 = x8

2. 2x3 + 4x3= (2)(4)(x3+3) = 8x6

3. (x4x3)(xy3) = (x4)(x3)(x)(y3) = x4+3+1y3 = x8y3

For any number x and any positive integer m and n(xm)n = xmn, copy the base and multiply the exponents.

Examples:1. (x3)5 = x(3)(5) = x15

2. (25)2 = 210

3. (ay)2 = a2y

For any number b and c and any positive integer x(bc)x = bxcx, raise each factor to the indicated power.

Examples:1. (ab)2(2a2)3 = (a2b2)(23a(2)(3)) = (a2b2)(8a6) = 8a6+2b2 = 8a8b2

2. (xy)2 = x2y2

3. (x2y)2 • 2xy = x(2)(2)y2 • 2xy = x4y2 • 2xy = (2x)(x4)(y2)(y) = (2x4+1)(y2+1) = 2x5y3

Polynomials Four or more terms 6x 5 + 3x 3 + 10x 2 + 2x - 3

Trinomial Three terms 3y 2 + 6y + 4

Binomial Two terms 18x – 9

Monomial One term 12y

The Product of a Monomial and a Polynomial

Use the Distributive Property to multiply the monomial with each term of the polynomial.

Examples:1. 3x2(2x3 + 2x2 + x – 5) = 3x2(2x3) + 3x2(2x2) + 3x2(x) - 3x2(5)

= 6x6 + 6x4 + 3x3 – 15x2

2. 2xy(x2y – 2y + 3x + 4) = 2xy(x2y) – 2xy(2y) + 2xy(3x) + 2xy(4) = 2x3y2 – 4xy2 + 6x2y + 8xy

= 2x3y2 + 6x2y – 4xy2 + 8xy3. 6x(x3 + x2 + x + 2)2 = 6x(x6 + x4 + x2 + 4)

= 6x7 + 6x5 + 6x3 + 24x

The Product of Two Binomials

Use the FOIL Method (First terms, Outer terms, Inner terms, Last terms)

Examples:1. (x + 4)(x – 1) = (x)(x) + (x)(-1) + (4)(x) + (4)(-1)

= x2 – x + 4x – 4= x2 + 3x – 4

2. (2x + 1)(3x + 1) = (2x)(3x) + (2x)(1) + (1)(3x) + (1)(1)= 6x2 + 2x + 3x + 1= 6x2

+ 5x +13. (2y + 5)(y – 6) = (2y)(y) + (2y)(-6) + (5)(y) + (5)(-6)

= 2y2 – 12y + 5y – 30= 2y2 – 7y – 30

The Product of Two Polynomials

Distribute each term of the Binomial to the Polynomial.

Examples:1. (2x + 5)(x2 – 3x + 7) = 2x(x2 – 3x + 7) + 5(x2 – 3x + 7)

= (2x3 – 6x2 + 14x) + (5x2 – 15x + 35)= 2x3 – x2 – x + 35

2. (x – 4)(x2 + 2x + 1) = x(x2 + 2x + 1) – 4(x2 + 2x + 1)= (x3 + 2x2 + x) – (4x2 + 8x + 4)= x3 – 2x2 – 7x – 4

3. (y + 8)(y2 + 2y + x + 1) = y(y2 + 2y + x + 1) + 8(y2 + 2y + x + 1)= (y3 + 2y2 + xy + y) + (8y2 + 16y + 8x + 8)= y3 + 10y2 + xy + 17y + 8x + 8

Special Products

Sum and Difference of the Same Two Terms(a + b)(a – b) = a2 – b2

Examples:1. (4x + 2)(4x – 2) = (4x)2 – (2)2 = 16x2 – 42. (2y – 1)(2y + 1) = (2y)2 – (1)2 = 2y2 – 13. (5x – 4)(5x + 4) = (5x)2 – (4)2 = 25x2 – 16

Square of a Binomial(a + b)2 = a2 + 2ab + b2

(a – b)2 = a2 – 2ab + b2

Examples:1. (x + 2)2 = x2 + (2)(x)(2) + (2)2 = x2 + 4x + 42. (2x – 4)2 = (2x)2 – (2)(2x)(-4) + (-4)2 = 4x2 + 16x + 163. (3x + 5)2 = (3x)2 + (2)(3x)(5) + (5)2 = 9x2 + 30x + 25

Cube of a Binomial(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a – b)3 = a3 – 3a2b + 3ab2 – b3

Examples:1. (x + 7)3 = x3 + (3)(x)2(7) + (3)(x)(7)2 + (7)3 = x3 + 21x2 + 147x + 3432. (2x + 2)3 = (2x)3 + (3)(2x)2(2) + (3)(2x)(2)2

+ (2)3 = 8x3 + 24x2 + 24x + 8

3. (x – 4)3 = x3 – (3)(x)2(-4) + (3)(x)(-4)2 – (-4)3 = x3 + 12x2 + 48x + 64

Prerequisite 5:

Factoring

“Whoah! That was tough! Two years had already passed since we had tackled this topic (If I’m not mistaken, I was in second year high school).

At first, it’s hard because I almost forgot how to multiply polynomials, but still at the end, it seems like I recovered a piece of puzzle in my mind. Though I’m always confused with the signs, especially when performing operations on it, I can still manage solving. I just have to determine first what method is to apply depending on the number of terms being multiplied.

In this lesson, I learned dealing with COMPLICATED integers and variables. But once you get familiar with these, they’ll just be nuts for you. How I wish, as we’re moving on with our lessons, I could still remember this topic.”

- Author

Getting Any Better?

Common Factorax + bx = x(a + b)

Examples:1. 2x2 + 6x = 2x(x + 3)2. 4x + 8x2 = 4x(1 + 2x)3. x2 + 4x) = x(x + 4)

Difference of Two Squaresa2 – b2 = (a + b)(a – b)

Examples:1. (4x – 16)2 = (2x + 4)(2x – 4)2. 4x2 – 42 = (2x + 2)(2x – 2)3. 16x2 + 362 = (4x + 6)(4x – 6)

Sum/Difference of Two Cubesa3 – b3 = (a – b)(a2 + ab + b2)a3 + b3 = (a + b)(a2 – ab + b2)

Examples:1. x3 + 27 = (x + 3)(x2 + 3x + 9)2. 8x3 – 64 = (2x – 4)(4x2 – 8x + 16)3. x3 + 8 = (x + 2)(x2 + 2x + 4)4.

Perfect Square Trinomiala2 + 2ab + b2 = (a + b)2

Examples:1. x2 + 8x + 16 = (x + 4)2

2. 4x2 + 20x + 25 = (2x + 5)2

3. x2 + 14x + 49 = (x + 7)2

Quadratic TrinomialUse the Trial and Error Method.

ORFactor the first term then find two numbers that when you multiply the answer is the

last term and when you add, the answer is the middle term.

x2 + 5x + 6 = (x + __)(x + __) Factor out x2.= (x + 2)(x + 3) Find two numbers that when multiplied

the answer is the last term of the given equation and when added the answer is the numerical coefficient of the middle term of the given equation.

Examples:1. x2 + 7x + 10 = (x + 5)(x + 2)2. x2 – 8x + 15 = (x – 5)(x – 3) 3. 4x2 + 5x – 6 = (2x + 6)(2x – 1)

“If we know how to multiply, of course we also have to know how to find its factors. This lesson shows how to and taught me as well. Actually, this was just a refresh on our previous learning that’s why it’s not too hard, just confusing. Anyway, I could say that I’m getting better in Mathematics. For the readers, be patient and you’ll learn!”

- Author

Getting Any Better?

Prerequisite 6:

Dividing One Polynomial by Another

Quotient RuleFor power of a quotient, both numerator and denominator are raised to the

indicated power, such that

( xy )m

= xm

ym , y ≠0.

Examples:

1. ( 25 )2

=22

52= 425

2. ( 3x7 y )4

= 81 x4

2401 y4

3. ( 2a2b3cd )2

=( 4 a4b29c2d2 )Quotient of Powers

When dividing powers having the same base, exponents are subtracted.xm

xn=xm−n ,m>n

xm

xn=1

xn−m ,m<n

xm

xn=x0=1,m=n

a+bc

=ac+ bc

If the denominator is a monomial, distribute it to the numerators then proceed with division.

Examples:

1. x2 y+x y2

xy= x2 y

xy+ x y2

xy=x+ y

2.6 x4−3 x3+18x2

9 x2=6 x

4

9 x2−3x

3

9x2+ 18 x

2

9 x2=2 x

2

3− x3+2=2 x

2−x+23

3.

2a4−8a2+6a2a

=2a4

2a−8 a

2

2a+ 6a2a

=a3−4a+3

“Among all the four fundamental operations, division is the hardest and possibly the longest one. Although that’s the case, I know I can understand more the division property.

In this lesson, I learned how to divide rational and exponential expressions. It’s just easy if you just keep your eye on it and concentrate so that you’ll not miss anything. This is the problem in dividing rational expressions, once you get out of the line, you’ll have to start at the beginning of your solution to arrive to the solution.

The only problem is the sign (positive and negative). There were times that I’m wrong in the signs but the terms in my answer are correct.”

Getting Any Better?

- Author

Prerequisite 7:

Rational Expressions

A rational expression is any algebraic expression that is in the form of a fraction wherein the numerator is a polynomial and the denominator is a nonzero polynomial.

{ pq |p∧qare polynomials , q≠0}To multiply, multiply numerator by numerator and denominator by denominator then simplify.

pq•rs= pr

qs(q≠0∧s ≠0)

Examples:

1.1215

•34=3660

=35

2.ax+x+5a+x+4

•2x+2ax

¿ (2x+2 )(ax+x+5)

ax (a+x+4)=2a x

2+2 x2+10 x+2ax+2x+10a2 x+a x2+4ax

=14 x+132a+2

To divide, use the cross multiplication process or multiply the dividend by the reciprocal of the divisor.

xy÷ab= x

y•ba=bxay

xy÷ab=bxay

Examples:

1.10x2 y8x y2

÷15 xy4 x

=10x2 y

8x y2•4 x15 xy

= 40 x3 y120 x2 y3

= x3 y2

2.

3x3 y+2 y2+x z2

3 xyz÷2 xyz+3xy−4

xy=3 x

3 y+2 y2+x z2

3 xyz•

xy2 xyz+3 xy−4

=¿ 3 x4 y2+2x y3+ x2 y z2

6 x2 y2 z2+6 x2 y2 z−12 xyz=−x2− y

7

3.x+4x2+5 x

÷x+4x+5 x

= x+4x2+5 x

•x+5 xx+4

= 6 x2+24 xx3+9 x2+20x

= 8x3+8

= 8x3

+1

When adding or subtracting rational expressions with different denominators, find the LCD (Least Common Denominator) for all denominators and change each rational expression that has LCD.

ab+ac=

db

(a )+ dc(a)

dab−ac=

db

(a )−dc(a)

d

LCD = d

Find the LCD. Divide it with the denominator of the first fraction then multiply with its numerator. Perform the same procedure on the next fraction. Then, add/subtract the products

to obtain the numerator of the answer. Reduce to lowest term.

When adding or subtracting rational expressions with the same denominator, copy the denominator then add/subtract its numerator. Then reduce to lowest term.

ab+ cb=a+c

bab− cb=a−c

b

Examples:

1.6

x−5+ 8 xx−5

=8 x+6x−5

2. 3x5

+2 x2−66 x

=18 x+(10 x2−30)

30x=10 x

2+18x−3030 x

=10 x2

30 x+ 18 x30 x

− 3030 x

= x+27 x

“Am I getting any better? Hmm. I think so.

Prerequisite 7 is not that tough. It’s just confusing. This is actually an introduction to rational expressions. I learned how to divide using the long method (like in dividing integers). At first, it’s hard especially when you just look at it. But once you tried to solve on your own, you’ll see that it’s not that hard.

Also, I recalled how to find the LCD or the Least Common Denominator which is necessary in dividing polynomials with different denominator. I just have to remember to factor out first the numerator and the denominator before proceeding to the process. My answers should always be in the simplest form. Simplifying expressions, perhaps, is the easiest one for me. HAHA !

Sometimes, it’s confusing when to use the cross-multiplication process. Now, in this lesson, I knew when to use it. You have to apply cross-multiplication process JUST IN DIVISION, if and only if the divisor is not on its reciprocal. The answer will still be the same.”

- Author

Getting Any Better?

Prerequisite 8: Radicals

Evolution is defined as a process of finding one of the equal factors of a given power (as in radicals).

Radical sign (√) is the symbol which indicates the roots of the number.Radicand is the number inside the radical sign or the number whose roots are to be considered.Index is the small which indicates the order of the radical. It tells what root is to be found.

index3√27 radicand

Radical sign

For any integer n, n > 1, and for any real number b for which n√b is defined, b1n is defined as n√b.

3√ x2 = x23 The index is the denominator of the fractional exponent and the

exponent of the radicand is the numerator of the fractional exponent.

b12=√b The denominator of the fractional exponent is the index of its

equivalent radical expression and the numerator of the exponent is theexponent of the radicand. If the index is 2, there’s no need to write 2 asthe index. And if the exponent of a certain term is 1, there’s no need towrite 1 as its exponent.

Examples:1. 2312=√23

2. (6 x2 )23=3√36 x4

3. xy−23 = 1

3√ x2 y2

Addition and Subtraction

Radical quantities may be added like rational quantities, by combining like terms. Like terms in radical expressions have the same indices and radicands. Before proceeding to the equation, reduce first the radicals.

Subtraction of radical quantities is to be performed in the same manner as addition, except that the signs in the subtrahend are to be changed.

Examples:1. √16 x2+√2=4 x+√2

2. √20 x y3+√5 x y3+√56=√4 •5 • x • y2• y+√5 x • y2• y+√14 •4 ¿2 y √5 xy+ y √5 xy+2√14

¿3 y √5 xy+2√14

Multiplication and Division

Radical quantities may be multiplied, like other quantities, by writing the factors one after another, either with or without the sign of multiplication between them. Quantities under the same radical sign or index may be multiplied together like rational quantities, the product being placed under the common radical sign or index. When radical quantities which are reduced to the same index, have rational coefficients, the rational parts may be multiplied together, and their product prefixed to the product of the radical parts. If the rational quantities, instead of being coefficients to the radical quantities, are connected with them by the signs + and -, each term in the multiplier must be multiplied into each in the multiplicand.

n√ab=n√a• n√ba∧bareboth nonnegative∧n is even∨odd .

The division of radical quantities may be expressed, by writing the divisor under the dividend, in the form of a fraction.

n√ ab=

n√an√b

Examples:

1. √20÷√64=√20√64

=2√58

2. √16 x÷√4 x2=√16 x√4 x2

=4 √x2 x

Rationalizing denominator

One of the “rules” for simplifying radicals is that you should never leave a radical in the denominator of a fraction. The reason for this rule is unclear (it appears to be a holdover from the days of slide rules), but it is nevertheless a rule that you will be expected to know in future math classes. The way to get rid of a square root is to multiply it by itself, which of course will give you whatever it was the square root of. To keep things legal, you must do to the numerator whatever you do to the denominator, and so we have the rule:

If the Denominator is Just a Single Radical

· Multiply the numerator and denominator by the denominator

Example:

· Note: If you are dealing with an nth root instead of a square root, then you

need n factors of that root in order to make it go away. For instance, if it is a cube root (n = 3), then you need to multiply by two more factors of that root to give a total of three factors.

If the Denominator Contains Two Terms

If the denominator contains a square root plus some other terms, a special trick does the job. It makes use of the difference of two squares formula:

(a + b)(a – b) = a2 – b2

Suppose that your denominator looked like a + b, where b was a square root and a represents all the other terms. If you multiply it by a – b, then you will end up with the square of your square root, which means no more square roots. It is called the conjugate when you replace the plus with a minus (or vice-versa). An example would help. Example:

Given:

Multiply numerator and denominator by the conjugate of the denominator:

Multiply out:

“Radicals are my favorite! They’re fun to solve.

In this lesson, the basics in dealing with radical expressions was taught, including the parts and its functions, four fundamental operations in radicals and rationalizing the denominator.

If you haven’t scanned the lesson, and you were given an equation involving radicals to be solved, at first you might forgot to rationalize the denominator. This is what happened to me before. This lesson was tackled when I was in second year high school, no wonder I forgot.

Like terms in radicals are the same with like terms in variables. They can be added and subtracted. But the unlike terms, cannot. This one also makes students confused, especially the beginners.”

- Author

Getting Any Better?

Prerequisite 9: Complex In the set of real numbers, negative numbers do not have square roots. A new kind of

number, called imaginary was invented so that negative numbers would have a square root. These numbers start with the number i, which equals the square root of -1, or i2 = -1.

i1=i i3=−i

i2=−1 i4=1

A complete number system, one that includes both real and imaginary numbers, was devised. Numbers in this set are called complex numbers. Complex numbers consist of all sums a + bi where a and b are real numbers and i is imaginary.

Addition and Subtraction(a + bi) + (c + di) = (a + c) + (b + d)i(a + bi) – (c + di) = (a – c) + (b – d)i

Examples:1. (6 + 8i) + (12 + 2i) = 18 + 10i2. (12 + i) + (6 + i) = 18 + 2i3. (4 + 2i) – (1 + 6i) = 3 – 4i4. (7 – 5i) + (5 + i) = 12 – 4i

Multiplication and Division

Multiplying complex numbers is the same as multiplying polynomials

When dividing complex numbers,

a+bic+di

=(a+bi ) (c−di )(c+di ) (c−di )

multiply thenumerator∧denominator

by the conjugateof the denominator

Examples:1. 5i(4 + 2i) = 20i + 10i2 = 20i + 10(-1) = -10 + 20i2. i(6 + 4i) = 6i + 4i2 = 6i + 4(-1) = -4 + 6i

3.6+2 ii

=(6+2i )(−i)

(i)(−i)=6 i−2 i

2

1=6 i+2

“Complex numbers weren’t taught to us before, even in our Geometry (2nd year Math). Now, I learned that only nonnegative integers can be radicands. Negative integers don’t have their roots, except to be extracted as an imaginary number (i). Complex numbers are numbers with real and imaginary numbers.

The four fundamental operations are also applied in dealing with radicals, which is an easy one. It’s easy for me because it doesn’t need long computations. It’s cool to solve such numbers.

There are only few answers for the in. It’s either +1 or -1 or; +I or –i. Cool, right?”

- Author

Getting Any Better?

Prerequisite 10: Ordered Pairs

Prerequisite 11:

Summation Notation

(a, b) x- coordinate (abscissa) y – coordinate (ordinate)

In the Cartesian Coordinate plane, there are two intersecting lines: the x-axis and the y-axis, which divide the plane into four quadrants. (a, b) is a set of ordered pairs with a as the x-coordinate and b as the y-coordinate. The x – coordinate indicates the distance to the right or left of the origin (0,0) while the y – coordinate indicates the distance above or below the origin(0,0).

“Ordered Pairs are coordinates. This is actually part of the lessons in Geometry. It’s easy for, you just have to know the coordinates, quadrants, and the Cartesian Coordinate Plane. The coordinate plane was invented by Mathematician Rene Descartes.

In this lesson, I learned how to graph equations, either linear or quadratic. But this lesson is only the basics. The x-coordinate is the input and the y-coordinate is the output.”

- Author

Getting Any Better?

∑i=1

n

x i=x1+x2+ x3+x4+…….+xn

The symbol ∑i=1

n

x i is called a sigma notation or summation notation which

abbreviates the sum of a certain numbers. The Greek sigma tells us to carry out the summation; i=1 at the bottom of the sigma indicates that the subscript of the first term in the summation is 1, and the n at the top indicates that the subscript of the last term in the summation is n.

Examples:

1. ∑i=1

6

i2=12+22+32+42+52+62=1+4+9+16+25+36=91

2. ∑i=1

3

(i+1)2= (1+1 )2+(2+1 )2+(3+1)2=4+9+16=29

3. X1 = 4, x2 = 8, x3 = 5, x4 = -2, x5 = 1

∑i=1

5

x i (2 y+i )=x1 (2 y+1 )+x2 (2 y+2 )+x3 (2 y+3 )+x4 (2 y+4 )+x5 (2 y+5 )=4 (2 y+1 )+8 (2 y+2 )+5 (2 y+3 )+[−2 (2 y+4 ) ]+1 (2 y+5 )=(8 y+4 )+(16 y+16 )+ (10 y+3 )+ (−2 y−4 )+(2 y+5 )

¿34 y+24

Getting Any Better?

“The scope of my knowledge is now getting broader!

This lesson taught me about summation notation. At first, I thought it’s hard because I saw the Sigma Notation, but I had a wrong perception on that. It only deals with addition and you just have to substitute for values to obtain the sum. If I’m not mistaken, it is also a part of the lessons in Geometry. Unfortunately, it wasn’t discussed to us before.

Sigma Notation is also fun to solve because it’s not too long and complicated. In math, we’re taught to make haste slowly and to be careful. We must think twice or more than before finalizing your answer or your decision. We have learned so many skills in Math; we just haven’t noticed it yet. It doesn’t only enhance our thinking ability. So we must thank thee for developing us.”

- Author

Lesson 1.1

Relations and FunctionsLetter-number correspondence

K 5E 3L 5L 5Y 9

Number-letter correspondence

5 L3 E4 I

A relation is a set of ordered pairs. The domain of a relation is the set of first coordinates. The range is the set of second coordinates.

(K,5), (E,3), (L,5), (L,5), (Y,9) Domain (x-coordinate) Range (y-coordinate) (ordered pairs)

(input) (output)

A function is a relation in which each element of the domain corresponds to exactly one element of the range.

Domain Range M 4 I 5 L 3 E 6 S 7

This is a function

Domain Range M 4 I 7 S 7 S 6

This is a relation

Characteristics of a function: Each element in domain X must be matched with exactly one element in range Y. Some elements in Y may not be matched with any element in X. Two or more elements in X may be matched with the same element in Y.

Examples: Determine if the following set of ordered pairs is a function or not.1. (1,2), (2,5), (3,10), (4,17) = A function.2. (3,10), (6,37), (4, 17), (4, 8) = A Relation. (not a function)3. (-6,6), (4,-4), (0,1), (3,4) = A Function.

Note: The rule or correspondence can be described by the equation y = x2 + 1.

The vertical line testIt is the simplest way of determining whether or not a graph represents a fraction. A

graph represents a function if and only if no vertical line intersects the graph in more than one point.

This is a function. This is a function.

This is not a function. This is not a function.

“There we go now.

Relations and Functions is also easy to understand. This is just correspondence. Almost all of us know how to use cell phones, so it would be easy for us to understand this lesson.

I learned about correspondence, relations and functions. One might be confused in determining a function and a relation. For me, the only clue is: see if an element in the domain, which is your x-coordinate or the input, is being repeated. If so, then it is a relation and if not, then the otherwise.

In graphs, on the other hand, it can be determined whether a function or not through the vertical line test. This one is easy. Just draw a vertical line across the graph and see if it intersects only one point. This indicates that the graph is a function and if two or more points of intersection are made, then it’s not a function rather, a relation.”

- Author

Getting Any Better?

Lesson 1.2

Evaluating Functions

y = f(x) “y is a function of x”

f(x) = x + 2 The name of the function is f.Other letters may be used to name It represents the value of the function at x.functions, especially g and h.

The domain of a function f is a set of values of x for which f is defined.The range of function f is the set of all values of f(x), where x is an element of the domain of f.

Remember that:a. f(x) means “the value of f at x”. It does not mean “f times x.”b. letters other than f such as g and h or G and H can also be used.c. F is the name of the function and f(x) is the value of the function at x.

If x = 2, thenf(x) = x + 4f(2) = 2 + 4f(2) = 6

Examples:1. f(6) = 2x + 2y = 2(6) + 2y = 12+2 y

2. g(2) = 2x2+x y3

xy = 2¿¿ = 8+ y2

3. f(4) = (x + 1)(x + 4) = (4 + 1)(4 + 4) = (5)(8) = 13

Even and Odd Functions The function f is an even function if and only if

f(-x) = f(x), for all x in the domain of f.The right side of the equation of an even function does not change even if x is replaced

with –x.

The function f is an odd function if and only iff(-x) = -f(x), for all x in the domain of f.

Every term in the right side of the equation of an odd function changes sign if x is replaced by –x.

Examples:1. h(x) = 2x2 + 1

h(-x) = 2(-x)(-x) +1h(-x) = 2x2 + 1 -- This is an even function.

2. f(x) = -x8

f(-x) = (-x)(-x)(-x)(-x)(-x)(-x)(-x)(-x) = x8 -- This is an odd function.

“To find f(x) for a given value of x is to evaluate a function. At first sight, this lesson seems difficult. But it’s not. You just have to substitute the values, and that’s it! There’s no need to make it more complicated.

This lesson taught me about even and odd functions. Functions are fun to deal with. They’re not boring unlike other problems and equations.

I also became familiar with functions, its parts and its functions.”

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Lesson 1.3

Operations on Functions

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

Difference(f – g)(x) = f(x) – g(x)

Product(fg)(x) = f(x) • g(x)

Quotient

( fg ) ( x ) = f (x )g (x)

, where g(x) ≠ 0.

Examples:1. Combine f(x) = 2x + 4 and g(x) = x – 1

= f(x) + g(x) = (2x + 4) + (x – 1) = 3 x+3

2. Multiply h(x) = x – 3 and f(x) = x + 4 = h(x) • f(x) = (x – 3)(x + 4) = x2+x –12

3. Multiply f(x) = 2x2 + x + 3 and g(x) = x + 12 = f(x) • g(x) = (2x2 + x + 3)(x + 12) = 2 x3+25 x 2+15x+36

4. Subtract f(x) = 5x + 4 to g(x) = (x2 – 5) = g(x) – f(x) = (x2 – 5) – (5x + 4) = x2 – 5x –9

5. Find the quotient of f(x) = 3x – 2 and g(x) = x2 + 2x – 3

= f (x )g (x)

= 3 x−2

x2+2 x – 3 = 3 x−2

( x+3 )(x−1) =

{x∨x ≠−3∧x≠1 }

6. Find the sum of f(x) = 6x + 2 and g(x) = 8x + 1 = f(x) + g(x) = (6x + 2) + (8x + 1) = 14x + 3

“To add and subtract functions is the same with adding and subtracting polynomials. You can add and subtract the like terms ONLY, not the unlike terms.

Division and multiplication is also the same with that of polynomials. In this lesson, I learned to deal with functions.

Functions are described by three letters, f, g, and, h. So, any of the following may be used.”

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Lesson 1.4

Composite and Inverse Functions

The Composition of Functions

The composition of the function f with g is denoted by f ο g and is defined by the equation

(f ο g)(x) = f(g(x)), where x is in the domain of g and g(x) is in the domain of f.

Given: f(x) = 2x – 1 g(x) = x2 + 2Find: (f ο g)(x)Solution:

(f ο g)(x) = f(g(x))f(g(x)) = 2(g(x)) – 1 Substitute the value of f(x) with g(x) as the value of x in the

f(x).f(g(x)) = 2(x2 + 2) – 1 Then, substitute the value of g(x). Proceed to the operation.f(g(x)) = 2x2 + 3 Therefore, (f ο g)(x) = 2x2 + 3.

Examples:1. Given: f(x) = x – 7 h(x) = x2

Find: a. (f ο h)(x) b. (h ο f)(x) Solution:

a. f(h(x)) = (h(x)) – 7= x2 – 7

(f ο h)(x) = x2 – 7

b. h(f(x)) = (f(x))2

= (x – 7)2

= (x – 7)(x – 7) (h ο f)(x) = x2 – 14x + 49

2. Given: g(x) = 2x2 + 5 f(x) = x2 + 5x + 2Find: a. (f ο g)(x)) b. (g ο f)(x) Solution:

a. (f ο g)(x)) = f(g(x)) f(g(x)) = x2 + 5x + 2 = (g(x))2 + 5(g(x)) + 2

= (2x2 + 5)2 + 5(2x2 + 5) + 2= 4x4 + 25 + 10x2 + 25 + 2

(f ο g)(x)) = 4x4 + 10x2 + 52

b. (g ο f)(x) = g(f(x))= 2x2 + 5= 2(f(x)) + 5= 2(x2 + 5x + 2) + 5= 2x2 + 10x + 4 + 5

(g ο f)(x) = 2x2 + 10x + 9

Inverse of a function

The function g is the inverse of the function f, and is denoted by f-1 (read as “f – inverse”.) Thus, f(f-1(x)) = x and f-1(f(x)) = x. The domain of f is equal to the range of f-1, and vice versa.

In finding the inverse of a function,

a. Replace f(x) with y in the equation for f(x).f(x) = 9x + 3 y = 9x + 3

b. Interchange x and y. y = 9x + 3

x = 9y + 3x = 9y + 3

c. Solve for y.x – 3 = 9yx−39

= 9 y9

x−39

= y

d. Replace y by f-1(x)

y = x−39

f-1(x)= x−39

Therefore, f(x) = 9x + 3 and f-1(x) = x−39

Examples:1. Find the inverse of f(x) = 3x – 2

f(x) = 3x – 2y = 3x – 2x = 3y – 2x – 2 = 3yx−23

= 3 y3

x−23

= y

y = x−23

f-1(x)= x−23

Thus, f(x) = 3x – 2 and f-1(x) = x−23

2. Show if f(x) = 12x + 2 and g(x) = x−212

is an inverse of each other.

f(g(x)) = 12( x−212 ) + 2

= 12x−2412

+ 2

= x – 2 + 2= x

g(f(x)) = (12 x+2)−2

12

= 12x+012

= x

Therefore, f(x) = 12x + 2 and f-1(x) = x−212

.

The Horizontal Line Test for Inverse Functions

A function f has an inverse that is a function, f-1, if there is no horizontal line that intersects the graph of the function f at more than one point.

It has an inverse function. It has no inverse function.

It has an inverse function. It has no inverse function.

“This lesson is hard, but just little. In the composition of functions, two functions were combined, or should I say, there are two functions describing the variable x. This is the reason why at first, it’s quite confusing.

Composition of functions, however, is also fun to solve. It’s not that boring.

On the other hand, the inverse of a function makes me confuse at first. I learned that it’s the inverse of the function being described and is in the form of f-1(x), read as f-inverse of x. This means that f(x) and f-1(x) will be both equal to 0 if they are inverse functions. Anyway, we can prove it by showing your solution. But if they’re not both equal to 0, then they’re not an inverse of each other.

Using the horizontal line test, you can also determine whether or not the graph has an inverse function. This is actually the same as in the vertical line test. If only one point became the point of intersection, then it has an inverse function otherwise, not. “

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Lesson 1.5

Linear

Linear equations in two variables are written in two forms: the standard form is written as Ax + By = C, where A, B, and C are real numbers; and the slope-intercept form is written as y = mx + b, where m is the slope and b is the y-intercept.

A Linear Function is a function that can be written in the formf(x) = ax + b, where a and b are real numbers with a and f(x) not both equal to

zero.

Constant functionf(x) = b, where a = 0.

Identity functionf(x) = x, where a = 1 and b = 0.

Graphing Linear Equations a. Point-plotting Methodb. Using the x- and y-intercepts Methodc. Using the Slope-intercept Method

Point-Plotting MethodIn this method, you have to find at least three or four solutions of the equation. Then

plot the points accordingly. Connect the points with a line.

Using the x- and y-intercept methodThe x-intercept is obtained by letting y be equal to 0 and solve for x. Thus, the

coordinate (x,0). It is the x-coordinate of the point where the graph intersects the x-axis.The y-intercept is obtained by letting x be equal to 0 and solve for y. It is the y-

coordinate of the point where the graph intersects the y-axis.To graph a linear equation using this method, determine first the x-intercept and the y-

intercept. Then plot the two points and connect with a straight line.

Using the Slope-intercept Method

Slope = m = riserun

The slope-intercept form of a linear equation in two variables is given byy = mx + b, where m is the slope of the line and b is

the y-intercept of the graph.

To graph a linear equation using the slope-intercept method, solve first for the equation y. Determine the slope and the y-intercept (y-intercept is denoted by b and the slope is denoted by m). Plot the y-intercept (0,y) on the Cartesian coordinate plane and label it. Then plot the next point by using the slope. The rise indicates the movement on the y-axis, down if it is negative and up if it is positive. The run indicated the movement on the x-axis, to the right if the variable is positive otherwise, to the left.

“This is just a recall of our past lesson in Geometry. I learned here how to graph a linear equation. The graph of a linear equation is always a straight line. You can use the Point-plotting Method, Intercepts Method and Slope-intercepts method in graphic a linear equation. This is just the same in Geomtrey.

There are also two kinds of function: the constant and the identity function.”

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Lesson 1.6

Quadratic

Quadratic Equation is an equation in the form of ax2 + bx + c = 0, where a, b and c are real numbers and a ≠ 0. It is a second degree equation since the highest exponent of the variable is 2.

Quadratic Function is a function in the form of f(x) = ax2 + bx + c, where a, b and c are real numbers and a ≠ 0.

ParabolaThe graph of any quadratic function is called a parabola.

Vertex

•

Parabola

Line of Symmetry

The vertex of the parabola is in its maximum point if the parabola opens downward. If it opens upward, then it is called the minimum point.

Graphing a Parabola

The most basic quadratic is y = x2. When you graphed straight lines, you only needed two points to graph your line, though you generally plotted three or more points just to be on the safe side.

a > 0, parabola opens upward.a < 0, parabola opens downward.

Here are the points:

Plot the points

Then connect the points with a curved line, forming a parabola.

Vertex Form of a Quadratic Functionf(x) = a(x – h)2 + k, a ≠ 0

To find for the vertex:(h, k) points

OR

h = −b2a

where h = vertex

Find the vertex of y = 3x2 + x – 2 and graph the parabola.

h = −b2a

= −1

(2 )(3) = −16

Find the value of h.

k = 3(−16 )2

+(−16 ) – 2 Find the value of k by evaluating y

at h.

= 336

– 16

– 2

= 112

– 212

– 2412

= −2512

Thus, (−16

,−2512

) is the vertex.

You can add more points for your graph. And this is your graph.

“Quadratic Function is the hardest lesson for me in the first quarter period. Quadratic Equation is an equation whose highest degree is 2, meaning its highest exponent would be 2. And the graph of a quadratic Equation is called a parabola.

The parabola has a vertex (minimum/maximum point). In order to graph a parabola, we must first find the x- and y-intercepts because they are needed in the plotting of points and formation of the parabola.

The method here seems too long, but once you became familiar with it, I’m sure solving it would be easy for you.

In this lesson, I learned about parabola, quadratic equations and most of all, I learned that even functions can be graphed through the linear and quadratic equations. Math is really CHALLENGING!

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