# Math 101

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Made by Harjaap Chahal

## contents

chapter 4.1 Investigating Non-Linear Relations

chapter 4.2 quadratic functions

Chapter 4.3 Transformations of Quadratics

4.4 graphing transformation of quadratic relation

4.5 finding roots/zeroes/x intercepts from vertex form

5.1 multiply polynomials

5.2 special products

chapter 5.3 common factoring

chapter 5.4 simple trinomial factoring

chapter 5.5 complex trinomial factoring

chapter 5.6 factoring special products

6.1 completing the squares

chapter 6.2 and 6.3 solving quadratic equations in factored form

6.4 quadratic formula

## chapter 4.1 Investigating Non-Linear Relations

Parabola

- a parabola can open down or up

- the vertex is the point where the axis of symmetry connect with the parabola

-its like the middle of the parabola and the highest point or if its a negative parabola which has a opening down, it would be the lowest point

- it also divides the parabola into two halves

- there is a term called zeroes where the parabola crosses the x axis

- other names for zeroes are roots or x intercepts ## chapter 4.2 quadratic functions

equation =y = ax2 + bx + c,

a,b,c are real numbers where a=0

e.g 4x2+10x+7

to make an parabola you need to make a chart it will be easier

y=x2

x y

-2 4

-1 1

0 0

1 1

2 4

if a is a negative number then the opening is down and if a is positive then the opening is up.

e.g y=-4x2+10x=7 - the opening will be down.

How do you know if its an quadratic relation or not?

if the second differences of the chart are constant then it is a quadratic relation and if the first differences are constant then it is an linear relation.

## Chapter 4.3 Transformations of Quadratics

Vertex Form of a parabola y=a(x-h)2+k h=x k=y

y and x do not change

terms:

a- determines how narrow or wide the parabola is

h- the x coordinate of the vertex

k- the y coordinate of the vertex

also:

a= the vertical stretch or compression by a factor of a/ reflected in the x axis if a is negative

h= the horizontal transition

k= the vertical transition

this is an basic equation:  ## 4.4 graphing transformation of quadratic relation

when a number is added to x2 the graph shifts upwards on the y axis.

when a number is subtracted from x2 the graph goes downwards on the y axis.

when x2 is multiplied with a number between 0 and 1 the width of the parabola becomes wider.

when the coefficient of x2 is negative the parabola opens downwards. making a frown.

the difference of y=a(x-3)2+4

1. opening upwards

2. translated 2 units to the right

3. translated 4 units up

y=(X+5)2=2 the vertex is (-5,2)

## 4.5 finding roots/zeroes/x intercepts from vertex form

to find the x intercepts you have to set the y to zero and solve and to find the y intercept you have to set the x to zero and solve.

vertex form y=a(x-h)2+k

## 5.1 multiply polynomials

1 term - monomial

2 terms- binomial

3 terms- trinomial

to simplify a polynomial you need use collecting like terms and distributive property.

(x+5) (x+6) 1. distributive property

x2+6x+5x+30 2. collect like terms

x2+11x+30

## 5.2 special products

squaring a binomial

1. square the first term

2. 2 times the product of the term

3. square the last term

example:

(x+6)2 x+6 x+6

x2+2(x)(6)+62

x2+12x+36

product of sum and difference

when the sum and difference of two terms are multiplied, the two middle terms cancel each other out so they are opposite

example:

(x+7) (x-7)

x2+-7x+7x+49

(x)2-(7)2

factoring

factoring is different than expanding, factoring looking for a expression to multiply while expanding has to do with just multiplying.

example ## chapter 5.3 common factoring

method #1: look for the greatest common factor

## gcf aka greatest common factor

-5y and -10y -4xy and -8y

gcf= -5y gcf= -4y

factoring a monomial

1. find gcf for both coefficient and variable

2. divide all the terms by gcf

example

2x+8y 5x+15x

2(x+4) 5x(x+3)

factoring a binomial

if there are 2 binomials that are the same then that is a binomial common factor.

2x(x+4)3(x+4) =(2x+3) (x+4)

4x(2x+3)2(2x+3) =(4x+2)

factoring by grouping

group the common factors into 2 terms to produce a binomial factor.

example

ax+ay+4x+8y

(ax+ay)+(4x+4y)

a(x+y)+4(x+y)

(a+4)+(x+y)

## chapter 5.4 simple trinomial factoring a simple quadratic is when a=1

a complex trinomial is when a b c are more than 1

if given a quadratic in standard form you factor it to make it factored form and the other way around

## standard form ## factored form ## chapter 5.5 complex trinomial factoring

not all quadratics can be factored.

how to factor a complex trinomial:

1. before anything look at the common factor first when factoring a trinomial.

2. find to numbers that multiplied together equal a and c but added together equal b

3. then check up the middle term and factor by grouping

example

6x2+5x+1

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

2x(3x+1)1(3x+1)

## chapter 5.6 factoring special products

formula:

(a+b)2 a2+2ab+b2

(a-b)2 a2-2ab+b2

example

(x-6)2

x2-2(x)(6)+62

x2-12x+36

r+s=b and (r)(s)=c

how to get a quadratic in standard form to factored form.

1. find 2 numbers that add up b and multiplied together equal.

2. if b and c are positive then r and s will be positive but if. if c is poistive and b is negative and one is positive then both r and s are negative but if c is negative then either r or s will be negative.

example

x2+7x+12

(x+4)(x+3)

x2+8x-12

(x+6)(x-2)

## 6.1 completing the squares

steps:

1.put equation into vertex form

2. group x terms

3. divide coefficient of middle term by 2 then square it then add and and subtract number inside bracket

4. remove subtracted term from brackets

5. factor brackets as perfect square trinomial

example

y=x2+8x+4

y=(x2+8x+16)-16+4

y=(x2+8x+16)-12

y=(x+4)2-12

completing the squares when the value of ''a'' is more than 1

step

1.group the x terms.

2. factor the value of ''a'' from the x intercepts.

3. divide coefficient by of middle term by 2 then square it and then add and subtract that number inside the bracket.

4. remove negative term from brackets and multiply it by the a value that you factored.

5. write the perfect square trinomial as a perfect square binomial.

example

y=2x2+8x+13

y=2(x2+8x+16)-16+13

Y=(x2+8x+16)-32+13

Y=(x2+8x+16)-19

Y=(x+4)2-19

you try

y=x2-6x+4 ANSWER-Y=(x-3)2-5

## chapter 6.2 and 6.3 solving quadratic equations in factored form

solving-finding the zeroes/x-intercepts/roots.

steps:

1. make one side zero and then make all the brackets to zero. = (x+r)=0 (x+s)=0

2. then solve for x

example

6x2-15=0

(6x2+9x)(-10x-15)

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

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

2x=-3 3x=5

x=-3/2 x=5/3

x=1.5 x=1.7

the two x intercepts are (1.5,0) (1.7,0).

## 6.4 quadratic formula the quadratic formula can be used all the time to find the x intercept/roots/zeros when the equation is in standard form. 