# Quadratic Relationship

### By: Samuel Roshan Thanabalasingam

## What is a parabola?

## Parabolas in real life

## Parts Of A Parabola

**Axis Of Symmetry :** The A.O.S. divides the parabola in 2 halves, it is a line of symmetry for a graph. It is when the two sides of a graph on either side of the axis of symmetry look like mirror images of each other. It can be found on the x-axis.

** Optimal Value:** The lowest or highest point the parabola can go. this is the y value in the vertex. The optimal value determines if the parabola goes upward or downward (max or min).

**Vertex:** The maximum or minimum point on the graph. The vertex consists of the axis of symmetry, which is the x coordinate and the optimal value, which represents the y value.

Y-**intercept****: **The y-intercept is point on the parabola that hits y-axis

**X-intercept: **The x-intercept is the point on the parabola that hits the x-axis

**Zeros and Roots:** these are the x values and is when you set the y=o

## Vertex Form: y=a(x-h)²+k

## Identifying Quadratic Relations in Vertex Form: y=a(x-h)2 +k

• *(h, k)* is the vertex of the parabola, and *x = h* is the axis of symmetry.

• the *h* represents a horizontal shift (how far left, or right, the graph has shifted from *x* = 0).

• the *k* represents a vertical shift (how far up, or down, the graph has shifted from *y* = 0).

• notice that the *h* value is subtracted in this form, and that the *k* value is added.

If the equation is *y* = 2(*x* - 1)2 + 5, the value of *h* is 1, and *k* is 5.

If the equation is *y* = 3(*x* + 4)2 - 6, the value of *h* is -4, and *k* is -6.

## Finding the Zeros in Vertex Form

**Step 1**:Find the vertex. Since the equation is in vertex form, the vertex will be at the point (h, k).

**Step 2**:Find the y-intercept. To find the y-intercept let x = 0 and solve for y.

**Step 3**:Find the x-intercept(s). To find the x-intercept let y = 0 and solve for x. You can solve for x by using the square root principle or the quadratic formula (if you simplify the problem into the correct form).

**Step 4**:Graph the parabola using the points found in steps 1 – 3.

## Step pattern

**The step pattern is a pattern that identifies your next points in each parabola.**

**The original step patter is 'Over 1 up 1 and Over 2 up 4'**

**But in quadratic relations if there is an 'a' value the step pattern changes.**

## Factored Form: Y=a(X-r)(X-s)

## What each variable represents

-The axis of symmetry: is the midpoint of the two zeroes

-The optimal value: is found by subbing the axis of symmetry value into the equation

## My videos

## standard form: y= ax²+bx+c

## Example of standard form and graphing

** ****Example 1:** standard form, *y*= *ax*2 + *bx* + *c*

Graph the equation y = x2 – 8x + 12 using factoring.

Factor the trinomial, x2 – 8x + 12. Identify 2 numbers whose sum is –8 and the product is 12. The numbers are –2 and –6. That is, y = (x-2) (x-6)=0 .

So, by the zero product property, either *x** *– 2 = 0 or *x** *– 6 = 0. Then the roots of the equation are 2 and 6.

Therefore, the *x*-intercepts of the function are 6 and 2.

The *x*-coordinate of the vertex is the midpoint of the x –intercepts (axis of symmetry). So, here the *x*-coordinate of the vertex will be 2+6/2=4. .

Substitute *x* = 4 in the equation y = x2 – 8x + 12 to find the *y*-coordinate of the vertex (optimal value).

y = x2 – 8x + 12

y = (4) 2 – 8(4) + 12

y = 16 – 32 + 12

y = -4

That is, the coordinates of the vertex are (4, –4).

Now we have 3 points (4, –4), (2,0) and (6,0) which are on the parabola. Plot the points. Join them by a smooth curve and extend the parabola.

**Example 2:** standard form, *y*= *ax*2 + *bx* + *c*

Graph the equation y = -x2 – 2x + 8 using factoring.

Factor the trinomial, -x2 – 2x + 8.

First, factor out –1.

= -1(-x2 – 2x + 8)

Factor the expression in the parenthesis. Identify 2 numbers whose sum is 2 and the product is –8. The numbers are 4 and -2. That is, y = (x+4) (x-2)

Then, the given function becomes y = - (x+4) (x-2)

So, *y** *= 0 implies, by the zero product property, *x*+ 4 = 0 or *x** *– 2 = 0.

Therefore, the *x*-intercepts of the graph are –4 and 2.

The *x*-coordinate of the vertex of a parabola is the midpoint of the *x*-intercepts. So, here the *x*-coordinate of the vertex will be -4+2/2= -1.

Substitute *x* = -1 in the equation y = -x2 – 2x + 8 to find the *y** *-coordinate of the vertex.

y = -x2 – 2x + 8

y = - (-1) 2 – 2(-1) +8

y = -1 + 2 + 8

y = 9

So, the coordinates of the vertex are (–1, 9).

Now we have 3 points (–1, 9), (–4, 0) and (2, 0) which are on the parabola. Plot the points. Join them by a smooth curve and extend the parabola.

## Examples of Quadratic formula

## Completing the square

**c**

**ompleting the Square**is a technique used to solve quadratic equations, graph quadratic functions, and evaluate integrals. This technique can be used when factoring a quadratic equation does not work or to find irrational and complex roots.

## common factoring

Common factoring should *always* be the first thing you look for when you have an equation in standard form. You should look to see if each of the terms in the equation has a similar component that each of the terms can easily be divided into. The common factor can either be a variable (x) or a coefficient (any number.)

Example 1: 8x + 6

Step #1: Find GCF ( what is the greatest # in common between 8 and 6)

GCF=2

Step #2: write solution with brackets

8x/2 + 6/2 = 4x + 3

= 2(4x + 3)

Example 2: 12x³ - 6x²

step #1: GCF= 6x²

Step #2: 6x²(2x-1)

check your answer!

6x²(2x-1)

=12x³ - 6x²

Example 3: 12x³y - 6x² + 18x² y

Step #1: GCF= 6x²y

Step #2: 6x²y(2x-1+3)

link to check out: http://www.mathsisfun.com/algebra/factoring-quadratics.html

## simple trinomials

## complex trinomials

*not*1. That makes it a little trickier, and more often the method

*trial and error*or

*guess and check*will have to be used.

## prefect squares

What is a perfect square?

Perfect Square Trinomial is the product of two binomials. But, both the binomials are same. When factoring some quadratics which gives identical factors, that quadratics are Perfect Square Trinomial.

Example of a perfect square:

x² + 10x + 25

= (x + 5)(x + 5)

= (x + 5)²

**Question 1: **Factor the trinomial x2 + 4x + 4**Solution:**

Given x2 + 4x + 4

The above trinomial can be written as x2 + 2x + 2x + 4

Take x as common from first two terms,

x(x + 2) + 2x + 4

Take 2 as common from last two terms.

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

(x + 2)(x + 2)

In this it is the product of two identical binomials

**(x + 2)2 ,** which is a perfect square.

**Question 2: **Factor the trinomial x2 + 8x + 16**Solution:**

Given x2 + 8x + 16

The above trinomial can be written as x2 + 4x + 4x + 16

Take x as common from first two terms,

x(x + 4) + 4x + 16

Take 4 as common from last two terms.

x(x + 4) + 4(x + 4)

(x + 4)(x + 4)

In this, it is the product of two identical binomials

**(x + 4)2**, which is a perfect square.

## difference of squares

**Difference of Squares** When working with a difference of squares, you must check to see that

1. Each term is a squared term and

2. There is a *minus* sign between the two terms.

a2 - b2

Useful because it can be factored into (a+b)(a−b)

## Examples of some word problems

## how to do Quadratic Word Problems

## For more word problem examples this website has a bunch!

## Reflection on the quadratic unit

Throughout the unit, I have learned a lot through and about quadratics. When I first got into the unit I seen what we need to now and my mind was going to explode. I immediately thought that quadratics was going to be hard and I was going to fail the whole unit. But once we started the unit bit by bit, (mini test by mini test), I started to understand it better/like it more and one of the reason I did is because I listen to my teacher with some distraction here and there and took notes and did my homework and reviewed. Thus my mini test reflected on my studying and it shows hard work pays of at the end of the day. I also learned how hard it is to fall behind and then catch back up, and from that I learned I need to work on asking people for help instead of getting frustrated about it and not getting anything done. This unit really showed me that it is not going to get easier, but harder and that if I what to do good I have to take more initiative in my work now and in grades to come and if I do the best of my ability in everything I do at the end of the day I can be proud of myself.

## connections

One cool thing about quadratics is that somehow all these forms and relations seem to relate to each other and in some ways can be made into one form to another. For an example: You have a standard form equation, which you can break down and simplify into a factored form equation and vice versa.