# Quadratics Relationships

## Introduction To Quadratics

The graph of a quadratic relationship will always be a parabola. A parabola has three main parts.

1) Vertex - Vertex determines where the graph changes direction. Its written in form of (x,y)

2) Optimal Value- The optimal Value is the y value of vertex. (x,y). This determines the maximum or minimum value of the graph.

3) Axis of symmetry- AOS is he x value of the vertex (x,y) and is an imaginary line that passes through center of graph splitting it into two equal parts. ## How to determine a quadratic relationship?

(A)Firstly the most easiest to determine is by looking at the graph (if provided) and see a curve of bit or a parabola.

(B)If not given a graph and given the table of values we could determine a quadratic by:

1) Calculating the first differences. This is done by subtracting the second y value with first, so it goes y2-y1. Doing this consecutively will give the first differences.

2) Then if the first differences aren't the same then subtract the first differences tog et 2nd differences. If second differences turn out to be the same then its quadratic relationship.

As you can see in the example the second differences are the same meaning graphing it will create a quadratic relation.

## Vertex Form: y=a(x-h)2+k

This equation tells us the form/shape of the parabola, where the parabola will change direction and where will it be located.

1) The (-h) value of the vertex form tells us where the parabola will move horizontally. If h value if negative then the parabola will move right ( meaning if h is negative then vertex will be h,k) whereas if value of h is positive then parabola will move left ( meaning if h is positive then vertex will be -h,k)

2) The value of k represents the movement of parabola vertically, meaning when k is negative the parabola will start at value of below 0 in y axis whereas if its positive then graph will start at value of above 0 in y axis.

3) The a value in vertex form represents stretch or compression of a parabola, basically the form/shape. It also determines whether parabola will open down or up. ( when a is negative parabola open down when a is positive parabola open up) ## How to graph vertex form?

There are two ways to graph vertex form, mapping notation and step pattern.

These videos show both ways how to graph vertex form.

Step pattern

Mapping notation

## Solving for x and y intercepts in vertex form The image gives a example of solved y and x intercepts.

These were steps followed:

(A)Y intercept sub x=0 to solve for y

1) Expand the integers inside the bracket to get 4

2) Multiply the 4 with negative to get -4 then

3) Add 25 to -4 to get 21 as y intercept.

(B) for x intercepts sub in y=0

1) Take k value to other then switch signs (25 to -25)

2) Divide the a value with both sides of equations (-25/-1=25)

2) Square root the both sides then separate to two parts because any positive number can be square rooted by negative numbers and positive numbers.

3) then solve the two separate sides to get x intercepts.

## Factored Form: a(x-r)(x-2)

A value of factored form is the form/shape of the parabola. It also tells direction of opening of the parabola.

R and S are the x intercepts/zeros/roots of the parabola.

## Multiplying binomials ( expanding)

We could multiply binomials by using FOIL.

(a+b)(c+d)= ac+ad+bc+bd Example1-

There are also special products :

(a+b)(a-b)=a2-b2 example 2-

There are also perfect square trinomials example 3-

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

(a-b)2=a2-2ab+b2   ## Factoring

First main step of factoring is to identify whether you can GCF first.

For example in equation: 2x+2b you can GCF the 2

Second step is to identify how many terms are left in bracket after GCF is complete.

## Two terms

If there are two terms in a equation use GCF or common factoring method which is:

Example1-

14m+21n------- GCF of 7

7(2m+3n)

Example2-

13ab-7ac------ Since coefficients cannot be factored try the variables.

a(13b-7c)

## Three Terms

If there are 3 terms identify what type of trinomial it is

1) Complex trinomial

2) Simple trinomial

Complex trinomial example

This equation is in standard form so we need to factor it to get factored form.

6x2-19x+10 Product (ac) Sum -19(b)

6x2-4x-15x+10 -4(-15) - 4-15

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

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

Simple Trinomial Example

Rules: If c is negative then one of r and s is going to negative

If both c and b are positive then both r and s going to positive

If b is negative then r and s going to both negative.

Example x2+4x-5

Product -1+5=4

Sum -1X5

Therefore (x-1)(x+5)

## Four terms

If you have four terms then you could factor by grouping

This means factoring first two variables and second two variables separately

This involves first using grouping to get factored form then using binomial common factoring to get final answer: ## Solving for X intercepts and Graphing the Relation. After we find the x intercepts and vertex we could then sub those into a graph to form a parabola.

## Standard Form: ax2+bx+c

1) Completing the square

2) Quadratic Formula

3) Discriminat

4) Graphing the relationship

These are four main parts of this unit.

In standard form the c value of the equation tells y intercept.

the a value of the form tells us the form/shape of the parabola.

## Completing the Square

Completing the square is a way other than factoring to convert Standard form to vertex form to find out the vertex and max/min value. This is an example of completing the square.

To complete the square

1) First check out for co efficient on a value if there is one divide the coefficient from both a and b values.

2) then the remaining the b vale do (b/2) squared.

3) Put the number in negative and positive value then take the negative number outside and you will get your k value in vertex form.

## Quadratic Formula

Quadratic is used to determine the x intercepts of the standard form equation. It is used to find roots/zeros/x intercepts. This is an example of quadratic formula. Here in the example the formula Is being used to determine the x intercepts of the equation.

## Discriminat b2-4ac

Rules:

If discriminat is equal to zero then there will be one x intercept to the equation.

Example

D=b2-4ac

D=(4)2-4(1)(4)

D=16-16

D=0

If discriminat is greater than zero then there will be two x intercepts to the equation

Example

D=(4)2-4(1)(1)

D=16-4

D=11

If discriminat is lesser than zero then no solutions.

D=(4)2-4(3)(3)

D=16-36

D=-20

## Graphing the relationship Here in the picture is an example of my unit test question where it asked to find x intercepts and vertex. To do this I used quadratic formula to determine the x intercepts then completed the square to find the vertex. Final part asked to graph the relationship I subbed in intercepts and vertex to graph parabola.

## Here are some examples of word problems in the unit. ## Examples

First question is a profit question where it asked to find the maximum profit. To complete this I used completing the square method to find the vertex to give the maximum profit.

Second one is a triangle question where it asked to solve for two sides when hypotenuse was given.

## Connections between different forms

First main connection between standard form and vertex form was that bx term in standard form was the axis of symmetry in vertex form meaning to check if I did vertex form correct I would check my answer from original equation. Now there were multiple ways to solve for x intercepts, because vertex form to standard then quadratic formula would give x intercepts but then vertex to standard then standard to factoring would give solutions as well. There were also now multiple ways to solve for vertex. In total all of the form are much needed for each other because one form alone wouldn't give all appropriate information required, so all of them are connected.

## Reflection- My personal thoughts

In our first day of quadratics I was very excited to learn about what we were teased in grade 9 (curve of best fit). Our first unit was one of easier ones where we learned how to graph vertex form of quadratic and use it to find max/min value and problem solving problems. Although it was very hard on the first few days and I wasn't getting anything as we progressed to learn how vertex connects to values etc. it started to make sense. Onto to the most lets not say the fun units factoring. This was a new learning curve as we now we had to convert these fully expanded equations into back to their original. It was very confusing and it was my worst scoring part in quadratics. Last unit was easiest of them all since we learned about standard form in factoring unit (standard to factored) it wasn't a surprise. One huge curve was to get used to learn quadratic formula to solve for mere two intercepts. It was a big pain since sometimes we didn't even an intercept. We learned about discriminat and using completing the square formula to convert standard into vertex form. Quadratics was a very fun unit since we got to learn how all these equations gave us one parabola.