### By: Mohini Kalia

The name Quadratic comes from "quad" meaning square, because the variable gets squared. Quadratic equations make a smooth curve.

## Reflection:

*In this parabola unit I learned many ways how to solve a quadratic relation and many new mathematical terminology. Something that I understood very well was how to describe a parabola, using appropriate terminology. The first document was my very first mini unit test for the parabola unit, and this is the communication part of the test. I scored 100% on this category, I understand how many units the parabola translates up, down, left, and right, if the parabola opens up or down. Basically, I know how describe a parabola with terms and words. The second document below, is the knowledge category of the mini test 2 of the parabola unit. It was how to solve for x when, you have an x square term. I scored 59% on this category, I think this was the most complicated thing that I had faced in the whole quadratics unit. But, the main reason why I failed in this category was because I didn't study for the test and I was also absent when Mrs.Wain taught this lesson to the class. But, now I know how to solve for x when you have a x squared term, so its all good. Other than that, this quadratic units had its up and down but, now I understand mostly of the quadratic formulas.

1. Vertex form: y=a(x-h)2+k
2. Standard form: y=ax2+bx+c
3. Factored form: y=a(x-r)(x-s)
4. Quadratic form: x=-b+-square root of b2-4ac/2a

*2 stands for squared!

## Quardratic Terms:

1. First Differences
2. Optimal value
3. Axis of Symmetry (A.O.S)
4. x-intercepts/zeros
5. Exact roots
6. Approximate roots
7. Discriminants

## finding the vertex:

The vertex is the made up of the h and k value of the vertex equation.

Example: y=(x+6)2+12

Vertex: (-6,12)

*If the h value is positive in the bracket, when plotting your vertex it turns negative. And if in the bracket it is negative, when plotting the h value turns positive (i.e. y=2(x-8)2-9, vertex= (8,-9)). The k value stays the same when plotting.

## A.O.S and Optimal value:

• A.O.S=the h value of vertex equation, it is also the point the divides the parabola in half.
• Optimal value=the k value of vertex equation

Example: y=-(x+5)2+9

• A.O.S= 5
• Optimal Value= 9

## How to find first and second differences:

1st and 2nd Differences

## note:

• if first difference the same=linear relation
• if second difference same=quadratic relation (parabola)
• if no difference the same=neither

a= if the parabolas opens up or down (negative=opens down and positive=opens up), compressed or stretched (if a decimal or fraction=compressed, if a whole #=stretched).

h=how many units the parabola horizontally moves (-#=moves to the left, +#=moves to the right).

k=how many units the parabola vertically moves (-#=moves down, +#=moves up).

Vertex= (h,k) the highest point of the curve(parabola)

## This paraobla:

• opens up
• is vertically stretched
• vertex (1,2)
• horizontally translates 1 unit to the right
• vertically translated 2 units up

## word problem:

A parabola has the equation of y=3(x+2)2+4

1. What would be the equation, if the parabola reflected over the x-axis?

• When looking for the reflection in the x-axis, you are looking for the reflection of the a and k value .
• So, it is the opposite integer of the a and k value (the a value of the original equation is 3 and the k value is 4, switch the operation signs)
• Therefore, the new equation is y=-3(x+2)2-4

2. What would be the equation, if the parabola travels 6 units to the right?

• Remember when translating left or right (horizontal position), it is controlled by the h value.
• Therefore, in the original equation the h value will change.
• Since moving to the right we add 6, in the equation it say positive 2 but remember when plotting we plot -2.
• So -2+6=4 (write -4 in new equation), new equation is y=3(x-4)2+4

## Finding equations in vertex fom

When working with vertex form, we are usually looking to solve for a. With many problems regarding this form the h,k,x,and y values will be given, as you have to do is sub in all the values bring the numbers to one side solve for a. Once get your values write your equation with the h,k,and a values.

## Example:

• vertex (3,-1) that passes through the point (1,7)

Label all the numbers:

• 3=h
• -1=k
• 1=x
• 7=y

Sub in the vertex form equation y=a(x-h)2+k:

• y=a(x-h)2+k
• 7=a(1-3)2-1
• 7=a(-2)2-1 (bring 1 to other side)
• 7+1=a(4)
• 8=4a
• 8/4=4a/4
• 2=a

Now complete the equation, by subbing 2 as a. Only sub your a,h,and k values

• y=2(x-3)2-1

## Example:

Find an equation for the parabola with the vertex (-3,-4) and passes through the point (2,6)

1. label all the numbers
• -3=h
• -4=k
• 2=x
• 6=y

2.sub in equation

• y=a(x-h)2+k
• 6=a(2+3)2-4
• 6+4=a(5)2-4+4
• 10=25a
• 10/25=25a/25
• 2/5=a

3. sub in a,h, and k values

• y=2/5(x+3)2-4

## Word problem:

The path of the soccer ball is represented by the following equation: h=-4(d-12)2+36. h represents the height in meters and d represents the horizontal distance.

1. What is the maximum height of the soccer ball?

• When asking for the max height (vertical position), it is basically asking for the k value.
• In this case, the k value is 36.
• So, the max height of the soccer ball is 36 meters.

2. What is the horizontal distance when this height occurs?

• Horizontal distance basically means the horizontal position, which means the h value.
• In the equation the h value is -12, but when plotting we plot 12.
• Therefore, the horizontal distance when the max height is 36 is 12 meters.

3. What is the height when the horizontal distance is 10 meters?

h=?

d=10

Sub whatever information you have in the equation:

• h=-4(d-12)2+36
• h=-4(10-12)2+36
• h=-4(-2)2+36
• -4(4)+36
• -16+36
• Therefore, the height of the soccer ball is 20 meters when the horizontal distance is 10 meters.

## Finding zeros/x-intercepts and y-intercepts in vertex form:

*when finding the y-intercept, always sub x=0

Example: y=-4(x+2)2+16

• y=-4(x+2)2+16
• y=-4(0+2)2+16
• y=-4(2)2+16
• y=-4(4)+16
• y=-16+16
• y=0

*when solving to find the x-intercepts, remember to always sub y=0

• y=-4(x+2)2+16
• 0-16=-4(x+2)2+16-16
• -16/-4=-4(x+2)2/-4
• 4=(x+2)2
• positive and negative square root of 4=x+2

Therefore, x=positive square root of 4-2 and negative square root of 4-2.

x=0, x=-4

## Step pattern:

The step pattern determines where to plot your points in the parabola. To determine you step pattern, you must know what a equals. Once you know your a value then, multiply your a value with 1,3, and 5 to get your step pattern.

Example:

a=2

• 1x2=2
• 3x2=6
• 5x2=10

From your vertex, move a unit to the left and 2 units up to plot you first point. From that point, move a unit to the left and 6 units up, and so on. Also, remember what ever you do on the left side you must do to the right side.

*If your a value is negative, then instead of moving upwards you have to move downwards to plot your points in the parabola.

## Linking all of vertex form:

*Once you have found the vertex, x-intercepts/zeros, y-intercept, and the step pattern (a value) now you can gather all your information and graph the parabola on a graph. The vertex is the first point you will plot on your graph, the next points to plot will be your x-intercepts (which you found by subbing y=0). Once you have plotted 3 point, now you can plot the rest by look at you step pattern (a value, a multiply by 1,3, and 5). You can also determine if your parabola opens up or down, by looking at the a value. Lastly, you can also plot your y-intercept (when you subbed x=0).

Factored form: y=a(x-r)(x-s)

*r and s represent zeros (x-intercepts)

1. find zeros
2. find A.O.S
3. find optimal value

Example:

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

1. Find zeros:
• x+4=0 x+6=0
• x+4-4=0-4 x+6-6=0-6
• x=-4 x=-6

2. A.O.S= -4+(-6)/2

• = -10/2
• = -5 (x value of vertex)
• 3.Optimal Value (sub -5 as x in original equation)
• y=-2(-5+4)(-5+6)
• y=-2(-1)+(1)
• y=2 (y/k value of vertex)

Vertex: (-5,2)

Vertex form: -2(x+5)2+2

## NOTE:

*Whenever working with factored form, you will always get the same number in the two brackets but one will be positive and the other will be negative that's how you know you are doing it right!

## example:

A parabola has the equation of y=(x+6)(x+5)

1. What are the coordinates of the vertex?

• Find the x-intercepts/zeros of this equation.
• x+6=0 and x+5=0, bring the 6 and 5 to the other side.
• x=-6 and x=-5 <----these are your x-intercepts.
• Find the A.O.S (h value of vertex).
• Add both zeros and divide by 2.
• (-6)+(-5)/2
• -11/2
• -5.5<----h value
• Find optimal value (k value).
• Sub x=-5.5, in original equation.
• y=(-5.5+6)(-5.5+5)
• y=(0.5)(-0.5)
• y=-0.25<----k value
• Vertex:(-5.5,-0.25)

2. Can write the equation in vertex form?

• When writing in vertex form, always switch the operation of the h value.
• y=(x+5.5)-0.25

## Word problem:

The predicted fight path of a plane is modelled by he following equation: h=-3(d-2)(d-12).

h represents the height in meters and d represents the horizontal distance.

1. When will the plane start its flight?

• Looking for the x-intercepts...
• d-2=0 and d-12=0, move the numbers to the other side.
• d=2 and d=12, the plane cannot start flying at 12 meters, therefore the plane has to start its flight at 2 meters.

2. When will the plane land on the ground?

• Still looking for the x-intercepts.
• It cannot be 2 meters, because that is when it start its flight, so it has to be the other x-intercept which is 12.
• Therefore, the plane lands on the ground at 12 meters horizontal distance.

3. What is the maximum height the plane can reach, and at what horizontal distance?

• Find A.O.S and Optimal Value.
• A.O.S, add both zeros/x-intercepts together and divide by 2.
• =2+12/2
• =14/2
• =7<---The h value
• Optimal Value, sub d=7
• y=-3(d-2)(d-12)
• y=-3(7-2)(7-12)
• y=-3(5)(-5)
• y=-3(-25)
• y=75<--- The k value
• Therefore, the maximum height of the plane is 75 meters, and it will occur when the horizontal distance is 7 meters.
3.13 Finding Vertex from Factored Form

Always remember:

First

Outside

Inside

Last

1. x*x=2x
2. x*5=5x
3. -5*x=-5x
4. -5*5=-25

• x2+5x-5x-25
• =x2-25

## Binomials to trinomials

*most of the time when multiplying two binomials, you will get a trinomial.

*trinomial will be in standard form (ax2+bx+c)

7x*2x=14x2

7x*4=28x

-3*2x=-6x

-3*4=-12

• 14x2+22x-12

## Word problem:

A square patio has a side length of x. One dimension is increased by 2 meters and the other is increased by 3 meters.

1. What would the area be of the square if all the sides where x?

• a=(l)(w)
• a=(x)(x)
• a=x2

2. What would the area be if the one dimension was increase by 2 and the other by 3.

• So, once side is (x+2) and the other side is (x+3).
• a=(l)(w)
• a=(x+2)(x+3)
• Use foil to expand.
• a=x2+3x+2x+6
• a=x2+5x+6<----new area

3. If x=4, how much will the area increase by.

• Sub x=4 in new area equation.
• a=x2+5x+6
• a=(4)2+5(4)+6
• a=8+20+6
• a=34
• Therefore, the area increased by 34 meters.

## common factoring: Simplifying

1. Find GCF
2. Divide the equation by the GCF
3. Write solution with brackets

## example: 8x+6

• GCF=2
• 8x/2+6/2
• 2(4x+3)

*To check if answer is right, expand!!

2(4x+3)

*multiply everything in brackets with 2

2*4x+2*3

=8x+6

## example: 2x3+4x+x *3 stands for cubed

*in this situation you can only factor a (X) because 2 and 4 can be divided be 2 but not x.

• GCF=x
• x(2x2+4+1)
Factor polynomials using the GCF

## Standard Form (ax2+bx+c):

The standard form is the most simplest of all parabola formulas. But yet it gives very little information about the parabola, that's why there are other formulas that's convert standard form into factored and vertex form. When you factor standard form you get information about the x-intercepts/zeros, which helps us in graphing the parabola. The formula is called factored form. Sometime factoring can be a pain so we use the quadratic formula, its useful when factoring gets complicated. And sometimes when we need to solve for vertex, we use complete the square which converts standard form into vertex form. Once we find our vertex form we also find the vertex, which helps when plotting points for the parabola.

## Factoring Trinomials: Standard form to factored form

When factoring a polynomial of the form ax2+bx+c (when a=1), we want to find:

1. 2 numbers that add up to get b
2. 2 numbers when multiplied get c

## Simple trinomials

x2-8x+12

__x__=12

__+__=-8

*list all the factors of 12

1,12

2,6

3,4

4,3

6,2

12,1

*in this case 6 and 2 work, but when you add 6+2 you get 8, our equation says -8 therefore both 6 and 2 have to be negative.

(-6)x(-2)=12

(-6)+(-2)=-8

(x-6)(x-2)

Factoring trinomials with a leading 1 coefficient

## word problem:

A rectangle has an area of 240cm2. The length is 5 cm more than the width. Find the dimensions of the rectangle?

1. Always use "let" terms, because when solving x there will be two term that include x.

• Let "x" rep width of rectangle.
• Let "x+5" rep length of rectangle, because length is 5 cm more than width.

2. Sub in all the values, in area formula.

• A=(l)(x)
• 240=(x+5)(x), doesn't matter which way you multiply.
• 240=x2+5x
• 240-240=x2+5x-240 *bring the 240 to the other side, now you have a simple trinomial. Factor out the trinomial.
• 0=x2+5x-240

__x__=-240

__x__=5

*In this case the values are 17 and -12.

• 0=(x+17)(x-12), find the x-intercepts.
• x+17=0 and w-12=0
• x+17-17=0-17 and x-12+12=0+12
• x=-17 and x=12, look at which x value fits our equation? -17 cannot work, because we cant have negative measurements for the dimensions of the rectangle. So x has to equal 12.

**Therefore the width is 12cm.**

3. Solve for the length, sub x=12.

• l=x+5
• l=(12)+5
• l=17

**Therefore, the length of the rectangle is 17 cm.

## Complex trinomials:

What make a trinomial complex is when the leading coefficient is not 1.

Example: 12r2-17r+6

1. Take the first and last terms and multiply them. (12x6=72)

2.Find two numbers that add up to -17 and, when multiplied those numbers you get 72.

• __x__=72
• __+__=-17

3. List all the factors of 72

• 1,72
• 2,36
• 3,24
• 4,18,
• 6,12
• 8,9

4. In this case the two numbers are 9 and 8, but when added they have to equal to -17, so then we have to turn both numbers negative.

(-8)+(-9)=-17

(-8)x(-9)=72

• (x-8)(x-9)<---Factored form
Factoring trinomials with a non-1 leading coefficient by grouping

## Difference of squares

When working with difference of squares, you are multiplying two binomials to get a binomial.

Formula: a2-b2=(a+b)(a-b)

when expanding (a+b)(a-b):

• (a+b)(a-b)
• a2-ab+ab-b2
• a2-b2

Example: x2-25

1. 25 is a perfect square, so square root x2 and 25.
2. once both are square rooted, you get x and 5.
3. sub x and 5 in the formula, and that's how you factor difference of squares.

• x2-25=(x+5)(x-5)
Factoring difference of squares

## Factoring special quadratics (factoring squares)

formula: (2 stands for squared)

• a2+2ab+b2=(a+b)2
• a2-2ab+b2=(a-b)2

When completing a perfect square, to get your answer you must square root first and last term. To know if the it is a perfect square the middle number must double the square root.

Example:

4x2+20x+25

*the middle number has to be double the square root of the first and last term (2x2x5=20), in order to be a perfect square trinomial.

square root first and last term( square root of 4 and 25= 2 and 5)

look at the two formulas that, which one applies to this equation?

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

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

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

Factoring Perfect Square Trinomials - Ex1

## Linking all the factoring formulas:

All these factoring formulas, help us find the x-intercepts. When graphing a parabola we are no always given vertex form, many time we are given standard form. It can be hard to find the vertex in standard form, so that's why we factor. When we factor, we can find the x-intercepts once we find the x-intercepts we can find the vertex too. All we have to do is add both x-intercepts then divide by 2, which gives us our h value(A.O.S) and subbing our h/x value in the equation we then can find our k value (optimal value). The same rules apply when we are given a equation where a binomial is squared (i.e. (x=5)2). All factoring formulas are another way of graphing a parabola, but not given the equation in vertex form.

## Completing the square:

• When completing the square, you are basically getting standard form into vertex form.

• Standard: y=ax2+bx+c
• Vertex: y=a(x-h)2+k

Convert to vertex form!

Example: y=2x2+8x+5 (the 2 stands for squared)

Steps:

1. Group the x2 and x terms together

y=(2x2+8x)+5

2. Common factor what is in the bracket (just the coefficients, not x)

*GCF between 2 and 8 is 2, so factor out 2.

y=2(x2+4x)+5

3. Complete the square inside the bracket.

(x2+4x)

(4/2)2=(2)2=4

y=2(x2+4x+4-4)+5

**we have to subtract the 4 because in math we don't make up numbers!**

4. Now as you can see in the bracket, there is a perfect square trinomial. So, now factor out the bracket but keep the -4.

y=2((x+2)2-4)+5

Tip: the number that you use to factor your perfect square is usually the number that you divide in your binomial to complete your square.

5. Now, bring the 2 in the first bracket and multiply 2 with whatever is in the first bracket. In this case it is -4.

y=2(x+2)2-8+5

6. Collect your like terms (-8+5), and now you have your vertex form.

y=2(x+2)2-3

Completing the Square - Solving Quadratic Equations

## Why completing the square is useful?

Completing the square is a useful technique for solving quadratic equations. It comes in handy when a equation in standard form cannot be factorized, its almost like a easy way out.

Sometimes standard form equations can be very messy, or cannot factor at all, or you just don't feel like factoring. While factoring may not always be successful, the Quadratic Formula can always find the solution.

## How to use quadractic formula:

*You can only use the quadratic formula, when you have a equation in standard form (ax2+bx+c).

Example: y=x2-13x+1

1. Label the a,b, and c values.

• a=1
• b=-13
• c=1

2. Sub the values in the quadratic formula and solve for x (t=x in the picture).

3. When solving for x, you can have negative and positive roots. Find the exact root, exact roots are the last bit of the equation before you solve for x and get you approximate root.

• x=13+square root of 165 divided by 2. Second root: x=13-square root of 165 divided by 2.

4. Now, once you have your exact root, which are equations to solve for x. You now can solve to find the approximate roots.

• Square root of 169 is 12.8, rounded up to 13.

x=13+13/2

x=26/2

x=13<----first root/x-intercept

x=13-13/2

x=0/2

x=0

**Therefore the roots are (13,0) and (0,0)**

## Discriminants:

Sometimes questions just ask how many root does a equation have. Instead of using the quadratic formula (because it can be a little complicated), we can use discriminates to determine how many roots an equation has. Discriminates are the values in quadratic formula in the square root (b2-4ac).

## note:

• If the discriminant is a negative number, then there are no roots/x-intercepts.
• If the discriminant is a positive number, then there are 2 roots/x-intercepts.
• If the discriminant equals to zero , then there is only 1 root/x-intercept.

## Example:

How many roots does the equation 0=3x2+4x-5 have?

1. Label the a, b, and c values.

• a=3
• b=4
• c=-5

2. Sub values in discriminant formula (b2-4ac)

• D=(4)2-4(3)(-5)
• D=16+60
• D=76

3. Check if the discriminant is positive or negative number.

• In this case D=76 (positive number), therefore there are 2 roots.

## word problem:

A hotdog is thrown in the air, the path of the hotdog is modelled by the following equation:

h=-4.9t2+9.2t+1.6. H reps the height in meters and t reps time in seconds.

1. How long does it take for the hotdog to hit the ground?

• Remember to label your a, b, and c values.

2. What is the time when the height of the hot dog is 4.5 meters above the ground?

• Sub 4.5 as the h value in the equation.
• Rearrange the equation to quadratic formula.

## finding vertex with standard form (esaier formula)

Sometime an equation doesn't ask for x intercepts but instead asks for the vertex. The following formula helps us in finding the A.O.S and optimal value faster and easier.

Formula;-b/2a

• The b and a values are from the standard form (ax2+bx+c)
• This will solve for the A.O.S.
• And to solve for the optimal value sub the A.O.S value as x in the standard form equation.

## Example:

y=3x2+6x+1

1. Label all the values

• a=3
• b=6
• c=1

2. Find A.O.S, with formula -b/2a.

• h=-b/2a
• h=-(6)/2(3)
• h=-6/6
• h=-1

3. Find the optimal value (k value), sub x=-1 in equation.

• y=3x2+6x+1
• y=3(-1)2+6(-1)+1
• y=3(1)-6+1
• y=3-6+1
• y=-2

**Therefore, the vertex is (-1,-2)**