# QUADRATICS

### By: Rhea sahoo

## INTRODUCTION

**Quadratics?**

Seems like a cool word,doesn't it?

Are you in grade 9 wanting to prepare for grade 10 or maybe your in grade 10 wanting to learn about Quadratics. Well you've come to the right place.

Quadratics can be difficult at first but you will get the hang of it.

Remember PRACTICE MAKES PERFECT!

**WHERE ARE THE QUADRATICS?**

Believe it or not, quadratics are all around you. you see them when you throw a ball to when your just looking at a normal arch bridge. Throwing a ball, kicking a ball, a rainbow, a bridge, and even the M in the McDonald logo use quadratics.

**WHAT ARE THEY USED FOR?**

Quadratics can be used for a lot of things like graphing a flight path or if you need a ball to land at a certain spot you can you use quadratics to make sure it does. Quadratics can also be use to make or find the maximum profit.

## Topics:

**Quadratics**

- First/second differences-Quadratic Relations and finite differences 4.2
- Key features of Quadratic Relations (vertex, AOS, zeros, optimal value)
- Transformations of Quadratics How does a, h and k transform the parabola? Can you write equations given the transformations?

**Equation**

- Vertex form
- Factored form
- Standard form

**Vertex form**

- 4.4 Graphing Quadratics in Vertex Form How do you graph using the step pattern? Mapping notation formula?

- Investigating vertex form
- Graphing in vertex form
- Finding x and y intercepts in Vertex Form
- Word problems for Quadratics in Vertex form - An object thrown in the air - Business Application (profit question) Quadratics in Factored Form

**Factoring**

- 5.1 & 5.2 Multiplying Binomials and Special Products
- 5.3 Common Factoring
- Factoring by grouping
- 5.4 Factoring Simple Trinomials- Factoring Complex Trinomials
- 5.6 Factoring Difference of Squares and Perfect Square Trinomials

- Equations in factored form
- Word problems

**Graphing quadratics/ solving quadratics**

- 6.2 Solving Quadratics by Factoring (finding the zeros)
- 4.5 & 6.3 Graphing Quadratics in Factored Form How do you graph using the x-intercepts? How do you find the vertex?
- Word problems for Quadratics in Factored Form - Dimensions of a rectangle given the area - Flight of an object - Shaded region of a shape Quadratics in Standard Form
- 6.1 Maximum and Minimum values (Completing the square) How do you go from Standard Form to Vertex Form?
- 6.4 The Quadratic Formula How do we get the x-intercepts when factoring is not possible?
- Word Problems in Standard Form (6.5) - Flight of an Object - Revenue problems - Consecutive integers and triangle problems - Maximum Area problems

## 1st and 2nd Differences

The Example Below Is A Quadratic Relation Since the second differences are constant.

## Key Features

**PARABOLAS INCLUDE:**

Vertex (Axis of symmetry and parabola meet)

A.O.S (axis of symmetry) ( divides the parabola into two equal halves)

Optimal value (highest or lowest point/ the Max or Min value)

X- intercepts ( Also known as "the zeros")

Y- intercept (Graph crosses the y-axis/Optimal Value)

## Rules of a Parabola

- Opens up or down
- Parabola is symmetrical
- Parabola has the axis of symmetry, the vertex, and a max or min value.
- x-intersepts are also called roots or zeros
- y-intersept is where the parabola crosses the y-axis

## Transformation

**The Vertex form formula; a(x-h)²+k, have values that are responsible for different transformations:**

- The "h" value moves the parabola left or right. Going left or right depends if the h value is positive or negative. When the H value is negative the parabola moves right and when the H value is positive the parabola moves left.
- The "k" value moves the parabola up or down. Going up or down depends on if the K value is positive or negative. When the K value is negative the parabola moves down and when the K value is positive the parabola moves up.
- The "a" value stretches the parabola. When the A value is negative the parabola opens down and when the A value is positive the parabola opens up.
- The Vertex is the h and k value. (h,k)

Here is an example:

## Step Pattern

The step pattern is one method to find out which points the parabola crosses.

Here are the steps required for Graphing Parabolas in the Form y = a(x – h)2 + k:

**Step 1**:Find the vertex.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.

## Equations

- Vertex form: a(x-h)²+k
- Standard form: ax²+bx+c
- Factored form : a(x-r)(x-s)

## Expanding and Factoring

## Expanding Binomials

EX:(3x+2)(3x-3)

To expand this binomial:

- First, you multiply the outer term with inner term. (3x)(3x)= 9x^2
- Second, you multiply the outer term with the inner term. (3x)(-3)= -9x
- Third, you multiply the two inner terms. (2)(3x)= 6x
- Fourth, you multiply the inner term with the outer term. (2)(-3)= -6
- Fifth, you group the like terms.

9x^2-3x-6

## Special Products

(x+4)^2

x is the value a and 4 is the value of b, so now you plug that into the formula:

x^2+ 2(x)(4)+ 4^2

x^2+8x+16

## Common Factoring

1) Binomial Common Factor

2) Factor by Grouping

**Binomial Common Factor:**

i) two binomials that are exactly same, are consider as a binomial common factor.

For example

x(x-4)+2(x-4)

- x-2 are the same. They get common factored and you would rewrite it as:

(x-4)(x+2)

**REMEMBER:** the common binomial goes first and what is left comes after.

**Factor by Grouping:**

When there is no common factor in all terms we can group terms togerther that have a common factor

Example:

df+ef+dg+eg

- STEP 1: df and ef have the common factor f and dg and eg have the common factor of g.
- STEP 2: Add brackets around the common terms:

- STEP 3: Take the common factor outside the brackets. if both of the terms inside the brackets are the same, then you are correct.

- STEP 4: Set them into a binomial common factor

## Simple Trinomial

ax^2 + bx + c = 0= (x+r)(x+s)

**Step 1:** Find the product and sum

- Find two numbers that multiply to equal the product or the c value
- Find two numbers that add up to equal the sum or the b value

**Example:**

x^2+12x+27

- Two numbers multiplied to equal the product (27) and also equal the sum (12) are 3 and 9.

**Step 2: **Look at the signs of b and c .

- when b and c are positive, both r and s are positive.

**Example:**

x^2+7x+12

(x+4)(x+3)

- When b is negative and c is positive, r and s are negative.

**Example:**

x^2-29x+28

(x-28)(x-1)

- If c is negative, r or s is going to be negative.

**Example:**

x^2+3X-18

(x-3)(x+6)

## Factoring Complex Trinomials

Complex trinomial's do NOT have a coefficient of 1 in front of the x^2 term

**Step 1:**Always look at the common factor first when factoring trinomial

-Ask first: **can you common factor?**

**Step 2:** To factor ax^2+bx+c, we first need to multiply the a value and the c value to get our product. Our sum which is the b value stays the sum.

**Step 3:** Factor by grouping or Decomposition or Trial and error

**Example:**

3x^2 +8x+4

- Two integers multiplied to give us our product which is 12 (3x4) and the same two numbers added together to get our sum which is 8 are 2,6

- Break up the b term which is our sum (8)

- Factor by grouping

(3x^2+2x)+(6x+4)

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

(3x+2)(x+2)

HERE IS A VIDEO TO HELP YOU DO THE "DECOMPOSITION" METHOD:

## Difference of squares

**Example:**

x^2-16

x^2-4^2

(x+4)(x-4)

## Maxima and Minima

Completing the squares: Standard form (x^2+bx+c) to vertex form (y=a(x+k)^2-h).

**Step 1:** Group the x and x^2 terms together

**Step 2:** Complete the square inside the bracket

**Step 3: **Write the trinomial as a binomial squared

**Step 4:** Simplify

**Example:**

y=x^2+8x+5

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

y=(x^2+8x+16)-16+5

y=(x+4)^2-16+5

y=(x+4)^2-11

vertex of this equation is (4,-11).

**Example 2:** Find maximum or minium point of the parabola y=2x^2+12x+11.

## Graphing Using the x-intercepts

**Step 1:**When graphing using the x-intercepts you may use the method of factoring to find the values of x.

**Step 2:** Use the values of x to find your vertex .

- Find A.O.S (axis of symmetry)
- Find Max/ Min (optimal value)

y=x^2-6x+8

-4,-2

y=x^2-4x-2x+8

y=(x^2-4x)(-2x+8)

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

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

x-4=0 x-2=0

x=4 x=2

(4,0) (2,0)

**x-intercepts:** 4,0) and (2,0)

**A.O.S:** 4+2/2=6/2=3

Sub into equation

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

y= (3-4)(3-2)

y=(-1)(1)

y=-1

**Vertex: **(3,-1)

**Y-Intercept: **8

**Step 3:** Graph the x-intercepts, vertex, and the y-intercept.

## Solution

There are 3 ways to find the solutions:

1. We can Factor (find what to multiply to make the Quadratic Equation)

2. We can Complete the square

3. We can use the Quadratic Formula:

## Quadratic fourmula

How to find the x-intersects using the quadratic formula:

**Step 1: **When given your standard form equation you determine your a value, your b value and c value so there is no confusion. **Step 2:** plug in all of you variables.

- First, you square your b and multiply your -4ac
- Multiply your 2a
- Thirdly, solve inside the square root
- Fourthly, square root that answer
- Lastly, the equation can go two ways;

- Add your -b with your square root divided by your 2a
- Subtract your -b with your square root and divide it with your 2a

**Step 3:**The two answers you get are your x-intercepts.

**Example: Solve 5x² + 6x + 1 = 0**

**Coefficients are:** a = 5, b = 6, c = 1

**Quadratic Formula:** x = −b ± √(b2 − 4ac)2a

**Put in a, b and c:** x = −6 ± √(62 − 4×5×1)2×5

**Solve**: x = −6 ± √(36 − 20)10

x = −6 ± √(16)10

x = −6 ± 410

x = −0.2 **or** −1

**Answer:** x = -0.2 **or** x = -1

You can also check your answers:

Check **-0.2**:5×(**-0.2**)² + 6×(**-0.2**) + 1

= 5×(0.04) + 6×(-0.2) + 1

= 0.2 -1.2 + 1 **= 0**

Check **-1**:5×(**-1**)² + 6×(**-1**) + 1

= 5×(1) + 6×(-1) + 1

= 5 - 6 + 1 **= 0**

## Word problem using quadratic formula

**Step 1:** Draw a diagram and label

**Step 2:** find what you know*Total width: x* + 12 + *x* = 12 + 2*x*

*Total length:* *x* + 16 + *x* = 16 + 2*x*.

Then the new area is given by:

(12 + 2*x*)(16 + 2*x*) = 285

192 + 56*x* + 4*x*2 = 285

4*x*2 + 56*x* – 93 = 0

**Step 3:** Quadratic formula

**The width of the pathway will be**

**1.5**

**meters.**

## Discriminant

**b2 - 4ac** comes from the quadratic formula. It is called the **Discriminant**, because it can "discriminate" between the possible types of answer:

- when
**D>0**,there are 2 solutions - when
**D= zero**, there is only one solution - when
**D<0**, there are NO solutions

## Consecutive word problem

**The product of two consecutive negative integers is****1122****. What are the numbers?**

*NOTE*consecutive integers are one unit apart (*n* and *n* + 1).

*n*(*n* + 1) = 1122 *n*2 + *n* = 1122*n*2 + *n* – 1122 = 0

(*n* + 34)(*n* – 33) = 0

*n* = –34 and *n* = 33.

*n* + 1 = (–34) + 1 = –33.

**2 #: ****–33**** and ****–34****.**

## Revenue Word Problems

a) Create an algebraic equation to model Elena's total sales revenue

Let (18.50+0.50x) represent the cost

Let (216-8x) represent the total number of people

R= (18.50+0.50x)(216-8x)

b) Determine the maximum revenue and the price at which this maximum revenue will occur

R= (18.50+0.50x)(216-8x)

18.50+0.50x=0 216-8x=0

0.50x= -18.50/0.50 -8x= -216/-8

x= -37 x= 27

-37+27/2= -10/2 = -5

R= (18.50+0.50x)(216-8x)

R= (18.50+0.50(-5))(216-8(-5))

R=(18.50-2.5)(216+40)

R=(16)(256)

R= 4096

Therefore the maximum Revenue $4096 will occur at the maximum price of $16

## Triangle problems

The shortest side of a right-angled triangle is 6cm shorter than its hypotenuse. The difference in length of other two sides is 3cm. If the shortest side is n-3, show that 2n2 = 12n. Hence, find n.

If the length of the shortest side is n-3, the length of the hypotenuse and the other side are n+3 and n respectively.

So, using Pythagoras Theorem,

(n-3)^2 + n^2 = (n+3)^2

n^2 - 6n + 9 + n^2 = n^2 + 6n + 9

2n^2 = 12n

2n^2 - 12n = 0

2n^2 = 12n

n^2 - 6n = 0

n = 0 or n = 6.

length cannot be zero, n = 6.

## Example 2

## Word problem

**Example:**

A ball is thrown upwards from a rooftop, 80m above the ground. It will reach a maximum vertical height and then __fall back__ to the ground. The height of the ball from the ground at time t is h, which is given by,

h = -16^t2 + 64t + 80.

- What is the height reached by the ball after 1 second?
- What is the maximum height reached by the ball?
- How long will it take before hitting the ground?

Follow the graph along with the calculation for a better understanding:

h = -16 x 1 x 1 + 64 x 1 + 80 = 128m

2) Rearrange by the completing the square, we get:

h = -16[t2 - 4t - 5]

h = -16[(t - 2)2 - 9]

h = -16(t - 2)^2 + 144

When the height is maximum, t = 2; therefore, maximum height = 144m.

3) When the ball hits the ground, h = 0;

-16t^2 + 64t + 80 = 0

Divide the equation by -16

t^2 - 4t - 5 = 0

(t - 5)(t + 1) = 0

t = 5 or t = -1

The time cannot be negative; so, the time = 5 seconds.

## Reflection

The test above wasn't really good. i could improve on communication an application by explaining my steps one by one. this test could help me study for other test and exams to improve better and learn from my mistakes.