# Quadratics:

### Unit 1,Unit 2, Unit 3

## Vertex Form

## Summary of Unit

**Finite differences:**

## How?

## Linear If the relation's first differences are all the same number, the relation is linear. | ## QuadraticIf the relation's second differences are all the same number, the relation is quadratic. | ## Neither If the relation's second differences are not the same number, the relation is neither. |

## Mapping Notation

Remember to substitute the variables into the mapping notation expression.

(x+h), (ay-k)

Also you must change h if it is a positive it must become negative and if it is negative it must become positive.

For example:

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

In order to find the points of the parabola, you must substitute the variables into the expression with each of the corresponding numbers in the vertex form equation.

(x+h)=(x+3), (ay+k)=(2y-4)

## Using this Information We Can Find The Coordinates

## Transformations

Looking at y= a(x-h)^2+k, we can analyze a few things.

## a:

If the variable 'a' is a negative the parabola is opening downwards, however if the variable "a' is positive, the parabola is opening upwards.

- If a is -1 < a < 1, then the graph is compressed
- if a is -1 > a > 1, then the graph is stretched

## h:

- If the value of h is negative the x-coordinate shifts right
- If the value of h is positive the x-coordinate shifts left

## k:

- If K is negative, then the graph of y=x^2 vertically translates the graph down.
- If K is positive, then the graph of y=x^2 vertically translates the graph up

## Graph

y=2(x-1)^2+4

## Word Problem

**45**.

**2**)^2+45.

**1**-2)^2+45.

Now we solve.

y=-4.9(**1**-2)^2+45

y=-4.9(**-1**)^2+45

y=-4.9(**1**)+45

y=-4.9+45

y=40.1

Therefore, the height of the ball after 1 second is 40.1 meters

Now we solve.

y=-4.9(**0**-2)^2+45

y=-4.9(-2)^2+45

y=-4.9(4)+45

y=-19.6+45

y=25.4

Therefore, the initial height of the ball is 25.4 meters

Now we solve.

**0**=-4.9(x-2)^2+45

-45=-4.9(x-2)^2

-45/-4.9= -4.9(x-2)^2/-4.9 *We divide by -4.9 to get rid of it*

√9.18= √(x-2)^2 *We square root it to get rid of the exponent*

3.03=x-2

3.03+2=x

5.03= x

Therefore, the ball was in the air for 5.03 seconds

## Factored Form

## Learning Goals For Unit

1. I will learn how multiply polynomials to get a simplified factor formed equation.

2. I will learn how to factor simple and complex trinominals.

3. I will learn how to factor by grouping.

## Summary of Unit

## Understanding the Factored Form Equation

**a**(

**x**-r)(x-

**s**)

The value of **a **is gives the shape and direction of opening for the parabola.

The value of **x **and **s** gives you the x-intercepts

## Section 5.1: Multiplying polynomials

2. Simplify by collecting like terms

y=a(x-r)(x-s)

**For Example:**

(X+2)(X+3)

=2x^2+3x+4x+6

=2x^2+7x+6

## Section 5.2: Special products Squaring Binomials

2. Twice the product of the terms/ add two middle terms after expansion.

3. Square the last term.

(a + b)^2

= a^2 + 2ab + b^2

or

(a - b)^2

= a^2 - 2ab + b^2

For Example:

(x+4)^2

=x^2+8x+16

## Product of a Sum and a Difference

(a - b)(a + b)

= a ^2– b^2

For Example:

(x-4)(x+4)

=x^2+16

## Section 3: Common Factoring

**Monomial Common Factoring:**

1. Find GCF of coefficients and variables.

2. Divide each term by GCF.

Example:

5c-10d

5c/5-10d/5

5(c+2d)

**Binomial Common Factor:**

1. If there are two binomials that are exactly same, consider that as binomial common factor.

Example:

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

(x+2)(x-2)

**Factor by grouping:**

1. Factor groups of two terms with a common factor to produce a binomial common factor.

Example:

__9x^2 + 15x__ + 3x + 5

3x

3x(3x+5)1(3x+5)

(3x+1)(3x+5)

## Section 4: Simple Trinomial Factoring

A simple trinomial is a quadratic where a = 1

**1. Find the Product and Sum**

- Two numbers that multiply to get C but add to get b.

x^2+10x+25

5+5=10

5x5=25

Therefore the product and sum in 5 and 5.

Answer:

(x+5)(x+5)

**2.Look at the signs of b and c to determine the factored form.**

x^2 + bx + c

(x-r)(x-s)

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

For Example: x^2+8x+16 = (x+4)(x+4)

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

For Example: x^2-16x+64 = (x-8)(x-8)

- If c is negative, one of r or s is negative.

For Example: x^2+10x-24 = (x+12)(x-2)

- If b and c both are negative, one of r or s is negative.

For Example: x^2-5x-23 = (x-8)(x+3)

## Section 5: Complex Trinomial Factoring

1. Multiply a by c.

2. Now decompose b so that it adds up to the value of b but multiplies to the product of ac.

3. Now factor by grouping to get answer.

Example: 3x^2+14x+8

3x8=24

3x^2+12x+2x+8

__3x^2+12x__ = 3x(x+2)

3x

__2x+8__ =2(x+2)

2

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

(3x+2)(x+2)

## How to Solve Complex Trinomial Factoring questions By Khan Academy on Youtube

## Section 6: Differences of Squares

How do you tell if differences of squares: Two terms are being subtracted and both of them are perfect squares.

Consider the Following: x^2-49

1. Square root both values

a=(x)^2 b=7

2.Sub a and b into a factor form. y=a(a-b)(a+b)

y=(x-7)(x+7)

When that becomes expanded it becomes x^2-49.

## Section 6 Cont: Perfect Squares

Consider the Following:

Verify that x^2+6x+9 is a perfect square trinominal.

1. Square root a and c

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

2. Check the middle term. It should be twice the products of the square roots of the first and last terms.

2(x)(3)= 12x

Now we know that x^2+6x+9 is a perfect square trinominal.

Next, the square rooted a and c go into the factor form equation to solve,

Therefore the answer is (x+3)^2

## Section 6.2 and 6.3: Solving Quadratic Equations in Factored Form

**Finding X-Intercepts**

With a factored form equation, make y equal to 0.

Consider the following:

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

0=(x-4)(x-2)

The factored terms inside the bracket should equal to 0.

x-4=0

x-2=0

Solve for x.

x-4=0

x=4

x-2=0

x=2.

Therefore the x-intercepts of y=(x-4)(x-2) are (4,0),(2,0)

**Finding the Vertex of a Factor Form Equation**

Consider the Following:

y=x^2-16x+60

y=(x-10)(x-5)

1. Find the x-intercepts

0=(x-10)(x-6)

0=x-10

10=x

0=x-6

6=x

Therefore the x intercepts are (5,0),(9,0).

2. Find the Axis of Symmetry

__x1+x2__ = AOS

__2__

__6+10__ = AOS

2

__16__ = AOS

2

8 = AOS

3. Now sub AOS into x

y=(8)^2-16(8)+60

y=64-128+60

y=-4

Therefore the Y-int is (0,-4)

The vertex is (AOS,Y-int)

Therefore the Vertex is (8,-4)

## Word Problem

a) Write the formula in factored form

__h=-2t^2+4t+48__

-2

h=t^2-2t-24

(t+4)(t-6)

b) When will the ball hit the ground?

0=(t+4)(t-6)

0=t+4 0=t-6

-4=t 6=t

Therefore the x-int are (-4,0),(6,0)

Therefore the ball will hit the ground at (6,0)

c) What is the maximum height the ball will reach in meters?

__6-4__ = AOS

2

__2__ = AOS

__2__

1=AOS

h=-2(1)^2+4(1)+48

=-2+4+48

=50

Therefore the maximum height is 50m

## Standard Form

## Learning Goals:

2.I am able to complete the square.

3.I can solve revenue, rectangle and triangle problems.

## Summary

## Completing the Square:

(a+b)^2 = a^2+2ab-b^2

Example:

(x+3)^2 = x^2+3x+9

How to make it a perfect square, if it's not given:

Consider the following:y=x^2+ 8x+5

1. Put '5' outside of the bracket

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

2. Divide 'b' by 2 then square it. In this case it's 8.

(8/2)^2 = 16

Then the equation becomes y=(x^2+8x+16-16)+5

3.Move the negative 16 outside the bracket.

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

4.Write as squaring a binomial

y=(x+4)^2-11

Therefore the vertex is (-4,-11)

## The Quadratic Formula:

Consider the Following:

All quadratic equations of the form ax^2+bx+c=0 can be solved using the quadratic equation.

This is the quadratic equation:

The values for the variables a,b, and c are taken from standard form quadratic equation.

For Example:

Use the quadratic formula to solve for the X-Intercepts

3x^2+5x+2=0

We know that a=3, b=5, and c=2

1. Substitute in the values.

2. Add the vaules together in the square root.

3. Square root the result.

4. Add and subtract the outside number with the product of the square root.

5. Then divide the number you get from subtracting and adding with 2a.

Here is a video by mahalodotcom on Youtube on how to solve the quadratic formula.

## The Discriminant:

The equation for the discriminant is:

b^2-4ac

If D>0 there are two x-ints

If D<0 there are no x-ints

If D=0 there is one x-int

## Word Problem: Completing the square

R=(price)(quantity)

let 'x' represent # of decreases

p=(12-0.5x)

q=(36+2x)

R=(12-0.5x)(36+2x)

=432+24x-18x-x^2

=__-x^2+6x__+432

-1

=-(x^2-6x)+432 (6/2)^2=9

=-(x^2-6x+9-9)+432

=-(x^2-6x+9)+432-9(-1)

=-(x^2-6x+9)+432+9

=-(x-3)^2+441

x=3

p=(12-0.5(3))

p=12-1.5

p=10.5

Therefore, Ms. Dhaliwal should change her price to $10.5 because she can maximize her profit to $441