The graph of a quadratic relation is called a "Parabola". It is a "U" shaped graph that can open up or down and change directions at the vertex. Linear systems have the same first differences while quadratic relations 1st differences are different but the 2nd differences are the same. Quadratics consists of three different types of equations: Vertex Form, Factored Form, and Standard Form.

## Identifying Quadratic Relations in Different Forms/Formulas

• Vertex Form: a(x-h)²+k

What does each variable represent?
- a is the vertical stretch (Either opens up or down).
- h moves the parabola left or right (The x coordinate & is always the opposite of the sign given for example -4 would be +4 & +3 is -3).
- k moves the parabola up or down (The y coordinate).

- The "²" sign is what makes the parabola a U-Shaped line.
e.g. y=(x-1)²+3
Three types of transformations:

1. Vertical Stretch
2. Vertical Compression
3. Vertical Reflection
Axis of Symmetry: Mid-point of parabola, can be found on the x-axis (Represents x coordinate).
Optimal Value: Highest point or Lowest point of the parabola, can be found on y-axis (Represents y coordinate).

• Factored Form: y=a(x-r)(x-s)

What does each variable represent?

X-intercepts: These are the "r" and the "s" variables in the equation
Axis of Symmetry: The mid-point of the equation
Optimal Value: The max or min of the equation

e.g. y=1(x+3)(x-9)

• Standard Form: y=ax²+bx+c
What does each variable represent?

Also solved as ax²+bx+c = 0

The vertex has the x-coordinate x = -b/2a

The y-coordinate of the vertex is the maximum or minimum value of the function.

The y-intercept of the equation is c.

e.g. y= 2x²+4x+10

## Finding the x-intercepts in Vertex Form

Determine the y-intercept and the x-intercepts

a) y= -4(x+2)²+16

To find the y-intercept, set x=0

y= -4[(0)+2]²+16
y= -4(2)²+16
y= 0
Therefore y-intercept is (0,0)

To find the x-intercepts, set y=0

0= -4(x+2)²+16

-isolate for x, BEDMAS backwards

(Bring the 16 to the left side)
-16= -4(x+)² (now subtract 16)

-16/-4=-4(x+2)²/-4 (now divide -4)

√4=(x+2)² (now square root)

x+2= +/- √4

x= +/- √4-2 (now subtract 2)

(Now you are going to find the x-intercepts)

x= + √4-2 or x= - √4-2

x=2-2 x=-2-2

x=0 & x=-4

## Graphing using the Step Pattern

Regular Step pattern:
Point one: Over one and up one
Point two: Over two and up four

*Always multiply the stretch given with the second step.

Examples

1. y=x²

Vertex: (0,0)

Step Pattern:

Point one: over 1 and up 1

Point two: over 2 and up 4

2. y=2(x-3)²+4

Vertex: (3,4)
Step pattern:

2x1=2 & 2x4=8
Point one: over 1 and up 2
Point two: over 2 and up 8

## Determine the Equation for Vertex form from a Graph

The vertex in the graph on the right is (2,-1), so h= 2 & k= -1

The parabola opens upward, so we know "a" will be positive.

y=a(x-h)²+k

y=a(x-2)²-1

The parabola passes through point (0,3). Substitute x=0 and y=3 and solve from there.

3=a(0-2)²-1

3=a(-2)²-1

3=a(4)-1 (Bring -1 to the other side and change the sign)

3+1=4a

4=4a

4/4=4/4a (Divide to cancel the "4" from the "a" value)

1=a

The equation is y=1(x-2)²-1

3.3 More Graphing from Vertex Form

## Graphing Factored Form

How to Find each Variable:

• X Value:

Ex. y=0.5(x+3)(x-9)

There are two x-intercepts in factored form, here they are -3 & +9.

Finding the x value a.k.a the axis of symmetry: Add the x values and divide by 2

Ex. x= -3 and x= 9

-3+9/2
x= 3

• Y Value:

To find the Y value a.k.a the optimal value: Sub the x value in the original equation

Ex. y=0.5(x+3)(x-9)
y=0.5(3+3)(3-9)
y=0.5(6)(-6)
y= -18

Therefore, the vertex for this graph is (3,-18).

## Common Factoring

When you are asked to factor, you have to find what numbers were multiplied to make up the value.

Examples:

1) Factor the number 6

6= 3x2

2) Factor (2x+6)

- You have to see what is common in this equation.

So you should get that 2 is common because it can be multiplied to 6 and 2x

3) 8x+6

- Find whats common or find GCF (greatest common factor)

GCF= 2

- Just write down the solution with brackets

4) 12x3y-6x2y+18x2y

GCF= 6x2y

Solution:

= 6x2y(2x-1+3)

= 6x2y(2x+2)

## Trinomials

What is a "trinomial"?

Trinomials are factors that have 3 values in them and can be followed by this formula (x²+bx+c).

EX. 1) x²+3x+4x

Ex. 2) 2x²+8x+9

Ex. 3) 3x²-5x+4

## Simple Trinomial

• Simple Trinomials are trinomials that have an "a" value of 1
• Many polynomials, such as 12x²+7x+2, can be written as the product of two binomials of the form (x + r) and (x + s)
• When factoring a polynomial of the form ax²+bx+c (when a = 1), we want to find:

(1) 2 numbers that ADD to give b

(2) 2 numbers that MULTIPLY to give c

To factor these trinomials, use a table to help you until you become familiar with the procedure

Example. Factor: x²-8x+12

=(x-6)(x-2)

Examples:

Factor the following trinomials using the procedure above.

a) m²-13m+26

13x2=26

13+2= -13

- not possible

b) h²-7h-18

-9x2= -18

-9+2= -7

=(h-9)(h+2)

## Complex Trinomials

Complex trinomial have a coefficient of more than 1 in front of the x^2 term

ax²+bx+c

(where a=1)

Can you factor the following?

3x²+17+10

- No common factor

- a=3

Example 1: Binomial Common Factoring

Factor: 3x(z-2)+2(z-2)

(Hint: a binomial can be the common factor)

= (3x+2)(z-2)

Example 2: Factor by graphing

Factor: (df+ef)+(dg+eg)

(There is no common factor in all terms)

(We can graph terms together that have a common factor)

= f(d+e)(f+g)

## How to Solve Complex Trinomials

The decomposition method:

A) factor

3x²+8x+5

Product of a multiplied c

= 3 x 5

=15

3x5=15

3+5=8

* Rewrite the middle terms with the 2 factors

3x²+3x+5x+5

Not 3x²+5x+3x+5

* Factor by grouping

(3x²+3x)+(5x+5)

= 3x(x1)+5(x+1)

* Think of the binomial (x+1) as

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

EXAMPLES OF COMPLEX TRINOMIAL

Example 1) 5x²-14x+8

P: 5x8= 40

-10 x (-4)=40

-10 + (-4)= -14

= (5x2-10x)(-4x+8)

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

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

Example 2) 8a²+10x-3

p; 8 x (-3)= -24

12x(-2)= -24

12+(-2)= 10

=(8a2-2a)+(12a-3)

=2a(4a-1)+3(4a-1)

=(4a-1)(2a+3)

HOW TO GRAPH COMPLEX TRINOMIAL

When solving an equation that requires factoring, ONE SIDE MUST EQUAL ZERO

Example 1: 5x²-14x+8

Step 1: Factor

P: 5x8= 40

-10 x (-4)=40

-10 + (-4)= -14

= (5x2-10x)(-4x+8)

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

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

Step 2: Find x-intercepts

(Now to solve for for x, you must let each bracket t zero)

First x-intercept

x-2=0

x=2

Second x-intercept

5x-4=0

5x=4 (Now divide 5 to find x)

5x/5=4/5

x=4/5

Example 2: 8a²+10x-3

Step 1: Factor

p; 8 x (-3)= -24

12x(-2)= -24

12+(-2)= 10

=(8a²-2a)+(12a-3)

=2a(4a-1)+3(4a-1)

=(4a-1)(2a+3)

Step 2: Find x-intercepts

First x-intercept

4a-1=0

4a=1

4a/4=1/4

a=1/4

Second x-intercept

2a+3=0

2a= -3

2a/2= -3/2

a=-3/2

## How to go from Factored form to Standard form

Example 1) y=(x+6)(x+4)

Expand and Simplify:

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

y=x²+4x+6x+24

y=x²+10x+24

Therefore the standard form of y=(x+6)(x+4) is y=x²+10x+24

Example 1) y=(x+9)(x+4)

Expand and Simplify:

y=(x+9)(x+4)

y=x²+4x+9x+36

y=x²+10x+36

Therefore the standard form of y=(x+9)(x+4) is y=x²+10x+36

## Graphing Standard Form

Have to make the equation into factored form

Check to see if the coefficient in front of the x² value can be divided equally and if you can divide the equation by that number

Now you should get a number like y=2(x2-2x-8) for example

You should now break the numbers in the bracket to make two numbers that multiply each other ex. y=2(x-4)(x+2) now find the x-intercepts, find the axis of symmetry & find the optimal value

And there you should get a parabola

EXAMPLES OF GRAPHING STANDARD FORM

Example 1) y=2x²-4x-16

Part 1

Make the equation into factored form

y=2x²-4x-16

y=2(x2-2x-8)

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

Part 2

Now graph it in the factored form

a) State the zeros

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

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

x-4=0 and x+2=0

x=4 and x= -2

b) State the axis of symmetry

x=4+(-2)/2

x= 2/2

x= 1

c) State the optimal value

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

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

y=2(-3)(3)

y=2(-9)

y= -18

d) State the direction of opening

The direction of opening is up

Example 2) y=2x²+8x-24

Part 1

Make the equation into factored form

y=2x²+8x-24

y=2(x2+4x-12)

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

Part 2

Now graph it in the factored form

a) State the zeros

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

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

x+6=0 and x-2=0

x= -6 and x= 2

b) State the axis of symmetry

x= -6+2/2

x= -4/2

x= -2

c) State the optimal value

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

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

y=2(4)(-4)

y=2(-16)

y= -32

d) State the direction of opening

The direction of opening is up

Gaphing from Standard Form by Factoring

## Perfect Squares

Whenever you multiply a binomial by itself twice, the resulting trinomial is called a perfect square trinomial

For example, (x + 1) × (x + 1) = x2 + x + x + 1 = x2 + 2x + 1 and x2 + 2x + 1 is a perfect square trinomial

EXAMPLES OF PERFECT SQUARES

Example 1) 9x²+24xy+16y²

=(3x)²+2(3x)(4y)+(4y)²

= (3x+4y)(3x+4y)

=(3x+4y)²

Example 2) 4m²-20mn+25n²

= (2m)²-2(2m)(-5n)+(-5n)²

= (2m-5n)(2m-5n)

= (2m-5n)²

Example 3) 49a²+42ab+9b²

= (7a)²+2(7a)(3b)+(3b)²

= (7a+3b)(7a+3b)

= (7a+3b)²

## Differences of Squares

Whenever you multiply a binomial that has a negative number and a binomial that has a positive number, the resulting trinomial is called a difference of square

For example, (x+8)(x+8) = x²- 64 and x²+64 is a difference of square trinomial

EXAMPLES OF DIFFERENCE OF SQUARE

Example 1) 9x²-49

= (3x)²-(7)²

= (3x+7)(3x-7)

Example 2) 81a²-4

= (9a)²-(2)²

= (9a+2)(9a-2)

Example 3) 4h²-49

= (2h)²-(7)²

= (2h-7)(2h+7)

Example 4) 144y²-169

= (12y)²-(13)

= (12y-13)(12y+13)

## How to solve for x-intercepts using quadratic formula

Solving for x intercepts using Quadratic Formula

## Axis of Symmetry H= -B/2A

WHEN CAN WE USE AXIS OF SYMMETRY:

You can use Axis of Symmetry when you are asked to find the vertex in a standard form. You have to use the Axis of Symmetry when you are using the quadratic formula and the discriminants values are negative.

## Reflection of the unit

When Quadratics was first introduced to us, I was told that at first that I was going to hate Quadratics but once I got the hang of it, it got pretty simple. That’s exactly what happened, at first I wasn't interested in Quadratics, I was interested when it came to solving linear systems or other units. I found Quadratics challenging and annoying just because the methods and graphing in such forms was very confusing. Throughout the unit when the teacher assigned homework to the class and doing some of the quizzes that was for practice. I understood the equations and started to understand Quadratics and I also knew that if I don’t understand these methods and formulas I wouldn't do well in class. But once I got used to the methods and formulas they became a part of my daily routine. In some aspects I still find somethings confusing but I reviewed and took the time to understand and practice for good marks on my tests.