## OVERVIEW OF THE UNIT

In the quadratic relationship unit I learned many ways and methods to graph and understand many types of equation.

## What is a parabola?

A symmetrical open plane curve formed by the intersection of a cone with a plane parallel to its side. The path of a projectile under the influence of gravity ideally follows a curve of this shape. You will know that it is a quadratic equation if it has a x².

## Parabolas in real life

Parabolas in the "Real-World"
Parabolas in real life

## Parts Of A Parabola

Axis Of Symmetry : The A.O.S. divides the parabola in 2 halves, it is a line of symmetry for a graph. It is when the two sides of a graph on either side of the axis of symmetry look like mirror images of each other. It can be found on the x-axis.

Optimal Value: The lowest or highest point the parabola can go. this is the y value in the vertex. The optimal value determines if the parabola goes upward or downward (max or min).

Vertex: The maximum or minimum point on the graph. The vertex consists of the axis of symmetry, which is the x coordinate and the optimal value, which represents the y value.

Y-intercept: The y-intercept is point on the parabola that hits y-axis

X-intercept: The x-intercept is the point on the parabola that hits the x-axis

Zeros and Roots: these are the x values and is when you set the y=o

## Identifying Quadratic Relations in Vertex Form: y=a(x-h)2 +k

(h, k) is the vertex of the parabola, and x = h is the axis of symmetry.

• the h represents a horizontal shift (how far left, or right, the graph has shifted from x = 0).

• the k represents a vertical shift (how far up, or down, the graph has shifted from y = 0).

• notice that the h value is subtracted in this form, and that the k value is added.
If the equation is y = 2(x - 1)2 + 5, the value of h is 1, and k is 5.

If the equation is y = 3(x + 4)2 - 6, the value of h is -4, and k is -6.

## Example of Vertex form

Example 1: y= (x+3)²

Vertex: (-3,0)

Axis of symmetry: X=-3

Stretch or compression factor relative to y=x²: None

Direction of opening: Upward

Value of x: X=-3

Value of y: Y=0

Example 2: y=(x+2)²+5

Vertex: (-2,5)

Axis of symmetry: X=-2

Stretch or compression factor relative to y=x²: None

Direction of opening: Upward

Value of x: X= -2

Value of y: Y=5

Example 3: y=3(x+7)²-2

Vertex: (-7,-2)

Axis of symmetry: X=-7

Stretch or compression factor relative to y=x²: 3

Direction of opening: Upward

Value of x: X= -7

Value of y: Y=-2

## Finding the Zeros in Vertex Form

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

Step 1:Find the vertex. Since the equation is in vertex form, the 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.

## Vertex Form With Step Pattern

The step pattern is a pattern that identifies your next points in each parabola.

The original step pattern r is 'Over 1 up 1 and Over 2 up 4'

But in quadratic relations if there is an 'a' value the step pattern change.

3.2 Graphing from Vertex Form

## Step pattern

The step pattern is a pattern that identifies your next points in each parabola.

The original step patter is 'Over 1 up 1 and Over 2 up 4'

But in quadratic relations if there is an 'a' value the step pattern changes.

here is a video to help with the step pattern

## What each variable represents

-The zeros/x-intercepts: are the r and s value you find them by setting each (factor) equal to zero.

-The axis of symmetry: is the midpoint of the two zeroes

-The optimal value: is found by subbing the axis of symmetry value into the equation

Example 1: y=0.5(x+3)(x-9)

1) find the zeroes

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

0=0.5(x+3)(x-9)

x+3=0 and X-9=0

x=0-3 and x=0+9

x=-3 and x=9

2) find the axis of symmetry

X=-3+9/2

x=6/2

x=3

3) state the optimal value

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

y=0.5(3+3)( 3-9)

y=0.5(6)(-6)

y=-18

Example 2: y=2(x+1)(x+4)

1) find the zeroes

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

0=2(x+1)(x+4)

X+1=0 and X+4=0

X=0-1 x=0-4

x=-1 x=-4

2) find the axis of symmetry

X=-1+-4/2

x=-5/2

x=-2.5

3) state the optimal value

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

y=2(-2.5+1)(-2.5+4)

y=2(-1.5)(1.5)

y= -4.5

## My videos

Graphing Parabolas in Factored Form Part 1
Graphing Parabolas in Factored Form Part 2
3.5 Graphing from Factored Form

## Example of standard form and graphing

Example 1: standard form, y= ax2 + bx + c

Graph the equation y = x2 – 8x + 12 using factoring.

Factor the trinomial, x2 – 8x + 12. Identify 2 numbers whose sum is –8 and the product is 12. The numbers are –2 and –6. That is, y = (x-2) (x-6)=0 .

So, by the zero product property, either x – 2 = 0 or x – 6 = 0. Then the roots of the equation are 2 and 6.

Therefore, the x-intercepts of the function are 6 and 2.

The x-coordinate of the vertex is the midpoint of the x –intercepts (axis of symmetry). So, here the x-coordinate of the vertex will be 2+6/2=4. .

Substitute x = 4 in the equation y = x2 – 8x + 12 to find the y-coordinate of the vertex (optimal value).

y = x2 – 8x + 12

y = (4) 2 – 8(4) + 12

y = 16 – 32 + 12

y = -4

That is, the coordinates of the vertex are (4, –4).

Now we have 3 points (4, –4), (2,0) and (6,0) which are on the parabola. Plot the points. Join them by a smooth curve and extend the parabola.

Example 2: standard form, y= ax2 + bx + c

Graph the equation y = -x2 – 2x + 8 using factoring.

Factor the trinomial, -x2 – 2x + 8.

First, factor out –1.

= -1(-x2 – 2x + 8)

Factor the expression in the parenthesis. Identify 2 numbers whose sum is 2 and the product is –8. The numbers are 4 and -2. That is, y = (x+4) (x-2)

Then, the given function becomes y = - (x+4) (x-2)

So, y = 0 implies, by the zero product property, x+ 4 = 0 or x – 2 = 0.

Therefore, the x-intercepts of the graph are –4 and 2.

The x-coordinate of the vertex of a parabola is the midpoint of the x-intercepts. So, here the x-coordinate of the vertex will be -4+2/2= -1.

Substitute x = -1 in the equation y = -x2 – 2x + 8 to find the y -coordinate of the vertex.

y = -x2 – 2x + 8

y = - (-1) 2 – 2(-1) +8

y = -1 + 2 + 8

y = 9

So, the coordinates of the vertex are (–1, 9).

Now we have 3 points (–1, 9), (–4, 0) and (2, 0) which are on the parabola. Plot the points. Join them by a smooth curve and extend the parabola.

Gaphing from Standard Form by Factoring

The quadratic formula is used to solve a very specific type of equation, called a quadratic equation. These equations are usually written in the following form:

Ax² + Bx + C = 0

## Example:Axis of symmetry (-b/2a)

Example: Axis Of Symmetry Equation

The axis of symmetry has its own equation as follows: -b/2a

Equation: y= x² + 12x + 32

a=1 b=12 c=32

x=-b/2a

x=-12/2(1)

x=6

## Completing the square

completing the Square is a technique used to solve quadratic equations, graph quadratic functions, and evaluate integrals. This technique can be used when factoring a quadratic equation does not work or to find irrational and complex roots.
3.14 Completing the square

## common factoring

Common factoring should always be the first thing you look for when you have an equation in standard form. You should look to see if each of the terms in the equation has a similar component that each of the terms can easily be divided into. The common factor can either be a variable (x) or a coefficient (any number.)

Common factoring:

Example 1: 8x + 6

Step #1: Find GCF ( what is the greatest # in common between 8 and 6)

GCF=2

Step #2: write solution with brackets

8x/2 + 6/2 = 4x + 3

= 2(4x + 3)

Example 2: 12x³ - 6x²

step #1: GCF= 6x²

Step #2: 6x²(2x-1)

6x²(2x-1)

=12x³ - 6x²

Example 3: 12x³y - 6x² + 18x² y

Step #1: GCF= 6x²y

Step #2: 6x²y(2x-1+3)

## simple trinomials

Simple trinomials have 3 terms, and can be written in the form y = ax² + bx + c, where x = 1
3.8 Factoring Simple Trinomials

## complex trinomials

Complex Trinomials follow the same rules as Simple Trinomials (see above), however the "a" term on a complex trinomial is not 1. That makes it a little trickier, and more often the method trial and error or guess and check will have to be used.
3.9 Complex Trinomial Factoring

## prefect squares

What is a perfect square?

Perfect Square Trinomial is the product of two binomials. But, both the binomials are same. When factoring some quadratics which gives identical factors, that quadratics are Perfect Square Trinomial.

Example of a perfect square:

x² + 10x + 25

= (x + 5)(x + 5)

= (x + 5)²

Question 1: Factor the trinomial x2 + 4x + 4
Solution:

Given x2 + 4x + 4

The above trinomial can be written as x2 + 2x + 2x + 4

Take x as common from first two terms,

x(x + 2) + 2x + 4

Take 2 as common from last two terms.

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

(x + 2)(x + 2)

In this it is the product of two identical binomials

(x + 2)2 , which is a perfect square.

Question 2: Factor the trinomial x2 + 8x + 16
Solution:

Given x2 + 8x + 16

The above trinomial can be written as x2 + 4x + 4x + 16

Take x as common from first two terms,

x(x + 4) + 4x + 16

Take 4 as common from last two terms.

x(x + 4) + 4(x + 4)

(x + 4)(x + 4)

In this, it is the product of two identical binomials

(x + 4)2, which is a perfect square.

Algebra - Perfect Square Factoring and Square Root Property

## difference of squares

Difference of Squares When working with a difference of squares, you must check to see that
1. Each term is a squared term and
2. There is a minus sign between the two terms.

Two terms that are squared and separated by a subtraction sign like this:

a2 - b2

Useful because it can be factored into (a+b)(a−b)
Algebra - Factoring Differences of Squares