Quadratic Relations

Math and Success add up!

Section 1: Parabolas and Vertex Form

Vertex- The vertex of a parabola is the point where the parabola crosses its axis. Always located halfway between the zeroes.

Optimal Value- The y-intercept of the vertex points< can be either maximum or minimum.

Axis of Symmetry- Goes through the vertex and is located at the x-intercept of the vertex.

Zeroes- Wherever the parabola hits the x-axis.

Y-intercept- Wherever the parabola hits the y-axis.

- The parabola can open upwards or downwards.


Vertex Form - y = a(x-h)² + k

- When the equation is in vertex form the vertex is always the negative reciprocal of (h, k) making this the easiest equation to find the vertex of the parabola.

Transformations with Vertex Form

Terminology:

- Stretch, Compress, Reflect (Flip).


Vertical Shift: The Vertical Shift is dependent on the "K" Value. As the "K" Value is one or higher, the parabola will move up. As the "K" Value is negative one or lower than negative one, the parabola will move down.


Horizontal Shift: The Horizontal is dependent on the "-H". When you write your equation, and when the "H" is positive, you move your parabola left when graphing, when your "H" is negative, you will move your parabola to the right.


Vertical Strech: The Vertical Stretch is dependent on the "A" Value. When the "A" Value is a fraction then that means parabola will become wider (stretch). When the "A" Value is a number that is one or higher, it will become thinner (compressed).

Step Patterns:

Basic Step Pattern - Over 1 Up 1, Over 2 up 4

This is a means to achieve where the points on the parabola lie on the graph starting from the vertex on both sides.

When a number is substituted for "a" and it is either stretched or compressed you will multiply it by the second step of both and then if it is positive it will be going up by that number and if it is negative it will be going down by that number.

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Quick Way of Graphing a Quadratic Function in Vertex Form

Finding Equations with Vertex Form

This is the basic example on how to find an equation with vertex form.
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Section 2: Factored Form y=(x-r)(x-s)

X-Intercepts - Points at which the parabola touches the x-axis. There can either be zero, one, or two. They are the negative reciprocals of r and s.
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Finding the Vertex with Factored Form

Axis of Symmetry - The x co-ordinate of the vertex.

To find the axis of symmetry you must find the zeroes, add them and then divide them by 2 to find the axis of symmetry.

e.g.

(X+2) (X+4)

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X+2=0 X+4=0

X=-2 X=-4

(R+S)÷2

(-2+(-4))÷2

-6÷2

=-3


You then insert that value into your equation to find the optimal value.

X=-3

(X+2)(X+4)

([-3]+2)([-3]+4)

(-1)(1)

=1

Types of Factoring

Common Factoring - You find the greatest common factor and divide it you of the equation to common factor.

e.g. 6x+12

6(x+2)


Factoring By Grouping - A factoring problem with four terms often is a grouping problem. The method is to factor in two stages

e.g. 6x+2x + 4y+3y

8x+7y


Simple Trinomials - You look for two values that add to bx and multiply to c.

e.g. x^2+9x+20

(x+5)(x+4)


Complex Trinomials - A trinomial of form ax^2+bx+c can be factored by finding factors of the product ac that sum to b. *List the factors and insert them to r and s in factored form, multiply the inner and then outer terms and add them to find b.

e.g. 2x^2+11x+15

(2X+ ) (X+ )

5 3

3 5

(2x+5)(x+3)


Special Factoring

Perfect Square - If a and c are squared integers, the trinomial can be a perfect square
9a^2-30a+25

- (3a)^2

- 5^2

If both are multiplied by 2 in an expression they must add up to b

- 2(3a)(5) = 30a

*Use the first sign in the expression to what sign to use in your answer.

(3a-5)^2


Difference in Squares - The difference of two squares is a squared (multiplied by itself) number subtracted from another squared number.

e.g. (x+4)(x-4)

x^2-16

Changing from Factored Form to Standard Form

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Section 3: Standard Form Ax^2+Bx+C

Completing the Square - This is a process in which you convert Standard Form to Vertex Form.

The Steps In Order

1. Group the X terms

2. Divide the coefficient of the middle term by 2, square it, then add and subtract it inside the brackets.

3. Remove the subtracted term from the brackets.

4. Factor the brackets into a perfect square and simplify the outside.

Example

y= x^2-6x+4

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

y=(x^2-6x+9)-9+4

y=(x+3)^2-5

Quadratic Formula

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It is used when trying to determine the x-intercepts when in Standard Form (Solving)

Once all variables are substituted you would solve like this -

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Optimal Value

When trying to find the optimal value you would find the equation through this process.

y=4x+8y+24

y=4(0)+8y+24

y=8y+24

y=8+24

y=32

Reflection

All three of these quadratic equations are not separate units rather they are on big unit since all three are either connected or similar in some way. When learning quadratics you should always remember that you can change the equation to another in some form or way when trying to answer different problems