Quadratic Relations

By: Bilal Baig

Table of contents



Types of Forms

Vertex Form

Factored Form

Standard Form


Vertex form y= a(x-h)² + k
Axis of symmetry
Optimal Value
Transformations
X-intercepts or zeroes
Step pattern

Graphing parabolas


Factored Form y= a (x-r)(x-s)
Zeroes or x intercepts
Axis of symmetry
Optimal value


Standard form y= ax²+ bx + c

Zeroes
Axis of symmetry
Optimal Value
Completing the square to turn to vertex form

Common factoring
Factoring to turn to factored form

Factoring to simple trinomial
Factoring to complex trinomial
Perfect squares
Difference of Squares


Word Problems


Reflection

Types of forms

  1. Vertex form - y = a(x-h)²+k
  2. Factored form - a(x-r)(x-s)
  3. Standard form - y = ax² + bx + c

Vertex form

Axis of symmetry

The axis of symmetry is the center point of the parabola on the x axis. In vertex form, the axis of symmetry is the opposite symbol of the h value. For example in the equation y= (x+5)² - 3 the axis of symmetry would be -5 because h is a negative so when you put two negatives together they become a positive.

Optimal Value

The optimal value is the maximum or minimum point of the parabola on the y axis. In vertex form the optimal value is K value. In the equation y= (x-5)² - 3 the optimal value would be -3 and minimum.

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Transformations

A transformation is how each letter causes the parabola to move.


The (-h) moves the parabola left or right. The parabola moves left if (-h) is a positive like (+5) and it moves right if it is a negative like (-5).


The (K) moves the parabola up or down. The parabola moves up if K is a positive and it moves down if K is a negative.


The (A) stretches the parabola. If the A is negative the parabola opens down, if A is a positive it opens up

Zeroes

Zeroes is the x-intercepts of a parabola. To find the x-intercepts you must set y=0 and solve.


For example:

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Step Pattern

The step pattern for a parabola is how much the parabola goes to the left/right and up/down. If your parabola has an (a) of 1, the three points on the parabola will be over: 1 up 1, over 2 up 4, over 3 up 9. If your (a) is a number higher or lower than 1 then the three points on the parabola will be: over 1 up 1x(a), over 2 up 4x(a) and over 3 up 9x(a).
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Graphing parabolas

3.2 Graphing from Vertex Form

Factored Form

Zeroes (X-intercepts)

The zeroes are the x intercepts of a parabola. To find the zeroes you must set y=0 and solve each pair of brackets separately. For example:
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Axis of Symmetry

To find the Axis of Symmetry you must solve for x by setting y=0 and solve each bracket separately then put the x intercepts together and divide by 2.

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Optimal Value

To find the optimal value you must solve for y by setting the Xs to the x-intercepts.
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Standard Form

Zeroes

To find the x-intercepts in a standard form, you must first figure out the a, b and c. Once you figure out the a, b and c you put it into the equation x = (-b ±√(b² - 4ac)) / 2a. In the example below we'll use the equation y = 6x² + 10x + 3.
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Axis of Symmetry

To find the axis of symmetry, you must solve the equation x = -b / 2a
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Completing the square

Completing the square is turning standard form into vertex form. In the example below we'll use the equation y = 5x² + 20x + 15. Here are the steps to completing the square.


  • First you must put ax + bx in brackets, then common factor the numbers and place the factor in front of the brackets.
  • After that you do (b/2)² and add it to bx, but since you can't just add to an equation you also have to subtract it too.
  • Next you'll notice in the bracket you have a perfect square (x² + 4x + 4) so now you square root the x² and 4 leaving you with x and 2. Then you put the x and 2 together in brackets and square it.
  • After that you multiply the - 5 with the - 4.
  • Then add the + 20 to +15 and you have your K value which is also the optimal value.

Optimal Value

To find the optimal value you need to complete the square by turning it into Vertex form so you can read the example above and look at the picture below to see how to find the optimal value of the equation y= 5x² + 20x + 15.
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Common factoring

You can only use common factoring when there is a number that you can divide out without getting a decimal. Once you find the common factor, you put the factored numbers into brackets and the factor in front of the brackets.
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Factoring to Factored Form

You can change the standard form into factored form by finding what adds up to bx and also multiplies to c then you place them in the equation y = (x - r)( x - s)

Factoring Simple Trinomials

Factoring simple trinomials is the same as factoring to factored form. Below there is an example of factoring simple trinomials.
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3.8 Factoring Simple Trinomials

Factoring complex Trinomials

Factoring complex trinomials is similar to factoring simple trinomials but the number before x in x² is anything other than 1.
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3.9 Complex Trinomial Factoring

Perfect Squares

To identify a perfect square, you must check if the ax² and c can be squared, if they can then you take the x and c and do 2 * x * c and if it is equal to bx then you have a perfect square.
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Difference of Squares

When doing difference of squares, you need something squared minus something else squared.

Word Problem

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Motion problem

A ball is thrown and follows the path h = 5t² + 9t + 7 when h = height in meters and t = time in seconds.


1.a) What is the max height? When does it happen?


1.b) When will the ball hit the ground?


1.c) Sketch a graph.

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3.12 Motion problems

Reflections

In the beginning of this unit, I found transformations to be a really easy concept. I was able move around the vertex very easily. As we got further into the unit, we started factoring and it wasn't too hard but it was quite a bit harder than transformations. After that we started graphing parabolas and finding the vertex, zeroes, axis of symmetry and optimal value through a formula which was much harder than the previous things we did. At first I really struggled due to lack of preparations and practice but once I practiced I found it much easier to do and solve than before. Next we did quadratic formulas, it was a bit confusing at first but once I tried it out, I found it much easier to do. Lastly we used everything we learned in word problems, after our teacher gave us an example and showed it to us, it seemed really easy to solve because we were just using everything we already learned and applying it to the problems given to us. Overall quadratics isn't too hard as long as you keep up with your work and practice or you'll fall behind.
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