All you need to know about parabolas!

By: Krishna Patel

Overview of the Unit

In this unit we learned how to graph different types of parabolas, using different types of equations. We learned how to convert from one equation to another, and were taught about the key features of a parabola. By the end of this unit we were able to solve problems that gave us equations and other where we had to make up the equations using the information given. We used all the skills we developed from Quadratics 1 and used them till the end of Quadratics 3.

Quadratics 1

· Intro to Quadratics

· Different parts of a parabola

· Vertex Form

· Finding equation and analyzing vertex form

· Factored form

Quadratics 2

· Multiplying Binomials

· Common Factors and Simple Grouping

· Factoring Trinomials with leading 1

· Factoring Trinomials without leading 1

· Factor Special Trinomials

· Solving by factoring

Quadratics 3

· Completing the square

· Solving from vertex form

· Quadratic Formula + Discrimnant

· More Quadratic and Discriminant

Quadratics 1

Introducing the Parabola

The graph of a quadratic relation is called a parabola
  • Parabolas can open up or down
  • The zero is also known as the x-intercepts or "roots"
  • The axis of symmetry divides the parabola into two equal halves
  • The vertex is the point where the axis of symmetry and the parabola meet displaying its maximum or minimum value
  • The optimal value is the value of the y co-ordinate
Zeros: also known as the x- intercepts are found on the x axis (vertical axis)

Vertex: Is always located half way between the zeroes, and it is where the parabola meets, there can either be a minimum of a parabola or a maximum.

Optimal Value: It is also known as the y coordinate and is the highest point of the parabola

Axis of Symmetry: The axis of symmetry goes through the x axis and is also known as the x value

This link below shows the different parts of the parabola, and also shows us how to find axis of symmetry, optimal value etc.

Vertex Form

Graphing Vertex Form

y= a (x-h)^2 + k

a= vertical compression or stretch

k= changes the y coordinate (up or down)

h= changes the x coordinate (left or right)

Always remember to describe a graph using SRT

Example: y= 2(x-3)^2+5

S: vertical stretch by a factor of 2

R; vertical reflection

T: Horizontal Translation to the right by 3 units (-3)

Vertical Translation up by 5 units (5)

Finite Differences

Finite Differences shows us whether a relation is linear, quadratic or neither.

If the first differences are constant it is linear, if the second differences are constant it is a quadratic relation.

1) Graphing using Step Pattern

Ex. y= 1/2(x-3)^2-5

Step 1: First plot y=x^2

Step 2: Plot the vertex of the equation (3, -5)

Step 2: 1,1 and 2,4 are the steps you use for step pattern

Step 2: Multiply the a value of the equation (1/2) by the y values of (1-1) and (2-4)

Step 3: Move accordingly

Instead of moving 1-1 and 2-4 from the vertex, you must move 1-1x1/2 and 2-4x1/2 which is: (1-0.5) and (2-2)

2) Graphing using Mapping Notation

  1. Write a mapping formula
  2. Complete a table of values for y= x^2
  3. Determine the transformed "key" points
  4. Sketch the base y= x^2
  5. Sketch the new graph
  6. State possible x and y values
Practice Questions (try these on your own):

Graph the following using mapping notation and step pattern:

1) y= -0.5 (x+4)^2 -3

2) h= -4 (t-6)^2 + 19

3) A parabola has an equation of y= 6 (x+3)^2 + 4

a) Graph this equation

b) State the direction of opening

c) Write the x intercepts

Finding Equations and Analyzing Vertex Form

Make an equation using the given vertex and points
Example: Find an equation for the parabola with the vertex (2, 6) and goes through the point (5, 3)

Step 1: y= a (x-h)^2 + k

Step 2 : y= a (x-2)^2 + 6

Step 3: 3= a (5-2)^2 + 6

Step 4: 3= a (3)^2 + 6

Step 5: 3= a (9) + 6

Step 6: 3= 9a+ 6

Step 7: 3-6 = 9a

Step 8: -3= 9a

Step 9: -3/9 = a

Step 9 (simplify the fraction): -1/3 = a

Therefore the equation is: -1/3 (x-2)^2 + 6

For more practice there are similar questions in the math workbook (chapter 4.4 page 42)

Below is another example of finding the equation of a parabola when given the vertex and a point

Factored Form

Find x-intercepts by setting y to 0 and the vertex is between the two x intercepts. For factored form the a is the stretch or compression, its a stretch if its a whole number and a compression if its a fraction.

Factored Form: y= a (x-r) (x-s)

Quadratics 2

Multiplying Binomials

There are many ways you can multiply binomials, by foiling, distributing, squaring binomials, and product of the sum and differences.

  • F: First
  • O: Outer
  • I: Inner
  • L: Last

Example: (x+1) (x+3)

  • F: (x) (x) = x^2
  • O: (x) (2)= 2x
  • I: (1) (x) = x
  • L: (1) (3) = 3

x^2 + 2x + x + 3

x^2 + 3x + 3


Example 1:

(r-8) (r+3)

(multiply the first r by the second bracket, and the -8 with the second bracket)

r^2 + 3r - 8r -24

r^2 - 5r - 24

Example 2:

5(x+4) (x+6)

(distribute 5 to the first bracket only)

(5x+ 20) (x+6)

5x^2 + 30x+ 20x + 120

5x^2+ 50x+ 120

Squaring Binomials:

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



= (x)^2 + 3(x) (4)+ (4)^2

= x^2 + 8x + 16 (the answer is called perfect square trinomial)

Product of Sum and Differences:

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


(x+5) (x-5)

= (x)^2 - (5)^2

= x^2- 25 (the answer is called differences of squares)

Below are written examples of all 4 types of multiplying binomials

Common Factoring and Simple Grouping

To common factor you must find a number that is common within the relation
Example: 2x^2 + 4x

2x^2------> 2. x . x

4x------> 2. 2. x

GCF: 2x

2x 2x^2+ 4x divided by 2

2x (x+2)

Factoring Simple and Complex Trinomials

Simple: Simple Trinomials are relations without a coefficient in front of the first value. We factor simple trinomials by using product and sum and decomposition

Product and Sum:

Example 1:

Factor x^2=7x+12

(Find a number that multiples to 12 and adds to give 7)

P= 12 (4 times 3)

S= 7 (4 plus 3)

Therefore: (x+3) (x+7)


Example 1:

Factor: x^2+7x+12


x^2+ 4x+ 3x+ 12

x (x+4) + 3 (x+3)

(x+4) (x+3)

Complex Trinomials


Example: 6x^2 + 11x + 3

The first 2 numbers should multiply to 6, and the other 2 should multipy to give 3

When you multiply across and add the numbers they should add to 11.

3x---------1 ------> (3 times 3 = 9)

2x ---------3 ------> (2 times 1 = 2)

9+2 = 11

(3x+ 1) (2x+ 3)


6x^2+ 11x+ 3

Sum= 11 Product= 6 x 3= 18 (9, 2)

6x^2+ 9x+2x+3

3 (2x+3) + 1 (2x+3)

(2x+3) (3x+1)

Factor Special Trinomials

Difference of Squares

  • a^2- b^2= (a+b) (a-b)
  • Both a and be should be perfect squares


81x^2 - 64y^2

(9x)^2 (8y)^2

= (9x+8y) (9x-8y)

Example 2:

144y^2 - 169

(12y)^2 (13)^2

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

Perfect Square Trinomials

  • The first and last terms are perfect squares, and the middle term is twice the product of the square roots
  • a^2- 2ab+b^2 = (a+b)^2

Example 1:

16m^2 + 24m+ 9

(4m)^2+ 2(4m)(3)+ (3)^2


Factoring Perfect Squares

This website helped me understand Perfect Square, I liked how they explained everything they did step by step and also gave reasons to why they're doing something in a specific way.

Solving by Factoring

Example 1:

(x+ 3) (x+4)= 0

x+ 3= 0 (x= -3)


x-4= 0 (x= -4)

therefore the x intercepts are x= -3 and x= -4

Example 2:

x^2+ 5x+6 = 0

P= 6 (3 times 2)

S= 5 (3 plus 2)

(x+3) (x+2) = 0

x+3= 0 (x= 3)


x+2= 0 (x= 2)

Therefore the x intercepts are x= 3 and x= 2

Example 3 (it gets harder!):

2x^2+ 14x+ 12= 0

(factor out the 2)

2 (x^2+ 7x+ 6)

P= 6 (6 times 1)

S= 7 (6 plus 1)

(x+6) (x+1)= 0

x+6= 0 (x= -6)


x+1= 0 (x= -1)

Therefore the x intercepts are x= -6 and x= -1

Quadratics 3

Completing the Square

moving from standard form (y=ax^2+bx+c) to vertex form (y=a(x-h)^2=k)

Simple Practice

Add in the missing amount of unit tiles to create a perfect square

1) x^2 + 2x + _____ (Answer= 2/2^2 = 1)

2) x^2 + 8x +1____ (Answer= 8/2^2= 16)

Below is a problem that I solved using Completing the Square

Quadratic Formula and Discriminant

This formula is used when asked to solve using a standard form equation
Before using the quadratic formula, it is important that you first find the discriminant using (b-4ac). This will help you to know whether you can use the quadratic formula or not.

If the answer is greater than 0 there will be two solutions (roots)

If the answer is less than 0 there will be no solution (roots)

If the answer is 0 there will be 1 solution (root)

Quadratic Formula

The quadratic formula can be used if you have a standard form equation and you are trying to solve (find x)


In Optimization questions you are to construct your own equations based on the information given, and using what you learned in quadratics so far to answer the questions that follow
Below is a video of a Quadratic Problem that I have created

Sorry if the video is a little blurry



I believe i did rather well on this test taking into consideration that it was the first one. I believe a couple of my strengths during quadratics especially something that can be witnessed through this test was being able to find an equation and analyze vertex form. I believe this was simple because it was basically a procedure that you had to remember, it was not like Quadratics 3 where we had to figure out what we would use to solve the problem. I believe I had difficulty with the word problem in this test, but I know I have improved because now i understand how to use what we've learned and I have learned when to use what in more depth. Something that I would work on within this whole unit is trying to understand the purpose for everything we're doing, so for example the purpose of changing an equation to standard, vertex or factored form to get a better understanding of what I am doing. Overall, this unit was not as hard as I had expected it to be and doing extra practice will make this unit very easy for me.
Below are some useful YouTube links that are not mine but I have found them helpful throughout this unit
8.5 Factoring Special Products - Algebra 1
3.5 Graphing from Factored Form