Quadratic Relationships Part 1

Sathun Shantharuban

Learning goals:

Summary of what we Learned in Quadratics

In this unit we have learnt all about the graphing techniques with little effort and limited information and to use vertex form, tables of contents. Also how to find linear and non-linear lines.


  • This unit you will learn Graphing using different methods step pattern, mapping formula, first and second differences, graphing using y=a(x-h)^2+k, and
  • We learn to determine how a graph is and whether it is linear or non-linear by using tables and first and second differences.
  • We learn how to use Mapping formulas from (X-H , AY-K) to graph parabolas.

Table of Contents:

1. Vertex Form:

2. How to determine an equation with a Graph.

3. What are parabolas

4. What are Quadratics

5. Finite Differences to find quadratic differences
6. Mapping notation

7. Word problems

8. Solving through step patterns.

9. Video!

What is Vertex Form?

Vertex form is the equation y=a(x-h)2+k. This equation is the most common and simplest way of solving an parabola with points given.

  • This formula can determine how the parabola is opening. The direction it opens either up or down. The way it opens is dependent on the (a) value.
  • The H value determines where on the graph it will be moving it vertically up or down times its value.
  • While the K moves the parabola horizontally left and right times its value.

How to Determine an Equation with A Graph

So how we will determine an equation can be simplified with a graph and a parabola with a few points given.

In a parabola there are very simple clues to look at that help you determine the equation.

Big image
  • In the Parabola given you have a Vertex which is -8 and -22
  • there are many points to be used as *X* and *Y*.
  • And using the plotted points and solve for a

Lets try it Ourselves!

First we find some plotted points that gives us a clue. The vertex is also known as a H and K value H being 8 and K as -22 so now we have the h and k. Now we look for the y and x value and it can be anywhere on the parabola so (-10,26) then sub these values in for y=a(x-h)^2+k.

What is an Parabola?

Parabolas are one form of an graphed quadratic equation. This does not go in a straight line like a linear equation. A quadratic makes a equal curve. The reason it bends is because it is an equation where it is doubled or squared. This causes it to bend and change its way because of it being squared.

What are Quadratics?

A Quadratic equation is a equal bending line up or down that creates a parabola. in the equation some of the variables are squared or multiplied by itself twice.

Finite Differences to find Quadratic Differences

Finite differences can find out if the graph is either linear or non-linear. If the line is straight that means it is linear and it cannot be a quadratic equation or function.


  • This graph is non-linear and it is show because the line has a flip and it turns and curves and has a curve of best fit. This means it is a parabola and it can be a quadratic function.
Mapping Notation


  • Mapping notation is when you separate (x-h, ay + k)
  • The mapping notations are shown through the charts and each chart being a value from the equation.
Big image
Lets make a word problem!

Joseph throws an baseball at an initial height of 6m. 6 seconds later it reached its maximum height of 25m and it later hits the ground at 14 seconds. Make an Vertex form solution.


A firework is shot at 2m off the ground. It reaches its highest point of 82 meters in the sky after 4 seconds. It comes down after 11 seconds after being shot. Make an vertex form equation to solve this.


A line passes through the vertex of (6,2) and passes through the coordinates of (2,6).

Answers to Questions:

1. Sub in values

0=a(11-6)^2+25

Solve for A

0=25a+25

Isolate so A is on its own

25=25a

1 =a

Use the final equation with all values

y=1(x-6)^2+25


2 .Sub in all coordinates given

0=a(14-2)^2+82

Solve for A

0=144a+82

Isolate A so its on one side

82=144a

82/144=a

Use all the found values for an equation

y=82/144(x-6)^2+25


3. Sub in the values given from question 3

6=a(2-6)^2+2

solve for A

6=a(-4)^2+2

6=16a+2

Isolate A on its own

4=16a

4/16|0.25=a

The final equation will be Y=0.25(x-6)^2+2

Solving through Step Pattern

This method is unique because it has its own simple way of plotting an graph. Each value apart of the equation plays a roll in the step pattern, step patterns go right by one then up one then right one up three, and one right five up etc.


Lets try it. For instance if you have an equation of y=4(x-8)^2 + 20 this can be graphed with just the three main points.


  • Take the two points of (h, k) 8 and 20 which is also the vertex and plot it for your own graph.
  • Use the A value as a multiplier for the step pattern. so for the amount the value goes up by can be multiplied by the A value.
  • now you have the method and you can plot it at anytime!

Quadratic Functions 1
Grade 11 University