# Graphing Vertex Form

## Learning Goals

Here are some of the learning goals for graphing using the vertex form quadratic equation.

1) You will learn how to find the points of a Parabola with the vertex form quadratic equation. y=a(x-h)^2+k

2) You will learn how to identify the transformations in a parabola only using the vertex form quadratic equation.

3) You will learn how to create a parabola using the mapping notation. (x+h, ay+k)

## Summary

First and Second Differences

First and second differences help determine if a line is linear or quadratic or neither.

If the first difference is constant, the line has a linear relation.

If the first difference is not constant this means that this is not a linear relation. This is when you check the second differences.
If the second difference is constant that means this is a quadratic relation.

If the second difference is not constant, than it is neither linear nor quadratic.

What is a linear equation, and what is a quadratic equation?

A linear equation is an equation of a line.

y=mx+b

(The line shows the linear equation of y=2x+1).

A quadratic equation is an equation of a parabola.

y=x^2

(This parabola shows the quadratic equation of y=x^2).

The difference is that a quadratic equation has at least one variable squared. A linear equation has no variables squared.

Parabolas

A parabola consists of multiple parts. These parts are the axis of symmetry, vertex, y-intercept, x-intercepts (up to two), and direction of opening.

Mapping Notation

Mapping notation is a method used to find all the points of a parabola on a graph. It allows you to use the vertex form quadratic equation to find the corresponding y values to x values in the parabola.

A major thing to remember is that you must substitute the variables in the mapping notation expression.

(x+h, ay+k)

For example, the parabola shown below follows the quadratic equation: y=2(x-6)^2+3.

In order to find the points of the parabola, you must first substitute the variables in the expression with the corresponding numbers from the vertex form equation.

(x+h, ay+k) = (x+6, 2y+3)

First, create a table of values and solve for y using the equation y=x^2.

Next, create another table of values, this time the x value is x+h. The x value that is substituted into x+6 (x+h) is the value from the first table of values.
Now, on the same table of values, the y value is 2y+3. The y value that is substituted into 2y+3 (ay+k) is the y value from the first table of values.
Transformations

The transformations in a parabola can be found in the vertex form quadratic equation.

You can find the reflection of the parabola, if it opens up or down.

If variable a is positive, the parabola opens up.

(Equation is y=ax^2)

If variable a is negative, the parabola opens down.

(Equation is y=-ax^2)

You can figure out if it is stretched or compressed.

If -1<a<1, the parabola is vertically compressed.

(Equation is y=-0.2x^2)

If a>1 or a<-1, the parabola is vertically stretched.

(Equation is y=2x^2)

You can identify the translations of the vertex from (0,0)

If k>0, the vertex moves up by k units.

If k<0, the vertex moves down by k units.
if h>0, the vertex moves to the right h units.
If h<0, the vertex moves to the left h units.

The Vertex Form Quadratic Equation is: y=a(x-h)^2+k

## Graph

The vertex form quadratic equation for the parabola shown below is:

y=2(x-6)^2+3

## Word Problem

The path a firework takes during a festival is modeled by the relation y=-5(x-3)^2+50. What is the maximum height of the firework? What is the horizontal distance from the starting when it reaches maximum height? How far off the ground is the firework fired?

First you can see from the relation that the maximum height (k) is 50m.

It is also shown that from the equation that the horizontal distance at which the firework reaches maximum height (h) is 3m.

Therefore the maximum height of the firework is 50m, and the horizontal distance at which it reaches 50m is 3m.

Now in order to figure out the height of the firework at which it was initially fired, you need to find the y-intercept of the parabola of the relation.

In order to do so, you must substitute the x value in the relation with a 0.

y=-5(x-3)^2+50

y=-5(0-3)^2+50

y=-5(-3)^2+50

y=-5(9)+50

y=-45+50

y=5

Therefore the height at which the firework was initially fired is 5m off the ground.

This is what the parabola would look like when plotted down.

As you can see the the vertex is (3,50)

3 being the horizontal distance at which the maximum height of 50 was reached.

You can also see that the y-intercept is (0,5)

0 being the starting point of the fireworks path and 5 being the height that the firework was initially fired at.

## Video - How to graph a parabola using the Vertex Form Quadratic Equation

Graphing a Parabola - Vertex Form