# Graphing Vertex Form

### Parabolas, Transformations, and Vertex Form Equations

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

**Linear and Quadratic Relation**

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

A linear equation is an equation of a line.

**y=mx+b**

**y=2x+1**).

A quadratic equation is an equation of a parabola.

**y=x^2**

**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.

**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.**

If variable a is *negative*, the parabola **opens down.**

You can figure out if it is **stretched** or **compressed**.

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

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

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

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

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

**k<0**, the

**vertex moves down**by k units.

**h>0**, the

**vertex moves to the right**h units.

**h<0**, the

**vertex moves to the left**h units.

**Vertex Form Quadratic Equation**

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

## Graph

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

## Word Problem

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.

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.