# Quadratics Unit 3

### Standard Form

## Unit Summary

In the standard form unit we learned how to change an equation from the various different forms into a standard form equation to find 3 points that can be used to plot a graph, also known as the 3 point method. We learned methods such as completing the squares, which is used to find the x's also known as the 0s of the parabola. we also learned how to find the maximum or minimum value of a quadratic relation

## Learning Goals

1)I am able to complete the squares to find the x's of a quadratic relation

2)I am able to solve quadratic equations using the quadratic formula

3)I am able to graph a parabola using the 3 point method

## Examples

## Quadratic form When using the quadratic formula you will find 2 different answers which are both of the x's or 0s of the parabola. In this method you sub in the values from the quadratic relation that you are given. | ## vertex form The easiest way to solve an equation written in this form is to complete the squares, to do this you need to add and subtract the square root of h times 2, after that yo take the negative value out of the bracket and subtract it from k. | ## Standard form When working with quadratic relations one of the most commonly used ways of displaying an equation is in standard form, which is: ax^2+bx+c=0 |

## Quadratic form

When using the quadratic formula you will find 2 different answers which are both of the x's or 0s of the parabola. In this method you sub in the values from the quadratic relation that you are given.

## vertex form

The easiest way to solve an equation written in this form is to complete the squares, to do this you need to add and subtract the square root of h times 2, after that yo take the negative value out of the bracket and subtract it from k.

## Standard form

## Completing the squares

## Completing the squares

## Word problem

A garden area is 30 ft long and 20 ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400 ft^2, what is the width of the path?

Answer:

length times width = 400 sq/ft

(30-2x) * (20-2x) = 400

**600 - 60x - 40x + 4x^2 = 4004x^2 - 100x + 600 = 4004x^2 - 100x + 600 - 400 = 04x^2 - 100x + 200 = 0; a quadratic equation**

**Simplify, divide by 4 and you have:x^2 - 25x + 50 = 0**

**Completing the square is not a very good way to solve this**

**If you us the quadratic formula: a=1 b=-25 c=50**

Two solutions:

x = 22.8, not a possible solution

x = 2.19 ft is the width of the path

## Self Reflection/connection

The unit of quadratics as a whole was overall an okay unit there where a few places where i faced challenges which i had to work through but other than that this unit was easy to understand. I would've liked to learn more more on the topic of standard form which was my favorite unit in quadratics. In the beginning of the quadratics unit i believed that it would be very challenging, but after learning all of the various ways of using/solving quadratic relations i learned that is wasn't all that hard to understand if you understand the basics of the unit. A connection i made between 2 of the units in quadratics is between graphing and factoring, I made this connection because going through both of the units i realized the two work hand in hand with each other. It is a lot easier for you to factor out/simplify an equation than it is to try to graph a quadratic relation that is not simplified.

Changing a Quadratic from Standard Form to Vertex Form