# Quadratics Made Easy

### The Ultimate solution to grade 10 quadratics

## The Topics Covered.....

## Introduction to Quadratics

One special type of polynomial equation that you’ll work with a lot is a Quadratic equation.

The word "Quadratic" comes from the Latin word meaning "to make a square."

A quadratic is a polynomial that has an x(squared) in it; it’s as simple as that. The degree of quadratic polynomials is two, since the highest power (exponent) of x is two. The quadratic curve is called a parabola. Technically, the parabola is the actual picture of the graph (shaped like a “U”), and the quadratic is the equation that represent the points on the parabola. But a lot of times we hear the words “quadratic” and “parabola” used interchangeably.

The main thing we have to remember about quadratics is that they either go up or back down, or they go down and then back up. (Actually, they can go sideways, too, but that discussion is in a more advanced section). So they either have a maximum point, or a minimum point; whatever this point is, it’s called the vertex. The Parabola of a quadratic equation is always symmetric. A straight line drawn through the vertex of the parabola is called the Axis of symmetry. The y-coordinate of the vertex is called the Optimal Value.

The picture shows the different forms of a quadratic equation.

## Using Tables of Values & First differences

We can create a Table of Values for Quadratic Equations. To learn how to create a table of values for a quadratic equation, kindly visit the link included.

It is interesting to note that if you were to create a chart with the ** x**’s and

*’s with a quadratic equation, if the*

**y****’s are all spaced evenly apart, the**

*x***second difference**(the differences of the differences) of the

**’s are equal. A linear equation has the first differences equal.**

*y*## Investigating Vertex Form

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

*y - k = a(x - h)^2.*

This is called the vertex form of a quadratic equation. It is called as the vertex form it is quite simple to find the vertex of the quadratic parabola from this form.

In the simplest words, the coordinates of the vertex of the parabola would be (h,k)

Vertex is the point in the parabola where the y-coordiante or the ordiante is at its maximum possible value. Then we have to adjust the value of x in the vertex form such that it yields the maximum possible value for y. The value of y is at it's maximum when x=h. The bracket would then evaluate to zero and the value of y will become it's maximum i.e k.

• (h, k) is the vertex of the parabola, and x = h is the axis of symmetry.

• the h represents a horizontal shift (how far left, or right, the graph has shifted from x = 0).

• the k represents a vertical shift (how far up, or down, the graph has shifted from y = 0).

If a > 0, then the parabola opens upward. If a < 0, then the parabola opens downward.

## The Factored Form

The video below will elaborate on converting from standard form to vertex form.

## Factoring Quadratics

__Common Factoring: -__To common factor a quadratic equation, we find of the HCF or the highest common factor of all the terms in the quadratic expression. The HCF is taken out and placed in-front of the brackets. The expression inside the brackets is obtained by dividing each term by the HCF.__Factoring by grouping: -__In this kind of method, we factor the quadratic equation quite easily by common factoring in groups and then express it as the product of two factors.__Simple Trinomial Factoring:-__This method is used when the quadratic expression is in standard form i.e y = ax^2 + bx + c and a = 1 only.__Complex Trinomial Factoring: -__This method is used when the quadratic equation is in standard form i.e y = ax^2 + bx + c and a > 1.__Difference of squares: -__This method of factoring is used when the quadratic equation is expressed literally as the difference of two perfect terms. For example- 16x^2 - 81. (both 16x^2 and 81 are perfect squares here).__Perfect squares: -__This method is used when the quadratic equation can be expressed as a perfect square.

## Solving by Factoring

- First we move all the terms of the quadratic equation to one side and leave nothing i.e zero on the other side.
- Then, The quadratic equation is completely factored using a suitable method.
- Now, we set each factor as zero and solve.
- Each solution from step 3 is listed as a solution the original equation.

## Completing the Squares

__Step 1__ Divide all terms by a (the coefficient of x^2).

__Step 2__ Move the number term (c/a) to the right side of the equation, leaving all other terms on the left side.

__Step 3__ Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation.

__Step 4 __ Square root both sides of the equation.

__Step 5__ Subtract the number that remains on the left side of the equation to find x.

Here is a link that will be helpful to understand the "Completing the Squares" more: -https://www.mathsisfun.com/algebra/completing-square.html

## The Quadratic Formula

The following video created by us shows more about the quadratic formula!

## Graphing Factored Form

## Solving from Vertex Form

To solve for x from vertex form, we set y = 0. Then, we have to isolate x on one side and all the constants on the other so the equation looks like -k/a = (x-h)^2. In the next step, we square root both sides then solve for x using a suitable factoring method.

## Graphing Vertex Form

- (h,k) is the vertex of the parabola
- k is the optimal value i.e y-value of the vertex of the parabola. It also tells us the value of the vertical shift
- h is the axis of symmetry i.e the x-value of the vertex of the parabola. It also tells us the value of the horizontal shift.
- a depicts the vertical stretch or compression of the parabola. It effects the step pattern. If a is positive, the parabola opens up. If a is negative, the parabola opens down.

The videos below will elaborate more on Graphing vertex form.

## Application Problems Based on Flight

## Application Problems Based on Area and Perimeter

## Application Problems Based on Revenue

## Reflection

The pictures also illustrate the process by which I tried to find an answer.

I assigned the length of the fence, the variable of "l" and "w" as a variable for the width. Then I used to given perimeter (total fence value) to express "l" in terms of "w" i.e l = 130 - 2w. Then i plugged the new value of l into the equation for area. I got y= 130w - 2w^2 (y is the variable for the area). Then I used trial and error method by putting in different values for w to get different values for l and the area i.e y. I roughly incremented w each time by 10. I got a value after which the area started decreasing. So, I took those values as the answer.

I learned the correct way of solving the problem. Instead of using trial and error method, I should have realized that whenever the question mentions "maximum", that i have to find the vertex. That would have given me the precise value for the maximum area and the corresponding length and width.

The pictures below illustrate my refined process and what I learned on how to solve it correctly