## The Topics Covered.....

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 y’s with a quadratic equation, if the x’s are all spaced evenly apart, the second difference (the differences of the differences) of the y’s are equal. A linear equation has the first differences equal.

## Investigating Vertex Form

A Quadratic Equation can be also expressed as 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 factored form of the quadratic equation expresses the quadratic equation as a product of it's factors. The standard form of the quadratic equation is y = ax^2 +bx +c. The Factored form would be y = a(x - r)(x - s) where r and s are the x-intercepts or zeroes of the parabola. An x-intercept is the point where the parabola cuts the x-axis.

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

Converting Standard Form into Factored Form 5.5 gr 11 college 10 16 13

Quadratics can be factored using six methods. The methods are: -

1. 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.
2. 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.
3. 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.
4. 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.
5. 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).
6. Perfect squares: - This method is used when the quadratic equation can be expressed as a perfect square.

## Solving by Factoring

We can solve a quadratic equation by factoring it. The steps are: -

1. First we move all the terms of the quadratic equation to one side and leave nothing i.e zero on the other side.
2. Then, The quadratic equation is completely factored using a suitable method.
3. Now, we set each factor as zero and solve.
4. Each solution from step 3 is listed as a solution the original equation.
The next video will explain 'Solving by Factoring' in more detail.

## Completing the Squares

This is a very essential method as it helps us to go from standard form of a quadratic equation to it's vertex form. We can also derive the quadratic formula using this method. In this method, we literally "complete the square". The steps for completing the square are: -

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 is basically a formula that directly gives us the x-intercepts. This Formula can be used when factoring a quadratic equation would be hard, long or inconvenient.

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

## Graphing Factored Form

To graph the parabola of a Quadratic equation, we need basically three points, The Vertex and the x-intercepts or the y-intercept( in some cases). By Factoring a quadratic, we find the solution or zeroes of the quadratic equation. Let the solutions of a specific quadratic equation be

## Solving from Vertex Form

y = a(x - h)^2 +k is the vertex form of a quadratic equation. For finding the x-intercepts from vertex form, we set y = 0 and find the value/values of x. For finding the y-intercept, we set x=0 and solve for y. The vertex would be (h,k).

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

The vertex form of a quadratic equation is y = a(x-h)^2 + k. Each part of the vertex form has it's own significance.

• (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.
Quick Way of Graphing a Quadratic Function in Vertex Form

## Application Problems Based on Flight

The following video created by us will explain an application word problem based on Flight

## Application Problems Based on Area and Perimeter

The following videos created by us will explain an application word problem based on Area and Perimeter

## Application Problems Based on Revenue

Below is a video which demonstrates an application problem based on Revenue
Maximizing Revenue Word Problem (Completing the Square): Straightforward Worked Example!

## Reflection

I had the question above for a Quiz in our Quadratics unit.

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