Everything about Quadratics
By Makilthan Thuraisingam
How to tell if it is a Quadratic Function?
What are the forms of Quadratics?
- Axis of Symmetry
- Optimal value
- Transformations (Translation, Vertical, or Horizontal, Vertical Stretch, and Reflection)
- X-Intercepts of Zeros (r and s)
- Step Pattern
- Zeros or X-Intercepts (r and s)
- Axis of Symmetry (x=(r+s)/2)
- Optimal Value (sub in)
- Zeros (Quadratic Formula)
- Axis of Symmetry (-b/2a)
- Optimal Value (sub in)
- Completing the Square to turn it into Vertex Form
- Factoring to turn into factored form
- Common Factor
- Simple trinomial
- Complex Trinomial
- Perfect Squares
- Difference of Squares
The Vertex Form of a quadratic equation would like this:
Axis of Symmetry
To find the axis of symmetry in the vertex form the formula would be: y=(x-h)² where x=h.
For example, in y= -2(x+3)² -6, you would make it so x-h, and since x-3 (where x is 0) is -3, your axis of symmetry is -3. Simple!
Transformations (Translation vertical or horizontal, vertical stretch, reflection)
X-Intercepts or Zeros (Sub y=0 and Solve)
Zeros or X-Intercepts (r and s)
On a graph, the X-Intercepts would be here:
Axis of Symmetry (x=(r+s)/2)
On a graph, it would look like this:
A ball is thrown from child one to child two, from left to right, where child one is left. The formula for the ball being thrown in the formula of y= a(x-8)(x+1). Where is child one located? Where is child two located? What is the distance between them?
In order to solve this equation, we need to look at two things. First, we must make the brackets equal 0. x-8=0 and x+1=0.
Since we know that the ball is thrown from left to right and that child 1 is left, child 1 is located at (-1,0) and child 2 is left at (8,0). The distance between them is 9 units, because no unit of measurement is given.
A child throws a ball over a river. The ball is thrown in the formula of y= a(5+4)(5+3). The 'x' value is already given here, so what is the highest point the ball reaches?
Since the 'x' value is given, we just need to solve for 'y'.
The optimal value, or the highest point the ball is thrown, is at (5, 17).
Zeros (Quadratic Formula)
Now all you need to do is substitute it into the formula, and solve!
x= -(-2)+/- √2²-4(-2)(24)/2(2)
x+ 2+/- √4+192/4
x= 2+/- √196/4
There are two ways to go from here.
Those are your X-Intercepts for your parabola. You still need to find you axis of symmetry though, which is next.
Axis Of Symmetry (-b/2a)
Optimal Value (sub in)
Then y= 2(0.25)-1+24
23.5 is your maximum point in this scenario!
Completing the Square (Standard to Vertex Form)
Factoring to turn to Factored Form
Common factoring is when you divide the terms with one number, and isolating the 'a' value, and then turning it into factored form. We will look at how to common factor in a step by step method with an example.
The equation we are using first is y=8t²-24t-144. First, you isolate the 'a' and put the rest of the equation into brackets, so it would look like this: y= 8(t²-3t-18)
Next, you would have to find two numbers that add to -3 and when you multiply those two numbers, it must be -18.
After you find the factors, you realize that +3 and -6 work, because +3-6=-3, and +3*(-6)= -18.
Next, you remember that 4 can be divided by two, so you can split 4t² to 2t and 2t. You already have +3 and -6, so you just put it into factored form, and it looks like this:
y= 8(t+3)(t-6). You can check it by distributing the property.
Same equation as before! Remember that you can only common factor if there are common factors in the trinomial. If you don't have any common factors, you need to solve by the way you solve a simple trinomial, which we will look at next.
Simple trinomial's do not have any common factors in them. When you solve simple trinomials, you must first look for factors of the 'r' and 's' values. Next, you must do what you do when common factoring, and that is finding two numbers that add to the 'b' value, but those same numbers multiply to the 'c' value. Finally, you make sure that it is in factored form and that your 'x' value is squared. Here is a step by step example:
6 and 5 are the numbers we need to solve for. They cannot be common factored, so the best way to solve it is by doing 5+1 for the 'b' value and 5*1 for the 'c' value, and it works.
Therefore, the factored form for the equation is y= (x+1)(x+5).
Remember, this would not work for standard equation with an 'a' value, so we will look at how to solve those next.
When solving a complex trinomial, you do the same thing as you do for solving simple trinomials, but only for the x²+bx+c part. But the 'a' value is a little more difficult. A simple way to do it is by using a little chart. You make a little 2 by 3 chart. The first thing you must do on the side is find two numbers that multiply that equals the 'a' value. We will use the expression 6x²+11x+3 for this. You put the two numbers in the first column.
Here is a video of a word problem solved.
When solving perfect squares, you must have a values that can be square rooted. All you need to do is square root the 'x' value (or 'ax'), and square root the 'c' value. However, the square root of 'c', as in the two numbers, must add and equal to 'b' value. Here is an example:
Here is a video to show how to do special factoring, which in this case, would be difference of squares, as well as some more help with perfect squares.