Quadratic Relations
By: Shivam Sharma
Table of Content
Vertex Form:
- Axis of symmetry
- Optimal value
- Transformation
- X-Intercepts or zeroes
- Step pattern
- Zeroes or x-intercepts
- Axis of symmetry
- Optimal value
- Zeroes
- Axis of symmetry
- Optimal value
- Completing the square to turn to vertex form
- Factoring to turn to factored form (Common factoring, Simple Trinomial, Complex Trinomial, Perfect Square, Difference of Squares
Introduction to Quadratics
A quadratic relation will always have first differences. You can calculate the first differences by subtracting the Y value by the closest minor Y value.
Introduction to Parabolas
Vertex Form: y=a(x-h)²+k
I will be explaining how to find an equation when given a vertex.
Example 1:
Step 1: Vertex: (2,5). From this we can tell that the x-intercept is 2 and the y-intercept is 5. Now we can plug the two numbers into the Vertex Form. y=a(x-h)²+k is the Vertex Form, and when we plug in the x and y intercept, we need to understand that h is x and k is y. So we would write y=a(x-2)²+5 (When we put in the x-intercept in the h, we have to multiply the number by -1 so we flip the positive/negative). When we graph the equation, we graph the x-intercept starting from 0 and move left/right according to whether the x-intercept is positive or negative. In this case, we will move 2 places to the right. Next we will graph the y-intercept starting from 0 and move up/down according to whether the y-intercept is positive or negative. In this case, we will move 5 places up.
Step 2: Now that we have plugged in the x and y intercept, we now substitute (-2,0) in the equation.
Y=a(x-2)²+5
Y=a(-2-2)²+5
Y=a(-4)²+5
Y=16a+5
-5 = 16a
16 16
-5 = a
16
Now sub in the a value that we just found into the equation to complete the vertex form. Y=-5/16(x-2)²+5
Finding the x-intercepts
Example 1:
Step 1: Y=2(x-5)²+10. First off, we have to put the k value to the other side. 10=2(x-5)².
Step 2: Divide the equation by the a value, which in this case is 2.
10 = 2(x-5)²
2 2
5=(x-5)²
Step 3: To get rid of the ², square root the equation
5=(x-5)²
√5=√(x-5)²
2.24=x-5
Step 4: Bring -5 to the other side of the equation
2.24+5=x
7.24 = x
Step 5: To find the other x-intercept, take your Y and multiply it by -1, then do Step 4 again
-2.24+5=x
2.76=x
If the a value is 1/-1, >1 or -1<, then the parabola will be vertically stretched (The parabola will be skinny). If the a value is between > -1 and < 1, then the parabola will be vertically compressed (The parabola will be fat).
Factored Form: Y=a(x-r)(x-s)
Expanding is used when you want to change Factored Form equations into Standard Form. This video will thoroughly explain how to Expand.
Standard Form: Y= ax²+bx+c
Completing the Squares
Example 1: Y=2x²+10x+4
Step 1: Put brackets on the first 2 terms
Y=(2x²+10x)+4
Step 2: Factor the terms in the bracket
Y=2(x²+5x)+4
Step 3: Get your b value, divide it by 2 and then square it. Place it in between bx and c
Y=2(x²+5x+6.25-6.25)+4
Step 4: Put the -6.25 outside the bracket and multiply it with the a value
Y=2(x²+5x+6.25)+(2)(-6.25)+4
Step 5: Solve whatever you can
Y=2(x²+5x+6.25)-12.5+4
Y=2(x²+5x+6.25)-8.5
Step 6: Divide the b value by 2 and get rid of "5x+6.25"
Y=2(x²+2.5x)-8.5
Step 7: Factor the ² out of the bracket
Y=2(x+2.5)²-8.5
The Quadratic Formula is used to find an equation's x-intercepts when factoring it is not easy with other methods.
The Quadratic Formula is:
Step 1: Find the a, b, and c value
a = 2, b = 6, c = 4
Step 2: Plug in the a, b, and c values into the quadratic formula
x = (-b ± √b² -4ac) / 2a
x = [-6 ± √6² - 4(2)(4)] / 2(2)
Step 3: Solve the equation
x = [-6 ± √(6² - 4x2x4)] / 2(2)
x = [-6 ± √(36 - 32)] / 4
x = (-6± 4) / 4
x = (-6 + 4) / 4 x= (-6 - 4) / 4
x= -2 / 4 x= -10/4
x = -0.5 x= -2.5
Discriminant
The Discriminant of an equation quickly helps tell how many x-intercepts the parabola will have. The Discriminant is the b²-4ac part of the quadratic formula. When you plug in your values into the quadratic formula and solve it partially, you will find your discriminant. If b²-4ac = a positive number, then you will have 2 x-intercepts. If b²-4ac = 0, then you will have 1 x-intercept. Finally, if b²-4ac = a negative number, then you will not have any x-intercepts.