Quadratics & You
By: Suraj Sudan
Quadratics Are Fun!!
Quadratics are comprised of 3 main types of equations;
- Standard Form: [y=ax^2+bx+c]
- Vertex Form: [y=a(x-h)^2+k]
- Intercept Form: [y=(x-p)(x-q)]
What Is A Parabola?
Commonly Used Terminology
- Vertex - The minimum or maximum point in the parabola.
- Optimal Point (Minimum or Maximum Point) - The highest or lowest Y value in the parabola; The Y value of the vertex.
- Y-Intercept - Where the parabola passes through the Y axis.
- X-Intercept (Zeros/Roots) - Where the parabola passes through the X axis.
- Axis of Symmetry - The middle point of the parabola; The X value of the vertex.
- Vertical Stretch - When the A value exceeds 1, or is below -1, the parabola is stretched along the X axis.
- Vertical Compression - When the value of A is between 1 and -1, the parabola is squished closer together along the X axis.
Some Commonly Seen Parabolas Include:
Roller Coasters
Bananas
Landmarks and Structures
Parabolas Are Everywhere
Data Table Usage
In order to find out whether data forms a linear, or quadratic relation, a data chart must be used. If the differences in the first bunch are the same, or follow a certain pattern, the data forms a linear relation. If the differences are the same, or follow a pattern in the second bunch, then the data forms a quadratic relation. If the data is not the same, or it doesn't follow a pattern in the first or second bunch, then the data has no relation.
Formulas of Quadratics
Standard Form Equations
Standard Form Equations Into Vertex Form Equations
X = -B+/-√B^2-4(A)(C)/2(A)
As one follows this formula, they will get to the step of square rooting a number that is inside of the square root, that number is called the discriminant. One is able to find how many results will come up at the end of the quadratic formula by looking at the discriminant.
- If the discriminant is a positive number, then there will be 2 results.
- If the discriminant is 0, then there will be only 1 result.
- If the discriminant is a negative number, then there will be no results.
Factoring Standard Form Equations
- Common Factoring
- Simple Trinomial Factoring
- Complex Trinomial Factoring
- Factoring With Perfect Squares
- Factoring With Difference of Squares
Common Factoring
Example 1:
3x²+6x+12 GCF=3
The GCF is 3, because each term is dividable by 3.
=3(x²+2x+4)
Simple Trinomial Factoring
Complex Trinomial Factoring
Factoring Perfect Squares and Difference of Squares
(a+b)²=a²+2ab+b²
(a+b)(a+b)=a²+ab+b²
Difference of Squares:
(a+b)(a-b)=a²-b²
Perfect Square Example:
y=9x²+24x+16
(The square roots of 9x²=3x, and the square root of 16=4)
y=(3x+4)(3x+4)
(So the final answer would be)
y=(3x+4)²
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Difference of Squares:
y=9x²-16
(The square root of 9x² is 3x, and the square root of 16 is 4)
y=(3x+4)(3x-4)
(Now if you expand and simplify, you will get your final answer of)
y=9x²-16
Vertex Form Equations
Vertex Form Equation Into Standard Form Equation
Example 1:
y=-3(x-2)²+6
y=-3(x-2)(x+2)+6
y=(-3x-6)(x+2)+6
y=-3x²+6x+6x+12+6
y=-3x²+12x+18
Vertex Form Equation Into Factored Form Equation
Example:
Y=2(X-3)^2-4
0=2(X-3^2)-4
(You then carry the -4 over to the other side)
4=2(X-3)^2
(Then we must get rid of the a value, so you divide both sides by the a value)
4/2=2(X-3)^2/2
2=(X-3)^2
(Now to get rid of the ², you have to square roots both sides)
√2= √(X-3)^2
±√2=X-3
(Now you solve ±√2=x-3)
1.41=X-3 -1.41=X-3
1.41+3=X -1.41+3=X
4.41=X 1.59=X
X=4.41 X=1.59
So in factor form, the equation would be:
Y=2(X-4.41)(X-1.59)
Factored Form Equations
'(X-Intercept)+(X-Intercept)/2'
allows for the axis of symmetry to be found. Once the axis of symmetry is found, it can be placed in the original equation, substituting X, which would allow the equation to be solved. Once the equation is solved, the number given at the end is going to be the Y value of the vertex, and the axis of symmetry would be the X value of the vertex.
Factored Form Equations Into Standard Form Equations
Example:
y=(x+2)(x+3)
y=x²+3x+2x+6
y=x²+5x+6
So in Standard Form, the equation would be:
y=x²+5x+6
Factored Form Equations Into Vertex Form Equations
Example:
y=4(x+4)(x-9)
x+4=0 x-9=0
x=-4 x=9
(r+s/2=x)
-4+9/2=x
5/2=x
2.5=x
(Now plug in X value into equation and solve for Y)
y=4(2.5+4)(2.5-9)
y=4(6.5)(-6.5)
y=-169
Vertex = (2.5, -169)
(4 will remain as the A value)
(So now we put the variables we found into the Vertex Form Equation)
y=4(x+2.5)-169
Word Problems
Example 1:
P=-60(x-4)^2+120, where X is the price of a bad of popcorn.
A) What should the price be to maximize the daily profit? What is the maximum daily profit?
This question is indirectly asking for the vertex, which can be determined from the H and K value.
P=-60(X-4)²+120
-4+X=0
X=4
Y=120
Vertex= (4,120)
Therefore, in order to maximize the daily profit, the price should be 4 dollars. The maximum daily profit is $120.
B) Determine the price needed for the vendor to break even.
This question is asking about what the price would be if the daily profit was $0
P=0
X=?
0=-60(X-4)^2+120
-120/-60=-60(X-4)^2/-60
√2=√(X-4)²
±1.41=(x-4)
Now we have 2 solutions:
1.41=X-4
1.41+4=X
5.41=X
-1.41=X-4
-1.41+4=X
2.59=X
Therefore, the price of the popcorn should be $5.41 or $2.59 in order for the vendor to break even.
A flare is released into the air following the path, H=-5(T-6)^2+182, where H is the height in metres, and T is the time in seconds.
A) What was the flare's maximum height?
In order to find the maximum height, the time should be 0.
H=?
T=0
H=-5(0-6)^2+182
H=-5(-6)^2+182
H=-5(36)+182
H=-180+182
H=2
Therefore, the maximum height that the flare will reach is 2 metres.
B) What was the flare's initial height?
This question is asking about what the K value is in the equation; H=-5(T-6)^2+182.
The K value in this equation is 182.
This means that the initial height of the flare would be 182 metres.
A concert promoter models the profit from a concert, P dollars, by the equation;
P=-11(T-55)^2+10,571, where T dollars is the cost per ticket.
A) What will be the profit if the tickets were free?
P=?
T=0
P=-11(0-55)^2+10,571
P=-11(3,025)+10,571
P=-33,275+10,571
P=-22,704
Therefore, if the tickets were free, the profit made would be $-22,704
B) How much should be each ticket cost to break even?
T=?
P=0
0=-11(T-55)^2+10,571
-10,571/-11=-11(T-55)^2/-11
√961=√(T-55)^2
±31=(T-55)
Now we have 2 solutions:
31=T-55 -31=T-55
31+55=T-31+55=T
86=T
24=T
Therefore, the tickets should cost either $86 or $24 in order to break even.
Reflection Of The Quadratics Unit
Quadratics have also taught me that working hard, studying, determination, and perseverance will always pay off in the end. They are always worth it when the end results are good, and high marks.