Quadratics:
Unit 1,Unit 2, Unit 3
Vertex Form
Learning Goals For Unit
2: I am able to determine if a relation is Linear, Quadratic, or neither using finite differences.
3. How the values of a,h, and k affect the shape of the parabola.
Summary of Unit
Finite differences:
First we learned about using finite differences to find out if an relation is linear, quadratic, or neither.How?
Linear
Quadratic
If the relation's second differences are all the same number, the relation is quadratic.
Neither
Parabola
Mapping Notation
Remember to substitute the variables into the mapping notation expression.
(x+h), (ay-k)
Also you must change h if it is a positive it must become negative and if it is negative it must become positive.
For example:
y=2(x-3)^2-4
In order to find the points of the parabola, you must substitute the variables into the expression with each of the corresponding numbers in the vertex form equation.
(x+h)=(x+3), (ay+k)=(2y-4)
Using this Information We Can Find The Coordinates
Transformations
Looking at y= a(x-h)^2+k, we can analyze a few things.
a:
If the variable 'a' is a negative the parabola is opening downwards, however if the variable "a' is positive, the parabola is opening upwards.
- If a is -1 < a < 1, then the graph is compressed
- if a is -1 > a > 1, then the graph is stretched
h:
- If the value of h is negative the x-coordinate shifts right
- If the value of h is positive the x-coordinate shifts left
k:
- If K is negative, then the graph of y=x^2 vertically translates the graph down.
- If K is positive, then the graph of y=x^2 vertically translates the graph up
Graph
y=2(x-1)^2+4
Word Problem
Now we solve.
y=-4.9(1-2)^2+45
y=-4.9(-1)^2+45
y=-4.9(1)+45
y=-4.9+45
y=40.1
Therefore, the height of the ball after 1 second is 40.1 meters
Now we solve.
y=-4.9(0-2)^2+45
y=-4.9(-2)^2+45
y=-4.9(4)+45
y=-19.6+45
y=25.4
Therefore, the initial height of the ball is 25.4 meters
Now we solve.
0=-4.9(x-2)^2+45
-45=-4.9(x-2)^2
-45/-4.9= -4.9(x-2)^2/-4.9 We divide by -4.9 to get rid of it
√9.18= √(x-2)^2 We square root it to get rid of the exponent
3.03=x-2
3.03+2=x
5.03= x
Therefore, the ball was in the air for 5.03 seconds
Factored Form
Learning Goals For Unit
1. I will learn how multiply polynomials to get a simplified factor formed equation.
2. I will learn how to factor simple and complex trinominals.
3. I will learn how to factor by grouping.
Summary of Unit
Understanding the Factored Form Equation
The value of a is gives the shape and direction of opening for the parabola.
The value of x and s gives you the x-intercepts
Section 5.1: Multiplying polynomials
2. Simplify by collecting like terms
y=a(x-r)(x-s)
For Example:
(X+2)(X+3)
=2x^2+3x+4x+6
=2x^2+7x+6
Section 5.2: Special products Squaring Binomials
2. Twice the product of the terms/ add two middle terms after expansion.
3. Square the last term.
(a + b)^2
= a^2 + 2ab + b^2
or
(a - b)^2
= a^2 - 2ab + b^2
For Example:
(x+4)^2
=x^2+8x+16
Product of a Sum and a Difference
(a - b)(a + b)
= a ^2– b^2
For Example:
(x-4)(x+4)
=x^2+16
Section 3: Common Factoring
1. Find GCF of coefficients and variables.
2. Divide each term by GCF.
Example:
5c-10d
5c/5-10d/5
5(c+2d)
Binomial Common Factor:
1. If there are two binomials that are exactly same, consider that as binomial common factor.
Example:
x(x-2) +2(x-2)
(x+2)(x-2)
Factor by grouping:
1. Factor groups of two terms with a common factor to produce a binomial common factor.
Example:
9x^2 + 15x + 3x + 5
3x
3x(3x+5)1(3x+5)
(3x+1)(3x+5)
Section 4: Simple Trinomial Factoring
A simple trinomial is a quadratic where a = 1
1. Find the Product and Sum
- Two numbers that multiply to get C but add to get b.
x^2+10x+25
5+5=10
5x5=25
Therefore the product and sum in 5 and 5.
Answer:
(x+5)(x+5)
2.Look at the signs of b and c to determine the factored form.
x^2 + bx + c
(x-r)(x-s)
- If b and c are positive, both r and s are positive.
For Example: x^2+8x+16 = (x+4)(x+4)
- If b is negative and c is positive, both r and s are negative.
For Example: x^2-16x+64 = (x-8)(x-8)
- If c is negative, one of r or s is negative.
For Example: x^2+10x-24 = (x+12)(x-2)
- If b and c both are negative, one of r or s is negative.
For Example: x^2-5x-23 = (x-8)(x+3)
Section 5: Complex Trinomial Factoring
1. Multiply a by c.
2. Now decompose b so that it adds up to the value of b but multiplies to the product of ac.
3. Now factor by grouping to get answer.
Example: 3x^2+14x+8
3x8=24
3x^2+12x+2x+8
3x^2+12x = 3x(x+2)
3x
2x+8 =2(x+2)
2
3x(x+2)2(x+2)
(3x+2)(x+2)
How to Solve Complex Trinomial Factoring questions By Khan Academy on Youtube
Section 6: Differences of Squares
How do you tell if differences of squares: Two terms are being subtracted and both of them are perfect squares.
Consider the Following: x^2-49
1. Square root both values
a=(x)^2 b=7
2.Sub a and b into a factor form. y=a(a-b)(a+b)
y=(x-7)(x+7)
When that becomes expanded it becomes x^2-49.
Section 6 Cont: Perfect Squares
Consider the Following:
Verify that x^2+6x+9 is a perfect square trinominal.
1. Square root a and c
(x)^2+6x+(3)^2
2. Check the middle term. It should be twice the products of the square roots of the first and last terms.
2(x)(3)= 12x
Now we know that x^2+6x+9 is a perfect square trinominal.
Next, the square rooted a and c go into the factor form equation to solve,
Therefore the answer is (x+3)^2
Section 6.2 and 6.3: Solving Quadratic Equations in Factored Form
With a factored form equation, make y equal to 0.
Consider the following:
y=(x-4)(x-2)
0=(x-4)(x-2)
The factored terms inside the bracket should equal to 0.
x-4=0
x-2=0
Solve for x.
x-4=0
x=4
x-2=0
x=2.
Therefore the x-intercepts of y=(x-4)(x-2) are (4,0),(2,0)
Finding the Vertex of a Factor Form Equation
Consider the Following:
y=x^2-16x+60
y=(x-10)(x-5)
1. Find the x-intercepts
0=(x-10)(x-6)
0=x-10
10=x
0=x-6
6=x
Therefore the x intercepts are (5,0),(9,0).
2. Find the Axis of Symmetry
x1+x2 = AOS
2
6+10 = AOS
2
16 = AOS
2
8 = AOS
3. Now sub AOS into x
y=(8)^2-16(8)+60
y=64-128+60
y=-4
Therefore the Y-int is (0,-4)
The vertex is (AOS,Y-int)
Therefore the Vertex is (8,-4)
Word Problem
a) Write the formula in factored form
h=-2t^2+4t+48
-2
h=t^2-2t-24
(t+4)(t-6)
b) When will the ball hit the ground?
0=(t+4)(t-6)
0=t+4 0=t-6
-4=t 6=t
Therefore the x-int are (-4,0),(6,0)
Therefore the ball will hit the ground at (6,0)
c) What is the maximum height the ball will reach in meters?
6-4 = AOS
2
2 = AOS
2
1=AOS
h=-2(1)^2+4(1)+48
=-2+4+48
=50
Therefore the maximum height is 50m
Standard Form
Learning Goals:
2.I am able to complete the square.
3.I can solve revenue, rectangle and triangle problems.
Summary
Completing the Square:
(a+b)^2 = a^2+2ab-b^2
Example:
(x+3)^2 = x^2+3x+9
How to make it a perfect square, if it's not given:
Consider the following:y=x^2+ 8x+5
1. Put '5' outside of the bracket
y=(x^2+ 8x)+5
2. Divide 'b' by 2 then square it. In this case it's 8.
(8/2)^2 = 16
Then the equation becomes y=(x^2+8x+16-16)+5
3.Move the negative 16 outside the bracket.
y=(x^2+8x+16)-16+5
4.Write as squaring a binomial
y=(x+4)^2-11
Therefore the vertex is (-4,-11)
The Quadratic Formula:
Consider the Following:
All quadratic equations of the form ax^2+bx+c=0 can be solved using the quadratic equation.
This is the quadratic equation:
The values for the variables a,b, and c are taken from standard form quadratic equation.
For Example:
Use the quadratic formula to solve for the X-Intercepts
3x^2+5x+2=0
We know that a=3, b=5, and c=2
1. Substitute in the values.
2. Add the vaules together in the square root.
3. Square root the result.
4. Add and subtract the outside number with the product of the square root.
5. Then divide the number you get from subtracting and adding with 2a.
Here is a video by mahalodotcom on Youtube on how to solve the quadratic formula.
The Discriminant:
The equation for the discriminant is:
b^2-4ac
If D>0 there are two x-ints
If D<0 there are no x-ints
If D=0 there is one x-int
Word Problem: Completing the square
R=(price)(quantity)
let 'x' represent # of decreases
p=(12-0.5x)
q=(36+2x)
R=(12-0.5x)(36+2x)
=432+24x-18x-x^2
=-x^2+6x+432
-1
=-(x^2-6x)+432 (6/2)^2=9
=-(x^2-6x+9-9)+432
=-(x^2-6x+9)+432-9(-1)
=-(x^2-6x+9)+432+9
=-(x-3)^2+441
x=3
p=(12-0.5(3))
p=12-1.5
p=10.5
Therefore, Ms. Dhaliwal should change her price to $10.5 because she can maximize her profit to $441