Quadratic Relations
By: Ishani Sharma
Table of Contents
Quadratics In Vertex Form
- Introduction to Parabolas
- Key Features of Quadratic Relations
- Types of Equations: Vertex form
- Types of Equations: Standard form
- Types of Equations: Factored form
- Ways to Represent a Quadratic Relation
4.2 Quadratic Functions
- Investigating Vertex Form
- Graphing in Vertex Form
- Finding an Equation in Vertex Form Using a Graph
4.3 Transformations of Quadratics in Vertex Form
- Transformations of Parabolas
- Writing Equations from Given Transformations
- Graphing Transformations of Quadratic Relations
- Mapping Notation
4.4 Graphing Quadratics in Vertex Form
- Graph using Step Pattern
- Graph using Mapping Notation
- Word problems for quadratics in vertex form
- Finding x and y intercepts in vertex form
Mini Test #1
Quadratics in Factored Form
5.1 Multiplying Binomials
5.2 Special Products
5.3 Common Factoring
5.4 Factoring Simple Trinomials
5.5 Factoring Complex Trinomials
6.2 Solving Quadratics by Factoring (finding the zeros)
6.3 Graphing Quadratics in Factored Form
- Graph using x intercepts
- Finding the Vertex
- Word Problems for Quadratics in Factored Form
Mini Test #2
Quadratics in Standard Form
6.1 Maximum and Minimum Values (Completing the Square)
- How to go from Standard Form to Vertex Form
6.4 The Quadratic Formula
- Finding the x intercepts when factoring is not possible
6.5 Word Problems in Standard Form
Mini Test #3
Reflection
Useful Links
- Khan Academy
- Math Homework Help
4.1 Intro into Quadratics
Introduction to Parabolas
About Parabolas....
- Parabolas can open up/down
- The zero is when the parabola crosses the x-axis
- Zeros can also be called x-intercepts or roots
- The axis of symmetry divides the parabola in half equally
- The vertex of a parabola is where the parabola and axis of symmetry meet
- The vertex is where the parabola is at its maximum or minimum value
- The optimal value is the value of the y co-ordinate of the vertex
- The y-intercept is where the graph crosses the y-axis
Key Features of Quadratic Relations
Ways to Represent a Quadratic Relation
1) Table of Values
2) Graphs
Types of Equations
4.2 Quadratic Functions
Investigating Vertex Form
Graphing in Vertex Form
Finding an Equation in Vertex Form Using a Graph
4.3 Quadratic Functions
Transformations of Quadratics in Vertex Form
y = 3(x-1)^2+2
y = a(x-h)^2+k
Translations:
The parabola will open up , it is vertically compressed by a factor of 3. It is horizontally translated to the right by 1 unit and vertically translated up by 2 units.
Vertex Form:
- The "a" determines weather it is vertically compressed or stretched (narrow or wide). If the "a" value is greater than 1, the parabola will be vertically stretched and if the value of "a" is less than 1 the parabola will be vertically compressed.
- The "h" value will gives us the horizontal translation of right or left.
- The "k" value will give us the vertical translation of up or down.
- The sign in front of the "a" value will determine if the parabola open ups or down, If the sign is a positive it will open up, and if the sign is a negative, it will open down.
- Remember that the "h" value will change signs when brought out from the bracket, if it is a negative in the bracket it will become a positive outside the bracket, therefore will move to the right, and vice-versa.
Writing Equations from Given Transformations
The graph of y = x^2 is opening downward, stretched vertically by a factor of 3, translated horizontally to the left by 4 units and translated vertically down by 5 units.
The equation would become:
y = -3(x+4)^2 - 5
Graphing Transformations of Quadratics Relations
Mapping Notation
In general, to go from the graph of y = x^2 to y = a(x-h)^2 + k you can use the mapping notation:
(x,y) = x + h, ay + 1
Example:
a) y =5(x-4)^2 + 2 will become x + 4, 5y + 2
b) y =(x+7)^2 will become x - 7, y
c) y =(x-3)^2 + 5 will become x + 3, y +5
4.4 Graphing Quadratics in Vertex Form
Graph using Step Pattern and Mapping Notation
Word Problems for Quadratics in Vertex Form
Finding x and y intercepts in Vertex Form
Mini Test #1
5.1 Multiplying Binomials
Simplifying polynomials includes both collecting like terms and use of the distributive property. It is important to be sure to use BED MAS at all times. Exponent rules must be used for both multiplication and division when simplifying.
When multiplying binomials there are two rules you must follow
- Distributive Property
- Collect like terms
Example 1:
(x+3)(x+4)
- We will use the distributive property to multiply the first two terms
- Then we will multiply the two inner terms with the outer terms and add the outcome, in this case the outer term is 1 so you would simply add 3 and 4
- Then we will multiply the two inner terms with each other
x^2 + 4x + 3x + 12
- Lastly we will add the like terms together
x^2 + 7x + 12
Reminder - Multiply coefficient + add power
2x^2 (2x)
4x^3
Example 2:
2x(3x-4) + (x+3) - (2x-7)
(6x^2-8x) + (x+3) - (2x-7)
(6x^2-7x+3) - (2x-7)
6x^2 - 7x - 2x + 3 + 7
6x^2 - 9x + 10
5.2 Special Products
Example:
(3y+7x) = a^2 + 2ab + b^2
(3y)^2 + 2(3y)(7x) + (7x)^2
9y^2 + 42xy^2 + 49x^2
5.3 Common Factoring
- Finding the GCF
- Common Factoring
- Grouping
1. When factoring by GCF, you simply find the greatest common factor of the terms , for example the terms 5c + 10d both have 5 in common, so you divide both terms by 5. You always keep the common factor outside of the bracket and place everything else inside of the bracket, SO 5(c+2d).
2. Common Factoring is when the two binomials are exactly the same, for example
5x(3x+2) + 4(3x+2), since the two binomials are the same you will put them as one bracket and place the remaining into another bracket so, (3x+2)(5x+4).
3. When grouping factors you group a set of two like terms into brackets
- ax + ay + 2x + 2y
- (ax+ay) + (2x+2y)
Then you will remove the like term to the outside of the bracket
- a(x+y) + 2(x+y)
Now since you can see that there is a common binomial you follow the rules for Common factoring
- (a+2)(x+y)
5.4 Factoring Simple Trinomials
Where x is a variable and "a", "b", "c" are constants.
You can factor a quadratic in standard form to get factored form:
x^2 + bx + c = (x+r)(x-s)
STD FORM Factored Form
Where "r+s" is" b" and " rs" is "c"
Step 1: Find the product and sum
- find the two numbers whose product is "c"
- find the two numbers whose sum is "b"
Example 1: x^2 + 6x + 5
- Find the two numbers whose product when multiplied is 5 (1)(5) = 5
- Find the two numbers when added together equals to 6 5+1 = 6
- The factored form will be (x+5)(x+6)
Look at the signs of "b" and "c" to make it easier to figure out the sign of the two numbers
- if "b" and "c" are positive, both "r" and "s"
- if "b" is negative and "c" is positive , both "r" and "s" are negative
- if "c" is negative , only ONE of "r" or "s" is negative
- if both "c" and "b" are negative, only ONE of "r" or "s" is negative
Example 2:
n^2 + 6n + 8
(4)(2) = 8
4 + 2 = 6
(n+4)(n+2)
5.5 Factoring Complex Trinomials
Remember we are trying to break down the middle term so that it equates to 4 terms at which we can proceed to factor by grouping.
- Always look for a common factor
- To factor ax^2 + bx + c, find the two integers whose product is "ac" and whose sum is "b", similar to simple trinomial factoring
Example 1:
3x^2 + 8x + 4
(3)(4) = 12
(2)(6) = 12
2 + 6 = 8
Break up the middle term
(3x+6x) + (2x+4)
Factor by Grouping
3x(x+2) + 2(x+2)
Collect the binomials
(3x+2)(x+2)
You follow the same steps when factoring a trinomial with two variables.
6.2 Solving Quadratics in Factored Form
- Make one side equal to zero
- Set each bracket equal to zero
- Solve for "x"
Example 1:
x^2 + 9x + 14 = 0
x^2 + 7x + 2x + 14 = 0
x(x+7) + 2(x+7) = 0
(x+2)(x+7) = 0
x+2=0 x+7= 0
x= -2 x= -7
(-2,0) (-7,0)
6.3 Graphing Quadratics in Factored Form
Using the example from above, you can than proceed find the axis of symmetry by adding the two x intercepts together and dividing by two
(-2)+(-7) / 2
= -4
The -4 is the "x" value
The AOS can used to find the optimal value (y intercept)
y = (x+7)(x+1)
y = (-4+7)(-4+1)
y = (3)(-3)
y = -9
This is the "y" value
Now since you have the "x" and "y" values you can put them together to make the vertex, so the vertex in this case would be (-4,-9).
At the end of this you have 4 points:
- (0,7)
- (-7,0)
- (-1,0)
- (-4,-9)
You can use the points above to plot your graph
Mini Test #2
6.1 Maximum and Minimum Values
Completing the Square Steps:
- Group the x terms
- Divide the coefficient of the middle term by 2, square it, then add and subtract that number inside the brackets
- remove the subtracted term from the brackets
- factor the brackets as a perfect square trinomial
Example 1:
y = x^2 - 6x + 4
y = (x^2 -6x) + 4
y = (x^2 - 6x + 9 - 9) + 4
y= (x^2 - 6x + 9) - 9 + 4
y = (x-3)^2 -5
Vertex = (3,-5)
AOS = 3
Max or Min value = min -5
Values that "x" may take = all real numbers
Values that "y" may take = y greater than -5
When there is a number in front of the "a" value you must first divide the middle term by that number and place then proceed with the steps. You must also multiply the square outside of the brackets.
When there is a negative in front of the "a" value you must multiply the middle term by the negative, if it is already a negative it will become a positive and if it is a positive it will become a negative. You must also multiply the square outside of the bracket by the negative.
6.4 the Quadratic Formula
The discriminant
In order to figure that out you must find the discriminant. The discriminant is the value under the square root in the quadratic formula.
The formula to find the discriminant is D = b^2 - 4ac
When the equation gives you a positive number or a number greater than 0, there will be 2 solutions (x intercepts).
When the equation gives you a negative number or a number less than 0, there will no solutions (x intercepts).
When the equation gives a number equal to 0, there will be 1 solution (x intercepts).