Quadratic Relationships
Jasmeen Dourka
Unit Summary
Topics
Introduction to Quadratics:
o Analyzing Quadratics
o Properties
o Transformations
Graphing
o Graphing from Vertex Form
o Graphing from Standard Form
o Graphing from Factored Form
Factoring
o Factored Form
o Multiplying Binomials
o Common Factoring
o Simple Grouping
o Simple Trinomials
o Complex Trinomials
o Special Factoring (perfect squares, difference of squares)
Equations (Solving, Finding the Equation,Changing Forms)
o Finding the Equation (given the vertex)
o Solving by Factoring
o Completing the square (going from standard form to vertex form)
o Solving from Vertex Form
o Quadratic Formula (solving standard form)
o Discriminant
Applications
o Motion Problems
o Geometric Problems
o Economic Problems
o Number Problems
Introduction to Quadratics
Analyzing Quadratics
To determine whether or not a graph/equation is a quadratic relation, a table of values needs to be created. Quadratic relations have unequal first differences but they have equal second differences whereas linear relations have equal first differences.
For example, for the equation y=x² we can create a table of values to determine whether or not it is a quadratic relation.
Properties
- Vertex (axis of symmetry, optimal value)
- Axis of Symmetry (x value of the vertex/divides parabola in two)
- Optimal Value (y value of the vertex/highest or lowest point)
- Values x may take (all real numbers)
- Values y may take (y ≤ optimal value or y ≥ optimal value)
Transformations
- Whether or not there is a reflection on the x-axis
- The horizontal shift (left or right)
- The vertical shift (up or down)
- Whether or not there is a vertical stretch or compression
For example, the following can be said about the graph/equation below:
- There is a reflection on the x-axis
- There is a horizontal shift of 11 units to the right
- There is a vertical shift of 7 units up
- There is a vertical compression of 0.5 units
Graphing
Graphing from Vertex Form
For both methods I will use the equation y=2(x-2)² + 8.
The first is graphing using the step pattern.
Graphing using the step pattern has 3 steps:
2. Use the step pattern. The basic step pattern is over 1, up 1 and over 2 up 4. If there is an a value, the step pattern is affected. The value of x² is multiplied by the a value. Using the step pattern, at least four points of a parabola can be found (two on each side of the vertex). For this example, the a value is 2, so the step pattern becomes over 1,up 2 and over 2,up 8.
3. Plot the points and connect with a curve.
1. Determine the vertex. This is done the same way as it is when graphing using the step pattern. Therefore, the vertex is still (2,8)
2. Write the mapping formula. The mapping formula is (x+h,ay+k) The values are found in the equation and are simply plugged into the formula. Once again, h is also positive in the formula if negative in the equation, etc. For this example, the mapping formula is (x+2,2y+8)
3.Create a table of values for the base quadratic equation and a table of values for the new equation. Creating a table for the base equation allows you to easily find the values for the transformed equation. For this example, the tables are:
Graphing from Factored Form
As an example, I will use the equation y=-2(x+3)(x-6)
1. Find the x-intercepts.To find the x-intercepts, set y = 0 and isolate x. Do this for both brackets.For this example the x-intercepts are:
x+3=0 and x-6=0x=-3 and x=6
x=(-3,0) and (6,0)
x=-3+6/2
x=3/2 or 1.5
3. Find the y value of the vertex.To find the y value (optimal value) of the vertex, substitute the x value into the equation and solve for y. For this example:
y=-2(x+3)(x-6)y=-2(1.5+3)(1.5-6)
y=-2(4.5)(-4.5)
y=-2(-20.25)
y=40.5
4. Plot the three points on a graph and connect with a curve.
Graphing from standard form
For both methods I will use the equation y=2x²+4x-6.
The first method involves going from standard to factored form.
For this method, factoring is required,and once the equation is factored, the x-intercepts and the vertex can be found using the same steps as graphing from factored form.
1. Factor. Factoring will be covered thoroughly in the next section, but when factoring, you are going from standard form to factored form. For this example, y=2x²+4x-6 becomes y=2(x+3)(x-1).
2. Find the x-intercepts.For this example, the x-intercepts are x=(-3,0) and x=(1,0) because:
x+3=0 and x-1=0
x=-3 and x=1
x=(-3,0) and x=(1,0)
3. Find the vertex. For this example, the vertex is (-1,-8) because:
x=-3+1/2
x=-2/2
x=-1
________________
y=2(x+3)(x-1)
y=2(-1+3)(-1-1)
y=2(2)(-2)
y=2(-4)
y=-8
4. Plot the points and connect with a curve.
The second method is from just standard form.I will still use the equation y=2x²+4x-6 for my examples.
For this method, h=(-b/2a) is used to find the x value of the vertex (h) and to find k (y value of the vertex) you substitute in the h value into the standard form equation. However, you still need to find the x-intercepts, which can be done by using the quadratic formula or solving by factoring.
1. Use h= (-b/2a) and solve for k. This formula is an quick way to find the x value of the vertex. For example:
h=(-b/2a)
h=-4/2(2)
h=-4/4
h=-1
k=2x²+4x-6
k=2(-1)² +4(-1) -6
k=2(1) -4-6
k=2-4-6
k=-8
2. Solve using the quadratic formula or by factoring. For this example, the equation can be factored so it makes sense to solve by factoring.
y=2x²+4x-6
y=2(x²+2-3)
y=2(x+3)(x-1)
x+3=0 and x-1=0
x=-3 and x=1
x=(-3,0) and x=(1,0)
3. Plot the points and connect with a curve.
factoring
multiplying binomials
For example:
link to online algebra tiles
all types of factoring
Below is a video that I made that explains:
- Common Factoring
- Factoring by Grouping
- Simple Trinomials
- Complex Trinomials
- Special Factoring
Another useful video is below:
solving,changing forms & finding the equation
Finding the Equation (given the vertex)
Below is a picture of an equation I found using this method:
Solving by factoring
x+5=0 and x+11=0
x=-5 and x=-11
x=(-5,0) and x=(-11,0)
To check my answer, I used the LS/RS check method.
y=(x+5)(x+11)
0=(-5 +5)(-5+11)
0=(0)(6)
0=0
__________________
y=(x+5)(x+11)
0=(-11+5)(-11+11)
0=(-6)(0)
0=0
Below is a link that I found useful which explains solving by factoring.
Completing the square
To complete the square, you:
- Put brackets around the x² and x terms.
- Remove the common factor, if there is one (remember to just factor the number)
- Find constant that must be added and subtracted. This is done using the formula (B/2)² (remember to add and subtract the number because you are essentially adding 0).
- Group the three terms that complete the square and move the subtracted value out of the bracket by multiplying it by the common factor.
- Factor the perfect square and collect like terms.
Solving from Vertex form
quadratic formula
discriminant
If D<0, there are no solutions because you cannot square root a negative number.
If D>0, there are two solutions.
If D=0, there is one solution.
Using the discriminant is an easy way to see how many solutions a parabola will have without using the entire quadratic formula.
I found using the discriminant helpful because it also gave me a better understanding about the quadratic formula.
Below is a helpful link that also fully explains the discriminant
Applications
At the end of the unit, we applied our knowledge of the three forms (standard, vertex and factored) in solving a variety of word problems. The word problems can be categorized as motion problems, geometric problems, economic (revenue) problems and number problems. Geometric and revenue problems are also encompassed within optimization problems.
Below are answers to a variety of word problems that we worked through.
Motion Problems
Geometric problems
economic problems
The equation that can be set up to be used to solve these problems is:
R=(#sold)(price)
It is also important to have a let statement (let x represent the number of $_ increases/decreases in price).