By: Manpreet Deol
What are Quadratics
The word quadratics comes from the world "quad" which means square because the variable gets squared. In Grade 9, we were taught about linear relationships in which the result of an equation (in the form of y=mx+b) would form a straight line if we were to graph it. Quadratics help show display the path of an object (or other things such as a dive) by using curves! In linear relations the line would go straight up or straight down. However, in quadratic relations parabolas are used. A parabola is a curved line created from all the x and y points. Parabolas can be seen everyday as shown in the examples below. We encounter parabolas everywhere, whether we realize it or not. This website will summarize the entire quadratics unit, using a variety of media and appropriate terminology throughout.
Parabolas are everywhere!
Types of equations used in Quadratics
Vertex Form: a(x-h)^2+k
Factored Form: a(x-r)(x-s)
Standard Form: ax^2+bx+c
Quadratics in Vertex Form
Table of content
- Everything to know about parabolas
- Key Features of Quadratic Relations
- Quadratic Relations and Finite Differences
- Parabolas and Transformation
- Transformations of Quadratics
- Transformations on a graph (from y=x^2 to vertex form equation) S.R.T.
- Graphing Transformations of Quadratic Relations (Vertex Form) (Mapping Notation)
- Graphing Transformations of Quadratic Relations (Vertex Form) (Step Pattern)
- Finding X and Y Intercepts in Vertex Form
- Finding an equation given the vertex and 2 points
- Word Problems for Quadratics in Vertex Form
- I am able to use finite differences to analyze a quadratic relation.
- I am able to use a table of values to create a graph and I am able to graph the base graph of a quadratic function without technology.
- I am able to graph quadratics using transformations, using the mapping notation and step Pattern.
- I am able to find x and y intercepts in vertex form.
- I am able to solve word problems for quadratics in vertex form.
Everything to know about parabolas
- Parabolas can open up or down
- The zero of a parabola is where the graph crosses the x-axis
- "Zeros" can also be called "x-intercepts" or "roots"
- The axis of symmetry divides the parabola into two equal halves
- The vertex of a parabola is the point where the axis of symmetry and the parabola meet. It is the point 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 of a parabola is where the graph crosses the y-axis
Key Features of Quadratic Relations
Vertex form: y=a(x-h)^2+k
Vertex- (h,k). The vertex is where the axis of symmetry and parabola meet. The maximum or minimum point on the graph. it is the point where the graph changes direction.
Minimum/ maximum value- highest/lowest point on the graph.
Axis of Symmetry (AOS) (x=h) - x values of the vertex. The axis of symmetry divides the parabola in the center. The AOS Is a vertical line that runs through the center of a parabola creating two perfect halves and is the X value to the highest or lowest point on the parabola.
Zeros- x intercepts (x,0). Zeroes are also known as x intercepts or roots. The zero of the parabola is where the graph crosses the x-axis.
Optimal Value (Max,Min Value) (y=k) - y value of the vertex. The optimal value is the value of the y co-ordinate of the vertex.
Y intercept- The point where the parabola where graph crosses the y-axis.
Quadratic Relations and Finite Differences
First and second differences
Using first and second differences can make it easy to figure out whether the equation is quadratic or linear. For it to be quadratic the second differences will be the same and the first differences will be different. For it to be linear the first differences would be the same.
Calculating First Differences:
- In order to calculate "first differences" you first need to subtract the second y value from the first y value.
- Then repeat with each pair of y-values.
- If all the first differences are constant then that means the pattern is linear.
- The first differences also tell you about the rate of change for the situation.
- If the second differences are constant then that means your pattern is a quadratic relation.
- Remember you must have all your x- values in order.
- If there is no constant in any then its neither.
Parabolas and Transformation
Transformations of Quadratics
The vertex form of a Parabola: y=a(x-h)²+k
The a in the equation stretches or compresses the parabola. If the the a is positive the parabola will open up but if the a is negative the parabola will open down/reflect about the x axis. When describing this you would say there is a vertical stretch or compression by a factor of whatever number a is.
The h is the horizontal translation. If h is negative the parabola will translate right and if it is positive the parabola will translate left. When describing this you would say there is a horizontal translation to the left or right by whatever number h is.
The k is the vertical translation. When k is negative the parabola will move down and when it is positive the parabola will move up. When describing this you would say there is a vertical translation up or down by whatever number k is.
(h,k) are the vertex. in the equation when stating the vertex make sure to switch the x sign
State the transformations that occurred to y=x^2 to become y=3(x+2)^2-8 using appropriate terminology
-Vertical stretch by a factor of 3
-Vertical translation 8 units down
- Horizontal translation 2 units to the left
Breaking it down once more:
The value of 'a' determines the orientation and shape of the parabola.
If 'a' is positive the parabola opens up.
If 'a' is negative the parabola opens down, reflection.
If 'a' is greater than 1 than the parabola is vertically stretched.
If 'a' is not greater than 1 the parabola is vertically compressed.
vertically stretched by a factor of 3
vertically compressed by a factor by 0.25
The value of 'k' determines the vertical position of the parabola.
If 'k' is positive than the vertex moves up by 'k' units.
If 'k' is negative than the vertex moves down by 'k' units.
translated 8 units up
translated 5 units down
The value of 'h' determines the horizontal position of the parabola.
If 'h' is positive then the vertex moves to the left.
If 'h' is negative then the vertex moves to the right.
Example: y= (x-6)²
moves to the right 6 units
moves to the left 7 units
Identifying the values of a, h, k
-vertically stretched by a factor of 15
-translates to the left 5 units
-moves up 7 units
-vertically compressed by a factor of 0.5
-translates to the right 1 unit
-moves down 8 units
Transformations on a graph (from y=x^2 to vertex form equation)
- Transformations on a graph (from y=x^2 to vertex form equation)
- There are 3 main things you should know while identifying the transformations of an equation/parabola; and the way to know it is by remembering a simple acronym...
- The " S " stands for stretch/compression, you would explain how much of a vertical/horizontal stretch or compression the parabola has.
- The " R " stands for reflections, so here you would state whether the parabola is upwards facing, or downwards facing, and whether or not it has a reflection.
- The " T " stands for translation, which means how far has the parabola moved horizontally (left or right) and vertically (up or down).
Graphing Transformations of Quadratic Relations (Vertex Form) (Mapping Notation)
There are 2 ways you can graph a relation: mapping notation and step pattern.
Mapping notation is a strategy used to accurately graph any quadratic relation.
Before using mapping notation, you first need to know the "key" points of the basic quadratic relation
(y=x²) - the simplest parabolic curve with no transformations applied.
Fill in the table of values below to find all the "key" points for y=x², the base function of a parabola.
Table of values for y=x^2
Graph the points
Mapping notation examples:
Here is a video made by me explaining: How you can Graph Quadratics using mapping notation
Video link: https://www.youtube.com/watch?v=zXnVf7b8Cvw&feature=youtu.be
Graphing Transformations of Quadratic Relations (Vertex Form) (Step Pattern)
- The step pattern tells us how vertically stretched or compressed the parabola will be.
- The standard step pattern would be when Y=X^2, meaning that on all the points of the parabola the Y coordinate will be the X coordinate squared.
- So the step pattern would be over 1 up 1, over 2 up 4, over 3 up 9.
- But the step pattern can be changed when a co-efficient is added before the X.
- For Example Y=2X^2 would be twice as vertically stretched.
- Not only does the 'a' value control how stretched or compressed the parabola is but also which way it opens out.
- If the 'a' value is less than 0 the parabola will open downwards and vice versa.
Step pattern is a pretty easy way to graph in vertex form. All you have to do is:
Step 1: 1 x a
Step 2: 4 x a
Step 3: 9 x a
and then plot your points going 1 to the left or right and then up or down by the product you got from each step.
Here is a video explaining: How you can Graph Quadratics using Step Pattern
Transformations and vertex form video
Graphing quadratics using step pattern examples
Finding X and Y Intercepts in Vertex Form
Step 1: Find the vertex. Since the equation is in vertex form, the vertex will be at the point (h, k).
Step 2: Find the y-intercept. To find the y-intercept set x = 0 and solve for y.
Step 3: Find the x-intercept(s). To find the x-intercept set y = 0 and solve for x. You can solve for 'x' by using the square root principle or the quadratic formula (if you simplify the problem into the correct form).
Step 4: Graph the parabola using the points found.
Finding the y intercept is much easier than finding the x. To find the y intercept you will need to set the variable x to 0 and then solve the equation for y.
y= (5) (9)-2
Therefore, your y-intercept would be 43 units (0, 43).
When isolating for your x the first step you want to do is set y=0.
Then move the k value over to the other side of the equation.
Then divide the entire equation by the value of a, to continue eliminating to find the value of x.
Remember: If the equation is negative on one side you are unable to square root the equation which means there are no x-intercepts. If the equation is positive on the other side then you must square root that side to get x (h) on its own.
When this has been done, you move the h from one side of the equation to the other. Lastly, you will add one side of the equation, and subtract the other in order the determine both your ‘x’ values.
Now set y to 0.
Move -5 to the other side. Changing it to positive 5.
Now, move 1 to the other side as well.
Square root the equation on both sides.
Add one side and subtract the other in order to determine the x’s
Now we have determined that the two x intercepts are (5, 0) and (-1, 0)
Finding an equation given the vertex and 2 points
Sub (-2,8) which are (h,k) into the vertex form equation y=a(x-h)^2+k
then sub in the (x,y) points which are (1,10)
Therefore the equation is y=2/9(x+2)^2+8
Word Problems for Quadratics in Vertex Form
Quadratic word problems typically ask for the following information:
- What was the initial height of the ball (rocket, firework, etc.)
- Set x=0 and solve for y (the height of the ball when time=0 )
- or simply find the y intercept
- How long was the ball in the air?
- Set y=0 and solve for x (the time when the ball hit the ground, when height=0 )
- find the x intercept
- when did the ball hit the ground?
- see above........ the time when the ball hit the ground. when height = 0
- What is the maximum height of the ball?
- Find the vertex, the y value is the maximum height y=k
- optimal value
- At what time did the ball reach the maximum height?
- Find the vertex, the x value is the time when the max height occurred x=h (x value)
- What is the height of the ball at 3 seconds ?
- Sub the x-value into the equation to solve for y (sub t into the equation to solve for h)
Object thrown in the air word problem:
During a Blue Jays baseball game a baseball is thrown at the height of the following the path of y=-0.5(t-3)²+5, where h is the height in meters and t is the time in seconds.
a) What is the height of the ball 3 seconds after it is thrown?
y=k in your equation which determines the maximum height, in this case it is 5m is your height.
b) What is the initial height of the ball?
t=0 determines the initial height of the ball, in this case it is 0.5m.
c) When will the ball reach the ground?
In order to find the time you need to set h=0.
0=-0.5(t-3)² + 5
√10 = √t-3²
The ball will reach the ground at 6.16s.
d) What is the height of the ball at 1 second?
You plug in the 1 second for t. Then solve to find the height of the ball at 1 second.
h=-5(t-3)² + 46.5
The height of the ball is 26.5m at one second.
Business Application of Quadratics:
- The relationship between selling price and profit can be modeled by a quadratic relation
- The relationship between selling price and the number of units sold can be modeled by a quadratic relation
a)What is the maximum profit and how much does a slice cost at the maximum profit.
To find the maximum profit and how much a slice costs at the maximum profit use the vertex. The vertex is (2,2500) so the maximum profit is 2500 at a price of $2 per slice.
b)At what slice price will the profit be $0?
To find the profit and $0 we have to set P=0 and solve for the zeros.
0=-25(s-2)^2+2500 move the 2500 to the other side
-2500/-25=(s-2)^2 divide both sides by -25.
100=(s-2)^2 Now square root both sides.
Since we cannot have a negative amount of money, selling a slice of pizza that costs $12 will result in a profit of $0
Business (profit) word problem:
Gina is selling handbags. Let P represent the profit and let s represent the selling price. The profit can be modeled by the equation:
a) What is the selling price that will maximize the profits?
y=k in your equation which determines the maximum profit, in this case it is 2500.
b) What is the selling price when profits equal zero?
This is also known as the break-even point.
The profits will equal zero when the selling price is $60 and $10.
Quadratics in Factored Form
Table of Content
- Some key background information to know about factoring
- Algebra Review
- Solving Quadratics by Factoring (finding the zeros)
- Multiplying Binomials: Expanding (FOIL)
- Multiplying Binomials and Special Products
- Monomial common factoring
- Binomial common factoring
- Factoring by Grouping
- Factoring Simple Trionomials
- Factoring Complex Trinomials (Decomposition)
- Factoring Special Products (Factoring Difference of Squares and Perfect Trinomials)
- Graphing Factored Form
- Graphing Quadratic Functions in Standard Form
- Determining the Equation From a Graph
- Factored Form Word Problems (Different Types)
- I am able to find zeros/x-intercepts/roots
- I am able to solve quadratics by factoring (finding the zeros)
- I am able to multiply binomials: expanding (FOIL), and special products
- I am able to do factoring [monomial, binomial, factoring by grouping, simple trionomials, complex trinomials (Decomposition), special products (factoring difference of squares and perfect trinomials)]
- I am able to graph factored form and graph quadratic functions in standard form
- I am able to determine the equation from a graph
- I am able to solve factoring word problems
The function for factored form is y=a(x-r)(x-s)
Some key background information to know about factoring
- The values of r and s are the x-intercepts/roots/zeros (r,0) (s,0)
- The value of a will give you the shape and direction of the parabola/opening
- Axis of symmetry, AOS: x= (r+s/2) Sub this x value into equation to find the optimal value
- To find the y-intercept, set x=0 and solve for y
- Solve using the factors
- Greatest Common Factor
- Simple factoring (a=1)
- Complex factoring
- Special case - Difference of squares
- Special case – Perfect square
The x intercepts are: (-6,0) (3,0)
The x intercepts are: (-16,0) (4,0)
Solving Quadratics by Factoring (finding the zeros)
Multiplying Binomials: Expanding
There are also special products:
Difference of squares:
Multiplying Binomials and Special Products
Basic Knowledge and Review:
We know that a binomial is a polynomial with two terms, and special products mean the result we get after multiplying. To figure out these special products, we can solve by using BEDMAS.
Sometimes there will be more than one binomial that must be multiplied, in that case we must multiply and simply those binomials.
If there is a number in front of the binomial, you must use distributive property and then multiply the binomials.
Not all binomials will be multiplied; some will be added with one another. This does not make a difference because as long as the highest exponent is 2 than the equation will remain quadratic
Monomial common factors
These two have a common factor of 7, so 7 is the number we will divide by making our new factored equation: 7(x+3y)
These two have a common factor of 6, so 6 is the number we will divide by making our new factored equation: 6(x+6y)
These two have a common factor of 8, so 8 is the number we will divide by making our new factored equation: 8(x+10y)
These two have a common factor of 5, so 5 is the number we will divide by making our new factored equation: 5(x+10y)
Binomial common factoring
The 2 binomials are the same they are both (w+3) (common factor) so they will become one. Now if you divide both terms by (w+3) you will be left with (2y+2x) making your factored equation: (w+3)(2y+2x)
The 2 binomials are the same they are both (w+6) (common factor) so they will become one. Now if you divide both terms by (w+6) you will be left with (4y+4x) making your factored equation: (w+6)(4y+4x)
The 2 binomials are the same they are both (w+10) (common factor) so they will become one. Now if you divide both terms by (w+10) you will be left with (8y+8x) making your factored equation: (w+10)(8y+8x)
Examples of Binomial common factoring
Factoring by grouping
Group together xy and 4y because they both have a common factor of y and group together 3x and 12 because they have a common factor of 3. The equation now becomes:
Examples of Factoring by grouping
Factoring Simple Trinomials
You can use this method to change an equation from standard form to factored form.
x^2+bx+c to (x+r)(x+s). The method that will be used is called product and sum where c is your product (rs=c) and b is your sum (r+s=b). In this method you need to find 2 numbers whose product will be equal to c and whose sum will be equal to b. There are a few rules when doing product and sum which are:
-When b and c are both positive, r and s will also both be positive
-When b is negative and c is positive both r and s will be negative
-When c is negative either the r or s is negative
What 2 numbers will give me a product of 6 and a sum of 5?
so the 2 numbers are 2 and 3
The new equation will be:
Just to recap and breakdown:
Something that is a Trinomial has three parts, for example, x2 +5x+3. Why is it called a simple trinomial? Because the coefficient 1 is in front of the x2 , 1x2. This involves factoring, because 2 numbers are multiplied in order to get us, x2 +5x+3.
1. Look at the equation x2 +4x+3, as ax2+bx+c. We need to find two numbers that will add into b, and multiply to equal c. The two numbers that add to 4, but multiply to 3.
So in this case, our two numbers are 3 and 1.
2. Now we know that our answer is (x+1)(1+3)
What to keep in mind:
- Which two numbers add to b and those same numbers multiply to c.
- If b and c are positive, both numbers will be positive (x2 +4x+3)
- If b is negative and c is positive, both numbers will be negative (x2 – 29x+28)
- If b is positive and c is negative, one of the numbers will be negative. (x2+3x-18)
Decomposition (complex trinomials)
Example of Factoring Trinomials Using Decomposition
Factoring Complex Trinomials Video
Factoring Special Products Handout
Factoring Difference of Squares and Perfect Trinomials
When looking at Special Cases there are two types:
Perfect Squares which can also be equivalent to (a+b) 2 or a2+2ab+b2
- The trinomial that results from squaring a binomial is called a perfect square trinomial.
- Perfect square trinomials can be factored using the patterns from squaring binomials.
- In a perfect square trinomial, the first and last terms are perfect squares, and the middle term is twice the product of the square roots of the first and last terms
Here are the following steps in order to solve Perfect Squares:
1. Starting off, we must look at our equation: x2+6x+9, in this equation we need to determine whether our b and our a can be squared.
x2 -> (x) 2
9 -> (3)2
2. We may be able to determine that our answer can be (x+3)2, but in order to check if we are right or not, we must use our 2ab, and plug in our values, to see if we are correct.
So since it does equal back to 6, we are correct.
Differences of Squares which can be equivalent to (a+b) (a-b) or (a-b) 2
Here are the following steps in order to solve Differences of Squares:
1. Using the equation, x2-16 we must find the squares in our equation.
X2 -> (x)2
-16 -> (4)2
2. We can determine that our answer potentially be (x-4)(x+4) or (x-4)2
3. To determine if we are correct, we can expand and check.
Graphing Factored form
Graphing Quadratics in Factored Form
Graphing factored form is easy! All you need to do is:
1. Find the x intercepts/roots/zeros
2. Use the x intercepts to find the axis of symmetry (the x in the vertex)
[Axis of symmetry, AOS: x= (r+s/2) Sub this x value into equation to find the optimal value]
3. Plug the axis of symmetry (x) into the original equation to find y
4. Plot on your graph!
Graphing Quadratic Functions in Standard Form
Video on Graphing From Factored Form
Determining the Equation From a Graph
Factored Form Word Problems
Area of a Rectangle Word Problem
Flight of an Object Word Problem
Dimensions of a Rectangle Word Problem
Word Problem: Flight of an object
The height of a baseball is h=-3t²+21t+54, where t is the time in seconds, and h is the height in meters.
a) When does the ball hit the ground?
h= -3(t² -7t-18)
Therefore, the ball lands on the ground in 9 seconds.
b.) What is the maximum height of the ball and when does it reach the maximum height?
h= -3(t² -7t-18)
h= -3(3.5² -7(3.5)-18)
Therefore, the max height is reached in 3.5 sec at 90.75m.
Measurement Word Problem
Table of Content
- Solving from vertex form
- Graphing Standard Form (quadratic formula) (complete the square)
- Word Problems in Standard Form
- Connections between all forms; vertex, standard and factored
- I am able to Graph standard form using the quadratic formula and by completing the square (I am able to find the maximum or minimum by completing the square).
- I am able to solve all types of word problems related to standard form (triangle, consecutive integer, etc...).
- I am able to connect all three of the forms together (making connections)
Standard Form Y=ax^2+bx+c
- The value of a gives you the shape and direction of opening
- The value of c is the y-intercept
- Solve using the quadratic formula, to get the x- intercepts
- Max or min? Complete the square to get vertex form
Solving from vertex form
How to solve an equation to find the x-intercepts and y-intercepts
- The first and foremost step to solve an equation to find the x-intercepts and y-intercepts would be to change that "y=" into a "0="..this step is what will allow us to even solve for the equation.
- After this we would solve for "x", by bringing it to the opposite side, and either adding or subtracting is to the zero.
- Then in order to remove the squared sign on the brackets, you would have to square root the opposite side of the brackets.
- But because we square rooted one side, there are now to possible answers ( this is because when a number is square rooted it could be positive and negative, so technically both answers are correct.
- The final step would be to isolate for x and get the final 2 answers.
"Solve for x"
3=x+3 ----> 0=3
Completing the square and solving from vertex form (during end of video)
Graphing Standard Form
1. Find the x intercepts (quadratic formula)
2. Find the vertex (complete the square)
Maximum and Minimum Values (Completing the Square)
Completing the square:
Not all quadratic functions are written in vertex form.You can convert an equation from factored form to vertex form, by using the method of completing the square.
1. Using the equation y=x²+8x+5, group the x-terms to, y= (x²+8x.....)+5
2. Divide the coefficient of the middle term, which in this case is 8, by 2, giving you 4. Now square the answer, equaling to 16. Then add and subtract the number inside the brackets y=(x²+8x+16)-16+5.
3. Remove the subtracted term from the equation by solving it, y=(x²+8x+42)-16+5, which will become y=(x²+8x+42)-11.
4. Factor the brackets (trinomial) to a perfect square trinomial, y=(x+4)²-11
Method: PART 2
We will be performing the same steps for “Completing the Square.”, except we now have an “a” value of 2 so now need to common factor it out.
Using the equation y=2x²+12x+13, group the x-terms to, y= (2x2+12x.....)+13
Factor the coefficient of x², which is the 2 from the first two terms. y=2(x²+6x...)+13
Divide the coefficient of the middle term, which in this case is 6 by 2, square the answer, 9. Then add and subtract that number inside the brackets. y= 2(x²+6x+9)-9+13
Now, multiply the value outside the brackets, the 2, to the value that was solved, -9. y=2(x2+6x+9)-9 +13 to y=2(x2+6x+9)-18 +13
Remove the subtracted term from the equation by solving it, y=2(x²+6x+9)-18 +13, becoming y=2(x²+6x+9)-5
Factor the brackets and turn them into a perfect square trinomial, y=2(x+3)²-5
Examples with steps on completing the square
The quadratic formula
The is how the Quadratic Formula was made:
Here is a great video by Khan Academy teaching you how to use the Quadratic Formula:
When D>0, there will be 2 x intercepts
When D<0, there will be zero x intercepts
When D=0, there will be only 1 intercept
Word Problems in Standard Form
Flight of an object word problem:
We can determine that the initial height is 1.5 metres since y= 1.5.
b) What is the maximum height reached by the basketball and at what horizontal distance does that occur?
Using the method of completing the squares, we are going to convert this equation from factored to vertex Form.
h= -0.25d² +2d+1.5
= (-0.25d² +2d) + 1.5
=-0.25(d² +2d) +1.5
=-0.25(d² -8d+16) -16 + 1.5
=-0.25(d-4)²+ 4 +1.5
Therefore, (4, 5.5) is the maximum height reached by the ball in 5.5m and the horizontal distance is of 4 units.
Here is an example of a revenue word problem:
Examples of consecutive integers problems:
Therefore, the 2 numbers are 57 and 58.
Here is a video of a triangle problem:
More Word Problems
Connections between all forms; vertex, standard and factored
Here's a chart made by me to help you out with quadratics
Here is how the three forms are connected
Connections Between All Three Forms
- To go from vertex form to standard form, all you need to do is expand and simplify
- To go from standard form to factored form, all you need to do is factor
- To go from factored form to standard form, all you need to do is expand and simplify
- To go from standard form to vertex form, all you need to do is complete the square
- To go from factored form to vertex form, all you need to do is find the x-intercepts, solve and then find vertex.