Quadratic Unit Review

By: Hasan A

The Beginning of Quadratics Y=A(X-H)²+K

Quadratics is an important unit in grade 10, as it is covered through grade 11. By reviewing this site, it will help you know more about this unit, as so on.

First, lets begin with the basics of quadratics, and take a look at the following sheet to the side

By looking at the sheet, it explains how a parabola works. In most cases, the terminology is defined as:

Vertex: The maximum or minimum point of the graph. How Do I label it? (x,y)

Minimum/maximum value: Also called optimal value. it is the highest or lowest point of the graph. How Do i label it? (y=#)

Axis of Symmetry: Divides the parabola into two equal halves. How Do I label it? (x=#)

Y-Intercept: Where the graph meets the y-intercept. How Do I label it? (0,#)

X-Intercept: It is where the graph crosses the x-intercept. It is also known as ''roots'' or ''zeroes''. How Do I label it? (#,0)

Moving on, another key point to remember is this formula called vertex form: Y=A(X-H)²+K. The H in this formula represents the Axis of Symmetry which is your x intercept and divides the graph into two. K, is your optimal value which is the highest or lowest point of the graph and also the y-intercept

Lastly, lets graph y=a(x-h)²+k

Some things to know, is when you graph this, you sub your h value to be 0, and leave K as it is.

So for example to find your H value, if a question is y=(x-2)+4, make your H Equal to 0 (Opposite)

h=0 h=+2

And make sure to leave k (4) As it is! it does not change.

Therefore, H will be (+2,0) and once you add K, which is 4, you put it together as one point and graph it. (+2,4)

Transformation is another key point of analyzing quadratic questions. By knowing your H value, it tells you if you shift left or right, depending if its negative or not, and same with your K value, but up and down on your y-intercept. Also, if your A value is anything from, -1 to 1 is a vertical compression and a number less than -1 and grater than 1 is a vertical stretch. Knowing that, if the A value is negative it is a reflection, so lets begin with the first example

Example #1: Y= -2(x-3)²+5

1. Reflection on the x-axis (Because its negative)

2. Vertical stretch by a factor of 2 (Because its greater than 1)

3. Horizontal Shift right 3 units (Because your H value becomes positive)

4. Vertical Shift up 5 units (Because its positive)

Now! Try doing this one, which is more easy by yourself: Y= (x-2)²-5

Furthermore, the harder part of vertex form is to find your zeroes. There are 5 key steps to remember while doing this. So I will list them

Step 1: Set Y Value = 0

Step 2: Move K Value to Y Side

Step 3: Divide your equation by A

Step 4: Square Root your answers

Step 5: Move H Value

So lets analyze this question. Y = -3(x+3)² + 75

Step 1: 0 = -3(X+3)²+ 75 (Set Y = 0)

Step 2: -75 = -3(X+3)² (Move K Value to Y Side, the sign becomes opposite as you move, so +75 would turn into -75)

Step 3: -75 = -3(X+3)² (Divide by A Value, -3 gets divided by 75 and canceled by -3)

--- -------------

-3 - 3

Step 4: 25 = (x+3)² (Square root both answers, so 25 squared is 5 and (x+3)² squared is (x+3)

Step 5: 5 = X+3 - (Do not forget to add your + - sign. So turns into -3 + 5 = x

-3 -3 -

Finally, now solve using both signs. So -3 + 5 = 2 (Dont forget to put in brackets,)( 2,0)

-3- 5 = -8 (-8,0)

Now your two zeroes are (2,0) (-8,0)

Watch the video below posted by Mr. Anusik if you don't understand

solving quadratics from vertex form

The Beginning of QuadraticS Y=A(X-H)²+K


The last thing to remember is your step pattern, which is normally 1,4,9. If your A value is not given, then your step pattern will be normal which is, up 1 over 1, up 4 over 2. BUT If a A value is given such as 2, then you multiply it to your original step pattern which is (1,4,9) and get your answer.

Example of normal step pattern: Above

The picture above shows us the example of a normal vs when there is a given A Value. by remembering your step pattern helps you when you graph. 1,4,9 is your step pattern, and when A is a number, just multiply is to 1 and 4 and plot it. This now leads into factored form because its graphing and you need to know your zeroes and axis of symmetry in order to move on.

If you have any questions, please watch the video below posted my Mr. Anusik

3.2 Graphing from Vertex Form

Factored Form: y = a (x-r) (x-s)

Factored form is more of the easy side of quadratics. By now, you should know your zeroes, axis of symmetry and optimal value.

So lets begin with your formula, which is very important to memorize. y = a(x-r)(x-s)

Unlike the step pattern in vertex form, we now have two x-intercepts we plot of the graph which is R and S. To find these values, it is the same as we did before where we make the X=0 we will use this question - y=(x-3)(x-2)

So for example, if R = 0 R= 3 (3,0)and R=0 R=2(2,0) we know have our two x-intercepts which we plot of the graph.

In order to find the axis of symmetry, which is what divides the graph into half, we can do it through this formula


So we add our R + S = 3+2 = 5/2= 2.5

Therefore, your axis of symmetry will be 2.5

Now, the last part to this is to find the optimal value. Lets say, in the question above we we working on y=(x-3)(x-2)once we find the axis of symmetry, finding the optimal value is simple. You Have to plug your axis of symmetry in the X'S, solve and multiply. If there is a outer A value, just multiply it to the equation. So for the question above we we're working on, we got the axis of symmetry to be 2.5. Now, we plug that into our original equation which is y = (x-3)(x-2) which the x's are and solve.

So.. Y= (2.5-3)(2.5-2) = (-0.5)(0.5)

Now you multiply and get your optimal value to be -0.25

To have a better understand: please watch the video below that explains it through graphing. (All credit goes to Mr. Anusik for making this video as I do not own it)

3.5 Graphing from Factored Form

Standard Form: y = ax² + bx + c

Now lets begin standard form, which is represented as: y = ax² + bx + c. The first thing in standard form is to know your quadratic formula, which is: -b+-√b^x² - 4(a)(c)



Now,this formula may look confusing at first, but lets break this into parts. The standard form equation states y = ax² + bx + c. So your ax, is your a represented in the formula, bx is your b represented in your formula, and c is c in your formula. Another thing you should know is your b value has a negative sign in the start of the formula only, and there is a + and - so there will always be 2 answers. Also, if your square root is negative, there is no solution and is an extract answer represented as it is. So lets take some example ups

#1. 5x² - 7x + 2

Step 1: Know your values (A= 5 B= 7 C= 2)

Step 2: Plug in your formula: - (-7)+- √(-7)² - 4(5)(2)



Step 3: Start solving within, so square your 7 then multiply your 4x5x2 ( 7+- √ 49 - 40)



Step 4: Subtract within the square root then square root it (7+- √9=3)



Step 5: Now, use the + and - with your b value, add it to the square root then divide it by your A value below ( 7+3/10 = 1, x = 1) (7-3/10= 0.4)

So we have two X values from the quadratic formula. Again if your answer within the square root was negative, it would be an exact answer, with no solutions because you cannot square root a negative, so you leave it as it is, like : 7+-√-3



Moving on, we need to find the axis of symmetry and it looks like this > -b/2a. This is a quick way to get your Aos, and you plug your numbers from the formula. So, for example. y = 6x²+3x+2. So turn your B value into negative, multiply your 2 by your A value and divide it. so,

it would look like -3/2(6)

-3/12 = -4

We can use this to optimal value by subbing it in your x's so, lets go back to our normal equation. y = 6x² + 3x + 2. Plug your Aos in the x value part, y = 6(-4)² + 3(-4) +2

Then solve: 6(16) - 12 + 2

96-12+2 = 86 making your optimal value to be 86

Now, we have arrived at completing the square. Completing the square is old method of converting quadratics from standard to vertex form so they can be solved. Notice, y = a (x-h)² + k contians a perfect square in it, but y = ax² does not. so we want to force a perfect square in our standard form equation.

Step 1: Remove the common factor (x² and x)

y= (2x² + 12x) -3

y = 2(x² +6x)-3

Step 2: Find the constant that must be added and subtracted to create a perfect square. Rewrite the expression by adding and subtracting the constant

Y= 2(x²+6x+9-9)-3

Step 3: Group the three terms that form the perfect square (move the subtracted value outside the brackets by multiplying the comment factor first)

y= 2(x²+6x+9-9)-3

y= 2(x²+6x+9)-18-3

Step 4: Factor the perfect square and collect like terms

Y= 2(x²+6x+9))-21


There are video posted below to help you understand standard form.

3.14 Completing the square

Common Factoring

Now, we arrive to factoring in quadratics. Much of this part is mostly done on paper, and less graphing. So lets begin with the first part of it.

What is factoring? if you remember back to grade 6 or 7, its when you multiply two numbers to get the answer of the question.

For example: Factor of 6 = 3x2

So when your common factoring in quadratics, you want to find __ x __ = to your answer. There are two ways to do this, which includes tiles and algebraically. So lets start of with algebraically, since its more easy.

Example #1: 2x + 4

Step 1: Find whats in common. In this case, we have 2 in common (2x2=4, so we carry the 2 down and put brackets looking something like this 2( x + x ) and so forth)

Step 2: Now divide your question by 2, since its the number you took out (2x/2 + 4/2 )

Step 3: 2(x+2) - Once you divided your question by 2 and carried your common factor of 2 down, this should be your final answer

Doing the table method is also another option which can be only explained through a video, so please watch my common factoring video to understand it.

This now leads into simple trinominal which also deals with factoring and working backwards with tiles.

Common Factoring - 4School

Simple Trinomial

By watching my video above, it is important to know how to expand with tiles, because once you reach simple trinomials you have to work backwards with tiles

For example: you can graph this by tiles (x+2)(x+4) = all you have to do is expand, so multiply x by x, x by 4 and same with the 2, which goes 2 x x, and 2x4.

So you get your answer of x² + 6x + 8. Now, with simple trionomials, we can do it through tiles and algebraically

So for the answer above, of x² + 6x + 8, you can show this through a big x squared, 6 x tiles and 8 small cubes, and the table above that files it in is your answer. Below are couple pictures through the method of tiles.

Another method, which is more easy and preferred is algebraically.

The way we do this is simple, for example : x² + 4 + 3

Step 1 : We have to find out what x what is equal to your first part of the problem(x²). In this case its x² so it would be x (x) x

So we can put the bracket in place : (x + ) (x + )

Step 2: We now have to figure out what x what is your last term and equals your middle term, so in this case the only way to get 3 is 3x1 which is equal to 4

Step 3: (x+3)(x+1)

Step 5: Check your answer: you can expand to verify that your answer is correct

Simple Trinomials - 4School

Complex Trinomial

Now, in complex trinomails, we have to re call a simple trinomial which is x² + 4x + 3. It is basically your x value being less than 1

However, in complex trinomials, your x value can be greater than one, for example"

2x² + 11x + 15

Now, in order to solve this, its the same way as simple trinomials.

What x What is equal to your first term and What x What is equal to your last term.

However, sometimes, you have to switch up your numbers as their are multiple ways to get to a number.

To solve this: The only what to get 2x2 is by multiplying 2x by 1x. So that fills in your first two brackets

(2x + ) (x + )

Once we got this far, we have to find out what x what is equal to 15. There are different numbers that multiply such as 3 x 5, 5 x 3, etc but the right one is 5x3

(2X + 5) (X+3) Now to check our answers, we expand and collect like terms.

This will all be explained in my video

Complex Trinomials - 4 School

Perfect Squares

Knowing how to do complex trinomial is important, since its almost the same as perfect square, other than finding what x what is equal to your first and last term, you square root them to find your answer.

But.... In order to know if your question is a perfect square or not, it has to have a positive sign indicating that it can be squared. It must also have 3 terms within the question

Example #1: x² + 2xy + y²

We can square root x squared which becomes into x, and y squared which becomes into y.

So we but them in brackets, (x + y ) (x + y)

For this last step, you can put a square sign on top of the bracket, to not show repetition


Example #2: 36a² + 60a + 25

Square Root of 36a² = 6a Square Root of 25= 5

Put them in brackets.. (6a + 5) (6a+5)



Perfect Squares - 4School

Difference of Squares

This is the last part of factoring, so let begin.

Difference of squares is the literal definition, which means there two different numbers that can be squared. In order to know it is a difference of square and not perfect square, there has to be 2 terms, and a negative sign in between

For example: 81x² - 64y²

Now to solve this, we have to square the first and second term, so 81x² square root is 9x and 64y² square root is 8y. We can now put them into a bracket. But what will our signs be, well theres a negative in-between, and the only way we can get it is through a positive and negative. so our final answer will be


And that is how to solve differences of squares.

Differences of Squares - 4School

Word Problems In Quadratics

There are various types of word problems in quadratics such as motions, shapes, area, etc. But throughout the unit, I found these 3 problems to be important to understand to lets start the first one, that is the shape one. Look at the picture above to see it first please. (Example was taken from practice sheet posted on edmodo)

Step 1: Cut the shape into two halves and label it with two parts.

Step 2: To solve this is simple. You want to find the area, and its a rectangle. So the formula for the area is lxw, so do that to the two side lengths and you get x(x

+2) and for the other part, you get (x-1)(2x+4).

Step 3: Expand = X(X+2) + (X-1)(2X-4)

x²+2x + 2x² + 4x - 2x- 4

Step 4: Collect like terms = 3x² + 4x - 4

b) If the area of the shape is 11m², determine the value of x

3x² + 4x - 4 = 11 (Sub in 11 and move it)


3x² + 4x - 15 = 0 (Add your like terms which is -4-11)

(3x - 5)(x+3) = 0 (Do your simple trinomial like step)

x = 5 x = -3 (Now you get your 2 x intercepts)



So your 2 values would be 5 over 3 and -3


(This question is taken from my test)

1. The height of a rock thrown from a walkway over a lagoon can be approximated by the formula h = -5t² + 20t + 60, where "t" is time in seconds, and "h" is the height in meters.

a) Write the above formula in factored form.

H = -5t² + 20t + 60

H = 5(t² + 4t + 12) (Take the common factor out, which is 5 and divide your equation)

H= 5 (t + 6) (t - 2) (Do your simple trinomial like step)

x = (6,0) x= (2,0) (You get your two x intercepts)

b) When will the rock hit the water?

At 6 seconds


Describe the transformation:

\: Y= -2(x-3)2+5

1. Reflection on the x-axis (Because its negative)

2. Vertical stretch by a factor of 2 (Because its greater than 1)

3. Horizontal Shift right 3 units (Because your H value becomes positive)

4. Vertical Shift up 5 units (Because its positive)

Reflection on Quadratics

Quadratics in general was good unit. There were 3 parts to it, and the first 2 were really fun! I personally did better on the second unit, but the application but wasn't the best and could have been done better. The hardest part of quadratics is the last section, where the world problems come in.

Here is a picture of a test I received back, which I did bad on the application part.

Big image