# Quadratics

### A guide for grade 10 quadratics, including 3 forms and more!

## Table of Contents

- What is quadratics?
- Parabolas & parts
- Quadratic vs Linear
- Vertex form parts
- Graphing from vertex form (+ step pattern)
- Finding the vertex given the equation
- Factored Form + roots
- Finding the vertex with the roots
- Graphing factored form
- Standard Form
- Standard form graphing
- Multiplying binomials + special products
- Algebra tiles
- Factoring (common, grouping, simple, complex, perfect square trinomial, difference of squares)
- Completing the square
- Solving quadratic equations (quadratic formula, factoring, isolating
*x*) - Word problems (motion, revenue, area, consecutive numbers, right triangle, optimization)
- Reflection + an assessment

## What are quadratics?

**curved**lines. The name

*quadratic*is used because

*quad*mean

*squared*, and in a quadratic equation, the variable is

*squared*. There are 3 forms of quadratic equations that we learn in grade 10:

**vertex form**,

**factored form**, and

**standard form**.

## Parabola!

## PARTS OF A PARABOLA

## The Vertex: The Main Character

## Optimal or Minimum/Maximum value

## X-Intercept, Zeros, and Roots

**But be carefu**l. There can be one, two, or even no x-intercepts. X-intercepts are also called

*zeros*and

*roots*, so don't worry if you see these words used interchangeably.

## Some Other Interesting Facts About A Parabola

- A parabola can open up or down. This will be shown in a section below.
- The vertex is the point where the axis of symmetry and the parabola touch. This is the optimal value (minimum or maximum).

## QUADRATIC EQUATION VS LINEAR EQUATION

## What's the difference?

- A linear equation creates a straight line, whereas a quadratic equation creates a curved line.
- In a quadratic equation, the variable is squared.
- In a linear equation, the
**first differences**are constant. In a quadratic equation, the**second differences**are constant.

"Wait, wait, wait! Did you just say second differences?"

Yes, I did. In grade 9, we worked with first differences, the differences between two consecutive terms. The second differences are the differences between the first differences. Confusing? Here's a picture to help:

## VERTEX FORM

## What Is Vertex Form

An example of vertex form is:

## "a" Value

*a*" in the vertex form equation changes the vertical stretch or compression of the parabola. The higher the

*a*value, the narrower the graph. The lower the

*a*value, the wider the graph.

The *a *value can change a few things in the graph:

- If
*a*is**greater than 0**(meaning**positive**), then the graph**opens upwards**. When a is positive and the parabola opens upwards, the vertex is a minimum value. - If
*a*is**less than 0**(meaning**negative**), then the graph**opens downwards**. When a is negative and the parabola opens downwards, the vertex is a maximum value - When
*a*is negative, it is called a**vertical reflection**, or**reflection in the x-axis**. This is because the graph seems to have reflected and is now facing downwards.

- If
*a*is**greater than 1**, then the graph is**stretched**. This makes the graph narrower. - If
*a*is**greater than 0, but less than 1**(meaning it's a decimal/fraction), then the graph is**compressed**. This makes the graph wider.

## "h" Value

*h*" in vertex form is responsible for the

**horizontal translation**of the graph. This means that

*h*moves the graph left or right. There are a few things to remember:

- The
*h*value moves the vertex left or right, and this is called a**horizontal translation**. - The translation always starts at (0,0).

**opposite**way of the sign.

- This means that is
*h*is**negative**, then the vertex will move*h*units**right**(which is the positive part of the x-axis). - If
, then the vertex will move*h*is positive*h*units**left**(which is the negative part of the x-axis).

- The
*h*is also the**x-coordinate of the vertex**, but with the**opposite sign**.

## "k" Value

*k*" in the vertex form is used to translate the graph up or down, meaning that it moves the graph up or down. There are a few things to remember about the

*k*value as well, but it's easier than h.

- The
*k*value moves the vertex up or down, which is called a**vertical translation**. - Like the
*h*value, the vertical translation should start from (0,0). You can put the horizontal and vertical translation together to find out where you must move from (0,0).

Unlike

*h*, the

*k*value moves in the

**same way as the sign**, meaning:

- If
*k*is**positive**, then the graph will move*k***up**. - If
*k*is**negative**, then the graph will move*k*units**down**.

- The
*k*is the**y-coordinate of the vertex**.

## Put them together!

*a*changes the

**stretch/compression**and/or

**reflection**,

*h*changes the

**horizontal translation**, and

*k*changes the

**vertical translation**, we are almost at the point where we can graph.

First, let's try out an example, and see what information we have. We'll use the equation:

Lets look at the *a* value:

The *a *value is -2.

- We remember that
*a*changes stretch/compression and reflection. - Because
*a*is negative, the parabola is vertically reflected. This means that the parabola opens down. - Also, we can see the 2 in
*a*. This means that the parabola is stretched by a factor of 2.

Now you might be wondering what stretched is, and how we use it. That will be explained later, when we learn to graph the vertex form.

Next we can look at the *h* value:

The *h *is -1.

- We remember that
*h*is responsible for the horizontal translation. We also remember that the vertex will translate the opposite direction of the sign. - You can see that the
*h*is negative. This means that it will translate to the right, and it will translate by 1 unit.

After that, we can look at the *k. *

The *k* is 3.

- The
*k*is used to translate the graph vertically. This will move the graph up or down. - Remember, unlike the
*h*, the k moves in the same direction of the sign. - Because the
*k*is +3, the graph will move 3 units up.

Finally, we know what will happen to our graph. It is reflected vertically, stretched by a factor of 2, translated 1 unit right, and translated 3 units up.

- Notice the last two points: translated 1 unit right, and translated 3 units up.
- Earlier, when describing
*h*and*k*, we said that the translation would start at (0,0). - If it moves 1 to the right (x-axis), and 3 up (y-axis), what would the coordinates of the vertex be in (x,y)?

The vertex would be (1,3). This is why this form is called vertex form, because the vertex is written in the equation. Look at y=-2(x-1)^2+3.

- See -1? That's the 1 of the vertex (x-axis). Remember, the sign is opposite.
- See +3? That's the 3 of the vertex (y-axis).

Try to make the equations for the following:

- Stretched by a factor of 2. Vertex: (-3,4)
- Reflected in the x-axis. Compressed by a factor of 0.5. Vertex: (2,-5)
- Reflected in the x-axis. Stretched by a factor of 4. Vertex: (1,1)

Write the properties for the following parabolas based on the equations. Include stretch/compression, whether or not it is reflected in the x-axis, the vertex along with the translations it went through to get there, and if the vertex is a minimum or maximum point.

- y=-3(x+2)^2-5
- y=0.5(x-5)^2+2
- y=2(x-1)^2-7

We are almost ready to graph using vertex form. All that is left is to learn of something called the step pattern:

## The Step Pattern

Take a look at the graph for y=x^2:

*x*is 1,

*y*is 1. But when

*x*is 2, y

*is 4, and when*

*x*is 3,

*y*is 9. What do you notice? The y is x^2, x squared.

This is what is meant by the variable is squared. The y is equal to *x *times *x*. So if *x *is 4, the *y *is 16. This is a step pattern. The step pattern is a way to describe what the y is when the x is a certain number, and in this case: 1 and 2. Remember, the step pattern for this parabola is "Over 1, Up 1. Over 2, Up 4".

You begin the step pattern at the vertex. You can use this step pattern moving both left and right of the vertex, because a parabola is symmetrical from the axis of symmetry. This means that you can go left 1, up 1, left 2, up 4 AND right 1, up 1, right 2, up 4. You use the step pattern to plot points on your graph so you can sketch the parabola.

Now you might be wondering "Okay, so I know how to graph y=x^2. Just start at (0,0), and go over 1, up 1, over 2, up 4. But how would I graph something like y=-2(x-1)^2+3?"

The step pattern will change as the *a *value changes. When the graph is stretched or compressed, the steps will increase or decrease, accordingly. Also, if *a *is negative, then the step pattern will be "Over, Down. Over, Down" instead of up.

To find a step pattern for an equation, you can use the basic step pattern of Over 1, Up 1, Over 2, Up 4. We can just multiply all of the *y* values by the *a *value. For our equation y=-2(x-1)^2+3, we will multiply the *y* values by -2. It will look like:

Over 1, up 1*a*. Over 2, up 4*a*.

Using this, the new pattern would be: Over 1, down 2. Over 2, down 8.

Do you notice how this changes? When the a value changes, the step pattern changes. To recap, just multiply the *y* value by the *a* value, and you found the new step pattern.

So now we have the vertex (because of the translations), the direction of opening (reflection), and the step pattern. we are ready to graph!

## Graphing With Vertex Form

Let's look at y=-2(x-1)^2+3, and how to graph it. Here are the steps:

1. Notice the stretch and the direction of opening.

In this equation, the vertical stretch is by a factor of 2. Also, the direction of opening is down because it is negative.

2. Find and plot the vertex.

We have already found the vertex for this equation. By looking at the equation, we know that the vertex is (1,3). We can plot this point on the graph.

3. Find points on the graph.

You can do this by finding the step pattern of the graph. Remember, to find the step pattern by an equation, multiply the original step pattern by the *a* value. It is Over 1, up 1*a, *Over 2, up 4*a*. The new step pattern for this equation is "Over 1, down 2, Over 2, down 8".

Begin at the vertex that you plotted, and use the step pattern to find 2 points on both the right and left of the vertex. This means go left 1, down 2, left 2, down 8. Also go right 1, down 2, right 2, down 8.

Another way to do this is to sub in a point on the x-axis into the equation as the *x*, and solve for y. The *x* would be the x-coordinate of this point, and the *y* that you solved for would be the y-coordinate. For example, if our vertex is (1,3), I can look for the *y* when *x* is 2. so it would be:

y=-2(2-1)^2+3

y=1.

So a point on my graph would be (2,1). However, you should probably use the step pattern, as it is faster and you wont need to do the equation to find every point.

4. Sketch the parabola.

Now that you have the vertex and 4 other points, you can draw the U-shaped parabola. Just sketch it like its connect the dots, but extend it a bit and add arrows to each side. Your graph should look something like:

## Practice Word Problem

*h*represents height and

*t*represents time.

1. Graph this equation.

2. For how long was the ball in the air?

3. What was the highest point the ball reached, in meters?

4. At how many seconds did the ball reach this height?

Because this is a practice problem, we will solve this for you, so you can take this as an example.

1. Graph this equation.

For this question, we can look at the 2 x-intercepts. The first one on the left would represent where the ball was kick. From there to the other x-intercept would be the time that the ball was in the air.

The ball was in the air for 12 seconds.

3. What was the highest point the ball reached, in meters?

Now we need to look at the vertex, as it is the maximum point in this graph. We can look for the y-coordinate of the vertex, and that will be our highest point.

The highest the ball reached was 9 meters.

4. At how many seconds did the ball reach this point?

For this, we can look at the x-coordinate of the vertex.

The ball reached 9 meters at 6 seconds.

## Determining the Equation of a Parabola When Given the Vertex

Let's say the vertex is (-2,8), and the parabola passes through the point (2,0).

Remember how we found the vertex by looking at *h* and *k*? Also, remember how it was mentioned that the vertex was actually in the equation and that's why the form was called vertex form. Well, if you're given the vertex in a question, you already have 2 variables: *h* and *k*.

*h* is the x-coordinate of the vertex, but with the opposite sign. In this case, it would be 2.

*k* is the y-coordinate of the vertex. In this case, it would be 8

You can put these points into the vertex form equation. So at this point, our equation is:

Next, we know that this parabola passes through the point (2,0). We can substitute these values in for *x* and *y* in our equation. Now, our equation is:

At this point, the only variable left is *a*, meaning that we can solve for *a* by isolating it.

Now that we have *a*, we can place it into our equation, and change *x* and *y* back to variables. Finally, our equation is:

Remember, when you're solving one of these questions, first, put the *h *and *k* values into the equation, then substitute *x* and *y* to solve for an *a* value, and finally, but *a* into the equation to finish the equation.

Let's try another one.

"Find the equation for a parabola that was a vertex of (2,6) and passes through the point (5,3)."

The solution would look like:

Therefore, the equation for the parabola is:

## FACTORED FORM

## What is Factored Form?

## x-intercepts

*r*and

*s*in the equation. Once again, the signs are flipped, similar to the

*h*in vertex form. This means that if the x-intercept is negative, it will be positive when graphing, and vice versa. Here are what the x-intercepts look like:

## a Value

*a*value from factored form, at the beginning of the equation. Once again, the

*a*value represents the stretch. However, the

*a*value is not needed for graphing. You can still use it to make sketching the graph easier, or to check your graph by using the step pattern to see if you sketched it correctly.

## Finding the Vertex with x-intercepts

*y*.

## Graphing with the x-intercepts and the Vertex

1. Plot the x-intercepts. We will put points at (5,0) and (-3,0). Remember, for x-intercepts, the y-value is 0.

Let's try another example:

## STANDARD FORM

## What is standard form?

## Graphing standard form

**factored form**, which we can then use to graph (because we learned how to do that above).

Factoring will be taught below, after multiplying binomials. However, here is an example of graphing standard form. Keep in mind what is happening so that you will understand how to graph.

## MULTIPLYING BINOMIALS

## What is a binomial?

**factored form to standard form,**and

**vertex form to standard form**

**.**A question that involves multiplying binomials could look like:

If you're wondering how to convert standard form to factored form, that is done by factoring, which is further down on this site.

## Perfect Square Binomials

These special binomials are perfect squares, meaning all of the binomial is squared. This is also what you would use to convert **vertex form to standard form**. Examples are shown below:

## Positive

## Negative

## Product of sum and difference

*b*and a negative

*b*. The equation also changes in this scenario as well. The equation for this special product is:

*ab*is cancelled out because 1 is positive and one is negative. This can be shown with the distributive property:

## Algebra tiles

## FACTORING

## What is factoring?

**standard form to factored form**, and even

**standard form to vertex form (completing the square)**. We are going to learn how to factor algebraic expressions in a few ways. We will cover:

- Common factoring
- Factoring by grouping
- Factoring simple trinomials
- Factoring complex trinomials
- Special products
- Factoring perfect square trinomials

## Common factoring

Now, let's try a harder example:

## Factoring by grouping

## Factoring simple trinomials

*a*= 1. An example of a simple trinomial is:

- The
*b*value of standard form is equal to*r*+*s*of factored form. - The
*c*value of standard form is equal to*r*x*s*of factored form.

*c*, but there is a catch: we need 2 numbers that have a product of

*c*but a sum of

*b.*In other words, the numbers need to add to

*b*and multiply to

*c*.

To do this, we need to find factors of *c*, and whichever pair equals to *b* is the factors we are looking for. Lets look back at our example.

*c,*2 in this case, in the first column. Next, you will multiply them to ensure that they equal

*c*. Finally, you will add them to check if they equal

*b*. If they equal

*b*, these are the factors you are looking for. If they do not equal

*b*, you can check the next factors of

*c*.

So now we have the numbers that have a product of 2 and a sum of 3. We can finally fully factor.

Since we already know that the factors of x^2 are x and x, we can place those.

*a*value is always 1. This let's us know that there is

*x*with no coefficient in each bracket. Let's try one more, but with a negative number:

*b*is negative and the

*c*is positive. If we remember integers, negative times negative equals positive. So let's use the chart to find our factors.

## Factoring complex trinomials

*a*does

**not**equal 1. Also, when factoring complex trinomials, you also need to find the factors of the first term. A complex trinomial can look like this:

- First look for common factors, and remove them. If possible, turn the complex trinomial into a simple trinomial by factoring the
*a*value from the trinomial. If you can do this, then factor it like a simple trinomial. - Write the factors of the first and last term. In a
**simple**trinomial, you only need the factors of the 3rd term, because the factors of the first term were always*x*and*x*. However, in a complex trinomial, the*a*value is not 1, so you must find the factors.

Once again, you will need specific factors. Each of the factors of the first term must multiply with one of the factors of the third term. Then, when you add it all up, it should equal the middle term.

*b*x

*,*or -15x in this case. If it does not equal that, then we will need to try again but with different factors.

## Factoring the Difference of Squares

*x*is squared, but is that it? Nope. 100 is also a square. It is the square of 10. So in the difference of squares, we have a square subtract a square.

Look back to MULTIPLYING BINOMIALS, and look at *The Product of Sum and Difference*. Remember:

*a*+

*b*)(

*a*-

*b*) turns into

*a*squared subtract

*b*squared. Now look at this expression once again:

This expression follows the the *a* squares subtract *b* squared expression. In this case, *x* is *a* and 100 is *b*. This must mean that it is also equal to (a+b)(a-b). If we want to factor this expression, we must get it back to (a+b)(a-b).

Another reason why this equation must be (a+b)(a-b) is because there are 2 squares, and the sign is *negative*. A sign becomes negative when it is positive and negative, meaning that one must be positive and the other must be negative.

*a*and

*b*, we must find the square root of both of these.

*a*and

*b*, we can fill it into (a+b)(a-b).

*x*is squared and 64 is squared. If we find the square roots of all of these, they are:

5

*x*

8

This means *a* is 5*x* and *b* is 8. We can fill this into (a+b)(a-b).

## Factoring Perfect Square Tinomials

*a*and

*b*are squared. We will now see how to factor this.

To factor this, you can just take *a* and *b*, put it in a bracket and square it. The factored form of this looks like:

*x*has to be equal to 2

*ab*. We already know that

*a*is 5 and

*b*is 6. If we do 2(5)(6), we get 60. This just confirms that the equations are equal.

## COMPLETING THE SQUARE

## Put it together

**standard form into vertex form**. Briefly, The steps for completing the square are:

1. Make the standard form equation look similar to vertex form.

This can be done by factoring our the *a* value and separating the *c* to make it look like *k* from vertex form.

2. Expand what's in the brackets to make it a perfect square trinomial.

You will need to add and subtract the *c* from the perfect square trinomial.

3. Factor the perfect square trinomial.

4. Apply the distributive property from *a*.

5. Add like terms.

This will be better shown with an example:

In this step, we will factor our the *a* value, and separate the *c* by brackets.

In this step, we need to turn what is in the middle into a perfect square trinomial. To do so, we need a *c* value. In order to get the *c* value we need for a perfect square trinomial, divide the *b* value by 2. (This is because in a perfect square trinomial, the middle value equals to 2*ab* from the binomial.) In this case, -4 divided by 2 is -2. -2^2 is our *c* value, because a perfect square trinomial is *a^2 + 2ab + b^2.*

We will need to add the *c* value, and then subtract it. This is because we cannot just add a number into the equation, as that changes the equation completely. However, + *c* - *c* = 0, so we can do that.

In the final part of this step, the perfect square trinomial is in a bracket in a bracket. This is only to help organize, and is not necessary. Everything inside the bracket will be multiplied by the distributive property of 3 soon, however, putting the perfect square trinomial in a bracket just helps you see what will be factored.

Now we will factor the perfect square trinomial. We need to find the square roots of *x^2* and 4, which is *x* and 2. However, the *b* term is negative, meaning that the 2 is negative. Our factored form of the perfect square trinomial is (*x*-2)^. We can keep this inside of the bracket, as it will still be multiplied by 3.

The -4 (which was used to counter the +4 from the perfect square trinomial) will remain inside the bracket, and wasn't factored because it wasn't a part of the perfect square trinomial. It will also be multiplied by the 3.

Next, we will multiply everything that is inside of the bracket by the *a* value, which is 3. The (*x*-2)^2 will be multiplied by the 3, and the -4 will be multiplied by the 3 as well.

Finally, we can finish this by adding like terms. We must the number we left outside of the bracket to the number that was multiplied by the *a* value. In this case, the number outside was -5, and the number we got after multiplying by 3 was -12.

The *a* is 3, the *h* is -2, and the *k* is -17.

Let's try another example. This example will have a negative *a* value, so we will need to factor out -1 from everything inside the brackets.

## SOLVING QUADRATIC EQUATIONS

*x*when

*y*is 0. By solving a quadratic equation, we can find the axis of symmetry, optimal value, and all together, the vertex. This will allow us to graph our equation.

## Solving Quadratic Equations by Factoring

In factored form, if the equation is equal to 0 (the *y*), that means that one or both of the factors is equal to 0. So solve for *x*, we will set each of the factors to 0, and isolate *x*. It should look like this:

*x*'s are the roots. This is the same thing that you do to graph form factored form, and is very simple. Let's try an example:

Next, let's try an equation with a coefficient in front of *x. *Since the *x* is multiplied by the coefficient, the only new step is to divide both sides by the coefficient. Let's try an example:

*x*. The solutions to this quadratic equation were -4 and 4.

## Solving Quadratic Equations Using the Quadratic Formula

*a*from standard form will replace the

*a*in the formula. The same thing with the

*b*and

*c*.

Also, look back at the formula. There is a plus and minus sign together. This means that the formula can be used twice: one with a plus in that spot and one with a minus in that spot. Since the formula can be used twice, each time will give a different root, for a total of 2 roots.

Let's try an example:

*a*,

*b*, and

*c*into the quadratic formula so that we can solve the 2 roots.

There is one more thing to bring to attention. Look back to the work for the solution. There is a square root in there. The number that is being square rooted is called a **discriminant**. In this quadratic equation, the discriminant was 52. The discriminant can help us know **how many** roots there will be.

- If the
**discriminant is positive**, there will be**2 roots**. This is because the number will be squared, and either added to or subtracted from the number before it. - If the
**discriminant is negative**, there will be**no roots**. This is because a negative number cannot be square rooted, as a square is a number times itself. To get a negative sign, the signs multiplied must be opposite, so you cannot square root a negative following this concept. - If the
**discriminant is zero**, there will be**1 root**. This is because the square root of 0 is 0, so nothing will be added or subtracted to the number before it, and it will just be the number*-b*divided by 2(*a*).

## Solving Quadratic Equations by Isolating X

*x*. This way can be used in vertex form, and is very simple. It is isolating the variable to find out what it is. This method also includes a plus-minus sign. This way is better shown with an example, as there is no real formula. Let's use this example:

*x*.

*x*. This can give us 2 roots: 1 when the sign is positive, and the other when the sign is negative.

Like the quadratic formula and the discriminant, there is also a way to find out how many roots we will have by looking at the equation. For vertex form, we must look at the *a* and *k* values:

- If
*a*is positive, and*k*is positive, there will be 0 roots. - If
*a*is positive and*k*is negative, there will be 2 roots.

- If
*a*is negative and*k*is positive, there will be 2 roots. - If
*a*is negative and*k*is negative, there will be 0 roots.

- If
*k*is zero, there will be 1 root regardless of what the*a*is.

## LINKING TOPICS

- Expanding binomials will help you get from factored and vertex form to standard form.
- Completing the square will get you from standard form to vertex form. This also requires factoring (perfect square trinomials).
- Factoring will also help you get from standard form to factored form.
- Factoring relates to graphing because it can help us get roots.
- We can use quadratic formula from standard form to get roots.
- Roots can help us get the vertex, and graph the parabola.
- Vertex form relates to graphing because it shows us the vertex and translations, and also the step pattern.
- Learning about discriminants will help us know how many roots there are.
- Algebra tiles can help with multiplying binomials and factoring.

## WORD PROBLEMS

## Motion Problems

*t*seconds is modelled by this equation:

b) Find the times when the tennis ball is at a height of 4.5 m above the ground. Round your answers to the nearest tenth of a second.

c) What is the maximum height of the tennis ball? At what time does it reach this height? Round your answers to the nearest tenth.

To answer these:

For a), we need to find the roots. For b), we need to find the roots when y=4.5. For c), we need to find the vertex. We can answer these questions using the quadratic formula.

## Revenue Problems

Calculators are sold to students for 20 dollars each. 300 students are willing to buy them at that price. For every 5 dollar increase in price, there are 30 fewer students willing to buy the calculator. What selling price will produce the maximum revenue and what will the maximum revenue be?

To answer this:

We will first need to define our variables. Next, we will need to make our equation. We can use factored form for this. Then, we must find our roots and vertex.

## Area Problems

a) Write an equation that can be used to find the length and width of the rectangle.

b) What are the dimensions of the rectangle?

To answer this:

First, define your variables. Next, create an equation (remember, area = *l***w*.) Then, solve for your variables. Here's an important tip: dimensions cannot be negative. This narrows your roots to only one. Also, drawing a diagram can help.

## Consecutive Number Problems

**product**of two consecutive numbers is 1482. What are the two numbers?

To answer this:

First, define your variables. Next, create an equation. After this, you can solve the quadratic equation.

## Right Triangle Problems (Pythagoras' Theorem)

To answer this:

Define variables. Use Pythagoras' Theorem. Base the lengths off of 1 variable. Solve for that variable. (Hint: a diagram can be helpful.)

## Optimization Problem (Area)

To solve this:

First, define variables and create an equation. Next, you must find roots, axis of symmetry and the optimal value. Also, a diagram might help.

## REFLECTION OF THE UNIT

Here is one of my assessments from this unit:

- For the matching part, I was able to identify the types of equations and which methods would be used from practice.

- Number 6 was multiplying binomials. I multiplied the 2 binomials, then added like terms.

- Number 7 was factoring. The first question was a complex trinomial. I found the factors by finding the factors of
*a*and*c,*and making sure that when multiplied and added, they would add to*b*.

- Number 8 was factoring, and then solving for the roots. To do this, i factored each equation, then made it equal to 0. After this, I isolated the
*x*'s to find the 2 roots.

- In number 9, I once again used factoring to find the roots. Once I had the roots, I found the axis of symmetry and the optimal value, which gave me the vertex. Finally, I graphed it by plotting the vertex and the roots.

- In number 10, I created an equation for each shape. I made an equation to find the area of the square and an equation to find the area of the rectangle. I then subtracted the area of the rectangle from the area of the square, and added like terms. This gave me the area of the shaded region. I then factored it in the next question using common factoring.

- In number 11/12, I was able to find the answer that would give one root, which was 30. I got this because the trinomial in the question was a perfect square trinomial. When there is a perfect square trinomial, there is only 1 root, so I solved for the
*b*of the equation.

All in all, this assessment was good. I got a great mark, and I learned a lot from this part. Factoring and multiplying binomials is an important concept to understand, and I'm proud that I did good on the test.

That's all that's left on this Smore. However, if there is something you need clarification on, or need more practice, you can visit: