Free Homework Help https://schooltutoring.com/help Our education staff publish regular articles, tips and tutorials to help students with their homework Thu, 29 Aug 2024 16:24:58 +0000 en-US hourly 1 https://wordpress.org/?v=6.5.5 How to Construct an Essay https://schooltutoring.com/help/how-to-construct-an-essay/ Thu, 29 Aug 2024 16:24:58 +0000 https://schooltutoring.com/help/?p=67367 Essay writing is often a challenging endeavor for students, but it doesn't have to be. Here we will outline some tips to make essays writing a breeze!

The post How to Construct an Essay first appeared on Free Homework Help.]]> For many students, constructing a concise, compelling, and coherent essay can be a daunting task. It doesn’t need to be. Here we will look at some strategies and concepts that can be applied in the essay writing process.

Essay Structure

Most of the essays that students have to write in high school are 5-paragraph, argumentative essays. The purpose of such an essay is to convince the reader of an opinion, or thesis. As its name suggests, a 5-paragraph essay has 5 paragraphs, as outlined below:

  1. Introduction
  2. Body paragraph 1
  3. Body paragraph 2
  4. Body paragraph 3
  5. Conclusion

Looking at each of these paragraphs will give us a good idea of how to approach writing an essay.

Introduction

Unsurprisingly, the introduction paragraph focuses on introducing the reader to the topic of our essay. Often this begins with a hook. A hook is something that should grab your reader’s attention in relation to the topic of your essay. It can be a quote, anecdote, statistics, question, or anything else that makes a reader interested in the essay.

After beginning with a hook, we move to introducing the topic of our essay with a thesis statement. This is often a single sentence that explicitly states what the essay is about. Remember that in this type of essay (argumentative), we are expression our opinion or point-of-view on something. As such, our thesis statement should resemble our perspective as opposed to a statement of fact. For example, “grass is green” is not a thesis statement, but “pizza is the best type of food” is a valid thesis statement.

The final component of our introduction is to briefly state our points that support our thesis statement. There should be 3 of these, each corresponding to one of our body paragraphs.

Body Paragraphs

In each body paragraph, we focus on one of our supporting points that we listed in our introduction. In these body paragraphs, we will expand upon these points, provide evidence, and show how the point contributes to our thesis. For many essays in English classes, it is good practice to include a quote to support each point in your body paragraphs.

A best-practice to make writing body paragraphs easier is to use the “sandwich” method. In this method, we focus on 3 components, the top slice of bread, the insides, and the bottom slice of bread. The top slice of bread is a sentence that introduces the reader to the supporting point your body paragraph is about. It is a miniature introduction. The insides (i.e. lettuce, tomatoes, sliced meat) are the quotes, details, and other evidence that you use to expand on the focus of this paragraph. This should usually be 3-5 sentences. Finally, the bottom slice of bread is a miniature conclusion. Here, we restate the supporting point that this paragraph is about, how it relates to our thesis, and introduce the next paragraph.

Conclusion

Finally, we end our essay with a conclusion. In this paragraph we restate our thesis and supporting points. Aside from having a hook, our conclusion will have much of the same information as our introduction. There are two main differences though.

  1. In our conclusion we are ending our essay as opposed to introducing it. Our word choice and tone should reflect that level of finality.
  2. We have already written an entire essay trying to convince the reader of our thesis. As such, we can be bold and confident as we restate our thesis and supporting points. The reader may still disagree with our thesis, but we should look to present our thesis as though it has now been proven.

Additional Tips

Writing an essay involves more than adhering to the structure outlined above. Some additional considerations to remember are as follows:

  • Consistent verb tense – avoid mixing the past, present, and future tenses
  • Consistent perspective – in some essays 1st person (I, me, etc) is preferred. In others, the 3rd person (it, them, they) is better. Pick one and stick to it.
  • Make sure your supporting points always link back to your thesis – don’t get too focused on presenting everything you know about the topic of the essay. Everything should relate to and contribute towards supporting your thesis

This has been a brief introduction to essay structure. For more information on this topic as well as assistance with homework and test preparation, feel free to reach out to an Academic Director toll-free at 1 (877) 545-7737 or via our Contact Us page.

The post How to Construct an Essay first appeared on Free Homework Help.]]> Mean, Median, and Mode https://schooltutoring.com/help/mean-median-mode/ Tue, 30 Jul 2024 21:16:35 +0000 https://schooltutoring.com/help/?p=67366 Examine 3 key components in statistics: the mean, median, and mode. We will cover what these represent, how to calculate them, and their various applications. Examples are included.

The post Mean, Median, and Mode first appeared on Free Homework Help.]]> Broadly speaking, statistics is a tool that allows us to describe and compare large amounts of information in a digestible way. Some of the most basic, as well as most useful metrics when analyzing a dataset or distribution are the mean, median, and mode. Let’s take some time to define each of these terms, use them in examples, and describe situations where they may be useful.

Mean

Often referred to as the average, the mean of a dataset is the sum of values of a dataset divided by the number of values in a dataset. Mathematically, this is often represented as:

    \begin{align*} \frac{1}{n}\sum_{i=1}^{n}x_i \end{align*}

If that looks too complicated, don’t worry, it just means that we add all of our values in the dataset (the x‘s) and then divide by the number of values in the dataset (n).

Example

Let’s look at an example. Suppose we have the following values from a dataset:

    \[1, 1, 2, 4, 5, 7, 9\]

To compute the mean of these 7 values we do the following:

    \begin{align*} Mean = \frac{1+1+2+4+5+7+9}{7}=4.14... \end{align*}

Application

We might now ask “why do we care about the mean?” Let’s answer that via an example. Suppose you are a baseball player trying to hit the fastball of a pitcher. To anticipate how to approach your at-bat, you look at how fast the pitcher throws their fastball. You look at the pitcher’s last 5 fastballs, which were 90 MPH, 91 MPH, 88 MPH, 93 MPH, and 89 MPH. You could calculate the average speed of these pitches (90.2 MPH) to predict how much time you have to react to that pitcher’s fastball. A useful tool indeed!

Median

Next, we will consider the median of a dataset. The median is simpler to calculate than the mean. We simply sort the values in a dataset from least to greatest and choose the middle one as our mean. If there are an even number of values in the dataset, we simply find the mean of the middle two values.

Example

Let’s look at an example. Suppose we have the following values from a dataset:

    \[1, 4, 9, 1, 5, 7, 2, 8\]

First, we need to sort the values from least to greatest, giving:

    \[1, 1, 2, 4, 5, 7, 8, 9\]

This sorted dataset has 8 values, so we start by picking the middle two: 4 and 5. We then compute the mean of these two values to get 4.5 as the median of this dataset. You may notice that this is the same dataset as the one we used in the mean, but that the median and mean are different. Don’t worry, we will get back to that.

Application

The median of a dataset is similar to the mean of a dataset, except it is not affected by outliers: extreme values that can skew the mean. For example, let’s say you are analyzing the traffic speed on a highway. You observe the speeds of 5 cars: 55 MPH, 61 MPH, 58 MPH, 59 MPH, and 103 MPH(!). If we took the mean of this dataset we might predict that most cars are travelling at 67.2 MPH, but this is much higher than the speed that most cars are actually travelling. By choosing the median speed (59 MPH), we reduce the impact of the one car that is speeding and obtain a more realistic prediction for the speed of most cars.

Mode

The mode of a dataset is the least commonly used, but has some niche applications. It is simply the most commonly occurring value in a dataset. If there are multiple values that occur the same number of times, then there will be multiple modes.

Example

Let’s look at an example. Let’s return to the same dataset from our other examples:

    \[$1, 1, 2, 4, 5, 7, 8, 9\]

The mode of this dataset is 1, as it occurs twice and all other values only occur once.

Application

The mode is most useful when data is expected to represent pre-defined categories. A good example is shirt sizes. It is difficult to conceptualize what the average shirt size is (what would a medium-and-a-half even mean?) and we know that the median shirt size will always be medium or large depending on the number of available sizes. This is where the mode is important. Perhaps we are a clothing store and we want to know what shirt size is most common. The 5 most recent customers purchased the following shirt sizes: M, M, L, M, XL. The mode of this dataset would be M, as it occurs 3 times. We can then predict that medium is the most common size and order shirts accordingly.

Comparison

The astute reader of this post will notice that we used the same distribution in an example of the mean, median, and mode. A natural question would be “how do I know which one to use?”. Unfortunately, there is no universal answer to this question, but we can suggest the following:

  • If you want to make observations and predictions about a group and include extreme outliers, use the mean
  • If outliers are not important to your conclusions, use the median
  • If you are analyzing categories or binned (grouped) data, use the mode

This has been a brief introduction to the mean, median, and mode. For more information on this topic as well as assistance with homework and test preparation, feel free to reach out to an Academic Director toll-free at 1 (877) 545-7737 or via our Contact Us page.

The post Mean, Median, and Mode first appeared on Free Homework Help.]]> What is the Unit Circle and Why is it Important? https://schooltutoring.com/help/the-unit-circle-and-its-importance/ Wed, 27 Mar 2024 17:40:49 +0000 https://schooltutoring.com/help/?p=67337 We explore how to use the unit circle to understand some properties of trigonometric ratios and why this circle is so important in math.

The post What is the Unit Circle and Why is it Important? first appeared on Free Homework Help.]]>

What is the Unit Circle?

The unit circle is a tool that allows us to define and calculate some trigonometric ratios (sin, cos, tan, etc.) and uncover some relationships between these ratios with ease.

As its name suggests, it is a circle with a radius of one. This means that if you take a string of length one (it could be 1 m or 1 cm, or any unit) and hold one end of the string fixed while rotating the other, you would have draw a unit circle.

To understand the relationship between the unit circle and trigonometric ratios, we must choose a random point in the circle. Let’s call this point (x,y). The angle formed between the positive x-axis and the line joining the center to the point (x,y) will be called \theta. As we change (x,y) around the circle, you can see that we can get any angle you want.

Nevertheless, this property is not unique to the unit circle (we can do this with any circle), so what is so special about this circle?

Why is the Unit Circle Special?

Now that we have our angle \theta and our point (x,y) we can build a right triangle by tracing a line perpendicular to the x axis, that passes throw the point (x,y).

We observe something quite interesting in this triangle, the hypotenuse of all of these triangles is one! It does not matter which angle you use, they all have this property.

This may not look like a big deal, but when you try to calculate trigonometric ratios this changes everything.

Calculating trigonometric Ratios

The formula to calculate the sine and cosine for an angle are:

    \begin{equation*} \begin{split} \sin(\theta) &= \frac{opposite}{hypotenuse} \end{split} \quad \quad \begin{split} \cos(\theta) &= \frac{adjacent}{hypotenuse} \end{split} \end{equation*}

Since the hypotenuse is one, and knowing that the lengths of the opposite side of the triangle is x and the adjacent is y, we obtain

    \begin{equation*} \begin{split} \sin(\theta)& = \frac{y}{1} \\ & = y \end{split} \quad \quad \begin{split} \cos(\theta) &= \frac{x}{1} \\ &= x \end{split} \end{equation*}

Notice that sin(\theta) is equal to y while \cos{\theta} is equal to x. Wait a second, those are exactly the coordinates of the point we chose to form theta. That means that our point can be written as (\cos(\theta), \sin(\theta)). With this notion now we can define the trigonometric ratios of any angle without using right triangles.

Trigonometric functions for angles bigger than 90°

Ever wonder how is it possible to find the sin(\theta) for \that>90^{\circ}? This is a valid question. Sine is in fact opposite over hypotenuse, but you can only have a hypotenuse in a right triangle. However, if the angle is bigger than 90^{\circ} then what is the hypotenuse of a triangle containing this angle. The answer is, it does not exist.

We can solve this using the unit circle. We can generalize the property of angles in the first quadrant (x,y)=(\cos(\theta, \sin(\theta))) and say that this is true for all angles. That is, the point (x,y) in the circle contains the \cos(\theta) in the position of x and the \sin(\theta) in the position of y.

Therefore, if you are asked for the sine of \theta=180^{\circ} then you draw the angle, draw the unit circle and find the point (x,y) where the angle connects to the circle. In the case of 180^{\circ} that point is (-1,0) based on our definition then \cos(180^{\circ})=-1 and \sin(180^{\circ})=0.

Amazingly, in this example, we have the cosine of an angle being negative and the sine being zero, something impossible under the usual definition of the trigonometric ratios.

In fact the unit circle definition of sine and cosine is the standard way that mathematicians use to define these ratios.

What about Tangent?

I know what you must be thinking now, sure defining sine and cosine like this is good, but what happened to tangent? This is still defined as:

    \[\tan(\theta)=\frac{Opposite}{Adjacent}\]

We could stop there, but notice that if we input this using the unit circle property (x,y)=(\cos(\theta),\sin(\theta))

    \[\tan(\theta)=\frac{y}{x}=\frac{\sin(\theta)}{\cos(\theta)}\]

And now you could extract the tangent from sine and cosine that are now well defined for all angles.

This is great, but there is a way to see the tangent directly from the unit circle.

Consider the following picture

Notice that this is the same unit circle as before, but now we draw a  line tangent to the circle and we expand the radius to form a bigger right triangle. Using this triangle now we can calculate the tangent of the angle

    \[\tan(\theta)=\frac{Opposite}{Adjacent}=\frac{t}{1}=t\]

So the length of t (that comes from the tangent line) in the unit circle is equivalent to the tangent o an angle. And that is where the name tangent comes from. It is literally the tangent.

The unit circle makes the trigonometric rations appear in front of your eyes.

What Else can I Learn from the Unit Circle?

Did you notice that we found a relationship between tangent and sine and cosine? When you relate trigonometric ratios, the relationship is called an identity.

For example, the identity we found before is a quotient identity, as you found a relationship by diving trigonometric ratios.

    \[\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}\]

With the unit circle we can find one more identity. If we go back to our first triangle, we can use Pythagoras’ Theorem to find the following relationship

    \[1^2=y^2+x^2\]

Using again the property (x,y)=(\cos(\theta), \sin(\theta)) we find

    \[1=\sin(\theta)^2+\cos(\theta)^2\]

Wow, if you add the square of the sine and the square of the cosine of any angle you get one! This property is unintuitive, but really useful in certain situations, allowing us to change from sine to cosine at will. This property is called a Pythagorean Identity.

Similar identities can be found if we use our second triangle (Hint: the hypotenuse is the secant of the angle) therefore:

    \[\sec(\theta)^2=1^2+\tan(\theta)\]

Isn’t that relationship interesting?

The Unit Circle is Amazing

As we have seen, the unit circle can offer you new information that sometimes is hard to see if you use circles of different radii. However, the best application of the unit circle is its intimate relationship with trigonometric ratios. If you master the unit circle’s properties, then you will be a master of trigonometric ratios and identities, so spend some time memorizing them. It is also useful to try exploring them by yourself by drawing all different types of triangles using an angle, a point in the unit circle, and a line perpendicular to the x-axis or the y-axis.

This has been a brief introduction to the concept of the unit circle. For more information on this topic, as well as assistance with homework and test preparation, feel free to reach out to an Academic Director toll-free at 1 (877) 545-7737 or via our Contact Us page.

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The post What is the Unit Circle and Why is it Important? first appeared on Free Homework Help.]]> Factoring Expressions Using a Common Factor https://schooltutoring.com/help/factoring-expressions-using-a-common-factor/ Wed, 06 Mar 2024 21:52:50 +0000 https://schooltutoring.com/help/?p=67323 Learn how to factor algebraic expressions by dividing by a common factor. See how this can be a powerful tool in simplifying expressions. Examples included.

The post Factoring Expressions Using a Common Factor first appeared on Free Homework Help.]]> What is Factoring?

One of the most intuitive ways to approach factoring is to think of it as the opposite of expanding. This is demonstrated below (working from left to right):

Expanding:

    \begin{align*} \textcolor{red}{4x^2(x^3+2x)}\textcolor{black}{\quad\rightarrow\quad}\textcolor{blue}{4x^5+8x^3} \end{align*}

Factoring:

    \begin{align*} \textcolor{blue}{4x^5+8x^3}\textcolor{black}{\quad\rightarrow\quad}\textcolor{red}{4x^2(x^3+2x)} \end{align*}

This means that if we know how to expand, we can always check if our answer to a factoring problem is correct by making sure that when we expand we get the expression that we started with. This is similar to using subtraction to check addition, or multiplication to check division.

How to Factor

Here we will be showing how to factor polynomials where each term shares a common factor. Other methods of factoring do exist, but will be covered in a later post. This method of factoring has four steps:

  1. Determine the greatest common factor (GCF) of the terms in the algebraic expression.
  2. Divide each term by the GCF.
  3. Enclose the result of step 2 in brackets.
  4. Place the GCF outside of the brackets from step 3.

Let’s look at an example!

Suppose that we want to factor the following expression:

    \begin{align*} 5x^3+20x \end{align*}

Let’s follow the steps we just outlined.

  1. We have two terms: 5x^3 and 20x. To find the GCF of these terms, we first find the largest common factor of 5 and 20. This is 5. We then want the GCF of x^3 and x. This is x. We then combine this information to conclude that the GCF of our terms is 5x.
  2. We now divide each term by 5x. 5x^3\div5x=x^2 and 20x\div5x=4
  3. We now place the result of step 2 in brackets to get: (x^2+4)
  4. Finally, multiply our result from step 3 by the GCF to get: 5x(x^2+4)

We have now successfully factored our expression. We can then multiply each of the terms by 5x to confirm that we did the factoring correctly.

Why do We Factor?

Now that we have seen how to factor, you might wonder why this is important. An example is instructive. Consider the following expression:

    \[\frac{2x^2+3x}{x}\]

This might initially look like a complicated expression, but it is important to notice that we can factor the numerator. The GCF of the terms in the numerator is x. If we use that to factor the numerator our expression now looks like this:

    \[\frac{x(2x+3)}{x}\]

Since we are multiplying and dividing by x (opposite operations) we can “cancel out” the x in both the numerator and denominator to get the simplified expression:

    \[2x+3\]

Note: This is true as long as x\neq0. See our post on asymptotes and holes if you’re interested in what happens if x=0.

This shows us that factoring can be a powerful tool to manipulate and simplify algebraic expressions, potentially allowing us to solve problems that would otherwise be complicated.

This has been a brief introduction to the concepts of factoring. For more information on this topic as well as assistance with homework and test preparation, feel free to reach out to an Academic Director toll-free at 1 (877) 545-7737 or via our Contact Us page.

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The post Factoring Expressions Using a Common Factor first appeared on Free Homework Help.]]> What are Radians and Why we Use them https://schooltutoring.com/help/what-are-radians-and-why-we-use-them/ Thu, 29 Feb 2024 01:29:23 +0000 https://schooltutoring.com/help/?p=67320 A radian is the natural unit of angles, but why do we use them instead of degrees. In this post we discuss the relationship between angles, radians and degrees.

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Why do we use Degrees?

When we think about angles, the first thing that comes to mind is degrees. Examples are 30°, 45°, 90°, 360°, etc. However, have you ever wondered why we define angles in this way?

The answer lies in the properties of 360°. This number is special as it can be divided exactly by no more than 24 numbers (ie. 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360). In consequence, if we use 360° to represent the angle that covers the full circle, a lot of pieces of the circle will have “nice” angles.

For example, half a circle is represented by 180°, a third by 120°, a quarter by 90°, a sixth by 60° and so on. However, this choice is arbitrary. Nothing is stopping us from making the angle that covers a full circle 50 new degrees and having half being 25 new degrees. Let’s represent new degrees using *

Let’s find the conversion factor to go from usual degrees \circ to new degrees * for the angle \theta:

    \[\frac{\theta_{*}}{\theta_{\circ}}=\frac{50^{*}}{360^{\circ}}=\frac{5^{*}}{36^{\circ}}\]

If we multiply this factor by 90° we’ll see how much this angle is worth in new degrees:

    \[90^{\circ}\left( \frac{5^{*}}{36^{\circ}}\right)=12.5^{*}\]

This value is exactly what we would expect.

What are Radians and Why do we use them?

Now that we have established that we can choose how to measure our angles arbitrarily, we should be wondering if there is a superior way of measuring angles. The answer is, that it depends on what you are working on. Let’s see an example. Let’s say you want to find a formula to relate an angle with the arc it forms in a circle of radius r.

In this graph we show the arc (in green) formed by the 45° angle (called \theta_{\circ}) in a circle of radius r=5 (in blue):

Example of the arc formed by an angle in a circle of radius r.

We can find a formula to calculate the arc algebraically like this:

    \begin{align*} \frac{\theta_{\circ}}{360^{\circ}}&=\frac{arc}{2{\pi}r}\\ arc&=\frac{2{\pi}}{360^{\circ}}\left( r\theta_{\circ} \right)\\ arc&=\frac{{\pi}}{180^{\circ}}\left( r\theta_{\circ} \right)\\ \end{align*}

Where arc is the arc that the angle \theta forms in a circle of radius r. This formula is extremely useful, but can this formula be simplified?

Let’s see what happens if we transform the angle \theta to new units using the conversion factor k that transforms the angles from degrees to a new unknown unit. In our new units, our angle will be:

    \[\theta_{\text{new}}=k\theta_{\circ} \implies \theta_{\circ}=\frac{\theta_{\text{new}}}{k}\]

Using this in our formula we obtain:

    \[arc=\frac{{\pi}}{180^{\circ}}\left( r\frac{\theta_{\text{new}}}{k} \right)\]

Now we remember that we are free to choose k to be anything as we are free to choose the unit we measure our new angle with. Why not make it such that it cancels the constant in the front? Let

    \[k=\frac{\pi}{180^{\circ}}\quad \text{Then}\]

    \begin{align*} arc&=\frac{\pi}{180^{\circ}}\left( r\frac{180^{\circ}\theta_{\text{new}}}{\pi} \right)\\ arc&=r\theta_{\text{new}} \\ \end{align*}

This formula looks cleaner! The arc length is just the radius of the circle times the angle in this new unit that transforms from degrees using the conversion factor k=\dfrac{\pi}{180^{\circ}}. This means that in this new units \pi is equivalent to 180^{\circ}, 2\pi is equivalent to 360^{\circ} and \frac{\pi}{2} is equivalent to 90^{\circ}.

This new unit is so important that it received its own name the radian. Using our new improved formula we can see that for 1 radian:

    \[arc=r\]

So 1 radian is the angle that forms an arc equivalent to the radius.

The relationship between 1 radian and the radius of a circle

Radians are really important in Calculus and Physics so it is imperative that students learn to do everything they can do with usual degrees with radians and to naturally move from one to the other. Let’s see some examples of how to transform some angles.

Transforming from Degrees to Radians:

To transform from degrees to radians, we simply need to multiply the angle in degrees times the transformation factor \dfrac{\pi}{180^{\degree}}

Transform 120^{\circ} to radians:

    \[\theta_{rad}= 120^{\circ}\left(\dfrac{\pi}{180^{\degree}}\right)=\frac{2\pi}{3}\]

Transform 60^{\circ} to radians:

    \[\theta_{rad}= 60^{\circ}\left(\dfrac{\pi}{180^{\degree}}\right)=\frac{\pi}{3}\]

Transform 45^{\circ} to radians:

    \[\theta_{rad}= 45^{\circ}\left(\dfrac{\pi}{180^{\degree}}\right)=\frac{\pi}{4}\]

Transforming from Radians to Degrees:

To transform from degrees to radians, we simply need to multiply the angle in radians times the inverse of the previous transformation factor \dfrac{180^{\degree}}{\pi}

Transform \dfrac{\pi}{6} radians to degrees:

    \[\theta_{\circ}= \frac{\pi}{6}\left(\dfrac{180^{\degree}}{\pi}\right)=30^{\circ}\]

Transform \dfrac{3\pi}{2} radians to degrees:

    \[\theta_{\circ}= \frac{3\pi}{2}\left(\dfrac{180^{\degree}}{\pi}\right)=270^{\circ}\]

The study of radians is extensive and we could talk about them for a long time like learning how to use trigonometric rations like \sin and \cos with them, but for now, knowing where they come from and how to transform from degrees to radians is enough to get started.

This has been a brief introduction to the concepts of quadratic equations. For more information on this topic as well as assistance with homework and test preparation, feel free to reach out to an Academic Director toll-free at 1 (877) 545-7737 or via our Contact Us page.

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The post What are Radians and Why we Use them first appeared on Free Homework Help.]]> Expanding Algebraic Expressions https://schooltutoring.com/help/factoring-and-expanding-polynomials/ Tue, 20 Feb 2024 20:00:26 +0000 https://schooltutoring.com/help/?p=67314 Expanding is a powerful tool that allows for the manipulation and simplification of algebraic expressions. Here we will introduce the distribution and foiling expansion methods and show some examples.

The post Expanding Algebraic Expressions first appeared on Free Homework Help.]]> Distribution

As we study mathematics, we are often presented with algebraic expressions that look complicated, but can actually be simplified to a more digestible form. Depending on the expression we are dealing with this can take many forms, but when dealing with polynomials we can often simplify an expression by expanding. Suppose we have the expression below:

    \[4x^2(x^3+2x)\]

Initially, this can look intimidating, but we can simplify it to remove the brackets and only have two terms! Remember that a term is number, variable, or product of numbers and/or variables. In this example, we would have two terms inside the brackets (x^3 and 2x) and another term (4x^2) outside of the brackets. To expand this expression, we can use distribution.  To do this, we multiply each term inside the brackets by the term outside of the brackets like so:

    \begin{align*} \textcolor{red}{4x^2}(\textcolor{blue}{x^3}+\textcolor{teal}{2x})&=\textcolor{red}{4x^2}(\textcolor{blue}{x^3})+\textcolor{red}{4x^2}(\textcolor{teal}{2x})\\ &=4x^5+8x^3 \end{align*}

Our final expression (4x^5+8x^3) now doesn’t seem so complicated, as we have only 2 terms!

Identifying When to Expand

In general, it is possible to expand an expression when there is multiplication of terms remaining to be done in the expression. For example 2x+4 can not be expanded as there is no multiplication of terms to done. However, we could expand (x+1)(x+2) as we could multiply what is contained in each of the brackets.

Foiling

If we look at the expression we just considered ((x+1)(x+2)), we notice that we can’t follow the same procedure as our first example (4x^2(x^3+2x)). To expand an expression like this, we need to introduce a process called foiling. Recall that to expand this expression we need to multiply (x+1) by (x+2), but we don’t know how to do that yet. Foiling is the solution! To foil, we multiply each term in one set of brackets by each term in the other set of brackets. We show this below with colour:

    \begin{align*} (\textcolor{red}{x}+\textcolor{blue}{1})(\textcolor{teal}{x}+\textcolor{orange}{2})&=\textcolor{red}{x}(\textcolor{teal}{x})+\textcolor{red}{x}(\textcolor{orange}{2})+\textcolor{blue}{1}(\textcolor{teal}{x})+\textcolor{blue}{1}(\textcolor{orange}{2})\\ &=x^2+2x+x+2\\ &=x^2+3x+2 \end{align*}

Foiling can be expanded to even more complicated expressions involving other mathematical operations (e.g. trigonometric functions, logarithms, exponentials, etc.), but the general procedure remains the same. As such, expanding and foiling are powerful tools in manipulating algebraic expressions.

This has been a brief introduction to the concepts of expanding. For more information on this topic as well as assistance with homework and test preparation, feel free to reach out to an Academic Director toll-free at 1 (877) 545-7737 or via our Contact Us page.

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The post Expanding Algebraic Expressions first appeared on Free Homework Help.]]> Slope and the slope formula for lines https://schooltutoring.com/help/slope-and-the-slope-formula-for-lines/ Thu, 15 Feb 2024 08:39:10 +0000 https://schooltutoring.com/help/?p=67310 The slope is an essential tool that we use every day. From calculating prices in the supermarket, to cooking we are constantly using it. In this post, we explain the concept of slope and we show some examples of how to calculate it.

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What is Slope?

Imagine you go to the store and want to buy your favourite chocolate. Your mom has given you 30 dollars to spend! Once you get there you notice that the price tag says 10 dollars each. You stop a second and wonder how many chocolates you can buy. Intuition (and math) tells you that you can buy 3 chocolates. in that way, you would spend exactly 30 dollars. The key to your quick calculation was knowing that each chocolate costs 10 dollars. Congratulations! You have successfully used the power of the rate of change.

The rate of change indicates how one variable changes with respect to another. In the case of the chocolate, we notice that the price changes with respect to the number of chocolates you buy. If you buy 1 chocolate, you would spend 10 dollars, if you buy 2 the price changes to 20 dollars and so on. When the rate of change is constant then it takes a special name and it is called slopes. We also noticed that in the case of chocolate, the price depends on the number of chocolates we buy, so we call the price the dependent variable and the number of chocolates the independent variable.

Mathematically we can represent the rate of change of price with respect to a number of chocolates like this:

    \[m= \frac{10 \text{ dollars}}{\text{chocolate}}; \text{ where m is the rate of change or slope}\]

Notice how the dependent variable is in the numerator and the independent variable is in the denominator. We can use slopes not only for chocolate but for many other things. Some examples are:

    \[m_{1}=\frac{2 \text{ miles}}{2 \text{ hours}} \quad m_{2}=\frac{1 \text{ cup of water}}{3 \text{ cups of flour}} \quad m_{3}=\frac{-10 \text{ customers }}{ 1 \text{ dollar increased}}\]

m_{1} indicates that if you drive 2 hours you cover a distance of 2 miles, so in 4 hours you would cover 4 miles. m_{2} indicates that for every 3 cups of flour you use you need to use 1 cup of water, so if you are using 9 cups of flour then you will need 3 cups of water. m_{3} shows that if you increase your price by 1 dollar you will lose 10 customers, so if you increase your price by 10 dollars, you would lose 100 customers.

Notice that even knowing m_{2}, it is not clear how many cups of water you would need if you use 2 cups of flour. This is because the denominator is given in terms of 3 cups of flour and not in terms of 1 cup. Therefore, sometimes it is useful to represent the slope having denominator one, this gives the change per unit. Let’s see how our previous examples look if we represent them with denominator one.

    \[m_{1}=\frac{1 \text{ mile}}{ \text{ hour}} \quad m_{2}=\frac{0.333... \text{ cups of water}}{ \text{ cup of flour}} \quad m_{3}=\frac{-10 \text{ customers }}{ \text{ dollar increased}}\]

Notice that now we can easily see that m_{2} tells us that for 2 cups of flour, we need 0.666.. cups of water.

How is the slope related to lines?

Let’s come back to the chocolate example. Can you know how much will it cost you (dependent variable)  to buy 5 chocolates (independent variable) if you know the slope is 10 \frac{\text{dollars}}{\text{chocolate}}? We have answered this question before, but let’s formalize the process.

    \begin{align*} cost &= \text{slope} *\text{ number of chocolates}\\ cost &= 10 \frac{\text{dollars}}{\text{chocolate}}(5 \text{chocolates})\\ cost &= 50 \text{dollars}\\ \end{align*}

So the cost is 50 dollars. This should look really familiar. If I replace the cost (dependent variable) with a y, slope with an m, and number of chocolates (independent variable) with an x, the relationship can be written as:

    \[y=mx\]

When the slope m is constant and x and y can be written in this way, we would say y is proportional to x. The graph of this proportional relation will be a line that passes through the point (0,0). Let’s see an example in a graph (Created using Desmos).

Let’s see what happens when we change the value of the slope. Please press the play button next to the m in the following graph.

Notice that the bigger the slope, the more steep the line is. The opposite is true if the slope approaches zero. When the slope is positive, as x becomes bigger y becomes bigger. When the slope is negative, as x increases y decreases.

From knowing just the slope you can learn a lot about how your line will look.

Finally, for the sake of completeness, let’s consider the following example. Imagine now that when you go to the store for chocolate your mom said you must buy some milk first and the rest you can spend it on chocolate. Milk costs 20 dollars (an expensive milk!). If we buy one chocolate our cost will be 10 dollars for the chocolate and  20 dollars for the milk. The total would be 30 dollars. If we buy 3 chocolates the cost will be 30 dollars for 3 chocolates and 20 dollars for the milk to a total of 50 dollars. In general;

    \begin{align*} cost &= \text{slope} *\text{number of chocolates}+\text{cost of milk} \quad \text{<em>or</em>}\\ y &= mx + b \end{align*}

Where b is the cost of the milk that is always present. If you buy zero chocolates you will still need to pay b dollars for the milk.

b is known as the y-intercept as it is the value y would take if x is zero. Therefore, it is the point where the line touches the y-axis. Moreover, any line can be written as:

    \[y=mx+b\]

This equation is so important that it has its own name. It is called the equation of a line or slope-intercept form.

How to calculate the slope of a line

To calculate the slope of a line we rely on the information we know from the line.

Equation of the slope given two points

If you know two points (x_{1}, y_{1}), (x_{2}, y_{2}) that belong to a line,  you can use the formula for slope given two points to calculate the slope:

    \[m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\]

To make sense of this formula remember that the slope is the rate of change between two variables and that the dependent variable goes in the numerator. Then you can see that the numerator is calculating the change in the dependent variable when you go from y_{1} to y_{2}. This is given by y_{2}-y_{1} and is called the rise of our line.

Similarly, the independent variable goes in the denominator and our formula calculates the change in the independent variable when you go from x_{2} to {x_{1}}. This is given by x_{2}-x_{1} and is called the run of our line.

By dividing rise over run, you obtain how much y has changed as a function of x and therefore you have successfully calculated the slope.

Example:

Calculate the slope of a line that passes through the points (8,-4) and (4,2).

Since we have two points we can use the the formula for slope given two points to calculate the slope. Choosing (8,-4) as (x_{1},y_{1}) and (5,2) as (x_{2}, y_{2})

(1)   \begin{align*} m &=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ m &= \frac{2-(-4)}{5-8} \\ m &= \frac{2+4}{-3} \\ m &= \frac{6}{-3} \\ m &= -\frac{2}{1} \\ m &= -2 \\ \end{align*}

We highlight a couple of things. First, it does not matter which point you choose to be (x_{1},y_{1}) or (x_{2}, y_{2}); the formula will work anyway. However, what is important is that once you have made your choice you do not change it. You must have y_{2} and x_{2} in the correct places indicated by the formula otherwise you will get the wrong answer.

Calculating the slope given a graph

If you are given the graph of a line, you can calculate the slope using the same technique that we described above.

You first need to choose two points. They can be any points. Since you have the freedom to choose the points you’ll use, you should choose wisely and try (if possible) to choose points that have exact integer values.

After that, you can use the slope formula from before and find the slope of the line using the points.

Let’s see an example using Desmos:

This has been a brief introduction to the concepts of slopes. For more information on this topic as well as assistance with homework and test preparation, feel free to reach out to an Academic Director toll-free at 1 (877) 545-7737 or via our Contact Us page.

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The post Slope and the slope formula for lines first appeared on Free Homework Help.]]> Proving and Using the Pythagorean Theorem https://schooltutoring.com/help/proving-and-using-the-pythagorean-theorem/ Tue, 06 Feb 2024 20:12:38 +0000 https://schooltutoring.com/help/?p=67301 Here we will look at one of the most important equations in geometry: the Pythagorean Theorem. We will begin by defining the Pythagorean Theorem and then we will demonstrate a short proof. Finally, we will look at two examples of how we might use the the Pythagorean Theorem in a real-world context.

The post Proving and Using the Pythagorean Theorem first appeared on Free Homework Help.]]> The Pythagorean Theorem

The Pythagorean Theorem is one of the most useful tools in geometry, allowing us to determine the length of a given side of a right triangle as long as we know the lengths of the other two sides. It is important to remember that this only works for right triangles (triangles where one interior angle is 90 degrees).

If we label the side lengths of a right triangle as a, b, and c as in the diagram above, with c being the hypotenuse (the length of the longest side), the Pythagorean Theorem states that:

    \[a^2 + b^2 = c^2\]

Proving the Pythagorean Theorem

Before we can see how the Pythagorean Theorem is useful, we must make sure that it is correct. Let us consider the diagram below, where we have a square constructed by joining 4 right triangles with side lengths of a, b, and c.

The area of this square could be calculated by multiplying its width by its height. This would give:

    \[A=(a+b)(a+b)\]

We could also calculate the area by adding the areas of the four green triangles to the area of the smaller white square. The area of one green triangle is \frac{ab}{2} and the area of the white square is c^2, so we get our second equation for the area of the large square:

    \[A=c^2+2ab\]

We can then equate these two equations for the area as they are both helping us find the area of the same square. This gives the following:

    \[(a+b)(a+b)=c^2=2ab\]

If we expand the left side of this equation, we get:

    \[a^2+2ab+b^2=c^2+2ab\]

Each side of the equation has the term 2ab, so if we subtract this term from both sides we are left with the Pythagorean Theorem:

    \[a^2+b^2=c^2\]

Examples

Now that we have proven the Pythagorean Theorem, it is instructive to see it in action. Broadly speaking, there are two types of problems that the Pythagorean Theorem helps us solve:

  1. We know the lengths of two sides of a triangle where neither is the hypotenuse.
  2. We know the length of the hypotenuse and one other side.

We will look at an example for each case.

Finding the Length of the Hypotenuse

Let’s say that you are walking through your neighborhood and you travel 400 m North. You then turn and travel 50 m East. You might wonder how far away you are from where you started. Even though you walked a total of 450 m, it was not all in the same direction, so you aren’t sure exactly how far away you are from where you began. How would we solve this?

We could construct a diagram as shown above to help solve this problem. The path we travelled forms two sides of a right triangle, with the third side (the hypotenuse) forming the last side of our triangle. Let’s use the Pythagorean Theorem to figure out how far we are from our origin. If a is 400, and b is 50, then we insert those values into the Pythagorean Theorem as shown below:

    \[400^2+50^2=c^2\]

Here we have used c instead of a question mark. We can simplify this equation to be:

    \[162,500=c^2\]

We then take the square root of both sides to find c:

    \[\sqrt{162,500}=403.11=c\]

We find that in our 450 m of walking, we have only ended up a little over 403 m from where we started!

Using the Hypotenuse and One Other Side

As another example, supposed that we are trying to find the dimensions of a certain rectangle. As shown below, its length is  4 and we know that the length from corner to corner is 5. What is the rectangle’s width?

We can see that the length, width, and blue line connecting the corners forms a triangle. We could insert the known values into the Pythagorean Theorem to get:

    \[4^2+b^2=5^\]

Here we have let b be the width of our rectangle. We can simplify our equation to get:

    \[16+b^2=25\]

We then subtract 16 from both sides to get:

    \[b^2=9\]

Finally, we take the square root of both sides to get:

    \[b=\sqrt{9}=3\]

We find that the width of our rectangle is 3!

This has been a brief introduction into the Pythagorean Theorem along with some examples of its usefulness. For more information on this topic as well as assistance with homework and test preparation, feel free to reach out to an Academic Director toll-free at 1 (877) 545-7737 or via our Contact Us page.

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The post Proving and Using the Pythagorean Theorem first appeared on Free Homework Help.]]> Solving Quadratic Equations https://schooltutoring.com/help/quadratic-equations/ Thu, 01 Feb 2024 09:59:00 +0000 https://schooltutoring.com/help/?p=67293 The quadratic equation is one of the most important topics in math, but sometimes it could look a bit complicated. We explore some of the simplest methods to solve quadratic equations and offer some examples of each.

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What are Quadratic Equations?

The general form of quadratic equations is

    \[ax^2+bx+c=0\]

where x is our unknown a,b,c are real numbers and a \neq 0. The last statement is important as if a was zero, the term ax^2 would be zero and the equation wouldn’t have a quadratic term.

Having a quadratic term is not enough though, you cannot have your unknown to any power bigger than 2 (i.e.. x^3, x^4, etc.) and also the expression must be a polynomial (there cannot be “weird” functions like \sin(x), \cos(x), \log(x), \frac{1}{x}, \sqrt{x}, x^{0.3} affecting our unknown) only non-negative integers are allowed.

In practice, quadratic equations can take a lot of forms and look quite different than their general form. Let’s see some examples:

    \[3x^2+2x+3=0, \quad 2y^2-y=0, \quad \pi z^2-3=0, \quad \frac{1}{2}x^2=-3, \quad -3x-4x^2=2\]

These are examples of equations which are not quadratic:

    \[x+2=0 \quad \text{is missing the quadratic term } x^2\]

    \[x^3+3x^2+x=0 \quad \text{has the unknown to a power bigger than 2} (x^3 \text{is the biggest})\]

    \[x^2+\frac{1}{x}=0 \quad \text{has a "weird" function affecting the unknown}\]

    \[x^2+x^{0.5}=0 \quad \text{has a term of the form } x^{0.5} \text{ which makes the expression not a polynomial}\]

Now we put our attention on how to solve these equations.

Solving Quadratic Equations

General Form

While solving quadratic equations we must take into consideration their form. To know the form of a quadratic equation we can put it in general form  (G.F). General Form is no more than the equation written as ax^2+bx+c=0. Notice that all the terms are on the left of the equal sign in order (x^2 first and x second), and with zero on the right. Let’s see some examples of how to take a quadratic equation to G.F.

    \[3x^2=2x-3 \implies 3x^2-2x+3=0\]

    \[\frac{1}{2}x^2=1 \implies \frac{1}{2}x^2-1=0\]

    \[-3x=x^2 \implies -x^2-3x=0\]

Solving the Quadratic Equation

The following diagram describes how to solve a quadratic equation, once it is in General Form.

Diagram on how to solve quadratic equations of all types.

Factor if you can

If you thought that factoring was not useful, you are wrong. Factoring is essential to solving equations of any type, especially for quadratic equations. We can see why by remembering a key feature of numbers.

“If a multiplication of numbers equals zero, then one of the factors must be zero”

For example, if x*y*z=0 then one of them must be zero x=0, y=0 or z=0. Let’s see how to exploit this principle to solve equations.

    \[x^2-4x+3=0 \implies (x-3)(x-1)=0 \quad \text{By Factoring}\]

Notice that this is a multiplication that equals zero then that means that

    \[x-3=0 \quad\text{ or }\quad x-1=0\]

Notice that both of the resulting equations are simple linear equations! And we know how to solve them, the results are:

    \[x=3 \quad\text{ or }\quad x=1\]

These are precisely the solutions to our equation.

If you can’t factor use the Quadratic Formula

Factoring is not always easy or obvious. For those cases, or any case in fact, you always have the quadratic formula at your service. At last here it is:

The solution to an equation to the form ax^2+bx+c=0 is given by

    \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

The quadratic formula can solve any quadratic equation, but it can take a lot of steps to reach a solution. This means that you should use it only when it is necessary. We must accept the formula can look a bit scary at first. However, all you need to do is identify the values of a,b and c from your equation. Let’s solve x^2-4x+3=0 using the quadratic equation.

First identify a,b and c

    \[\boxed{1}x^2\boxed{-4}x\boxed{+3} = 0 | \quad & \boxed{a}x^2+\boxed{b}x+\boxed{c}=0\]

Use them in the quadratic formula

    \[x&=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(3)}}{2(1)} \quad | \quad x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Simplify

    \begin{align*} x&=\frac{4\pm\sqrt{16-12}}{2}\\ x&=\frac{4\pm\sqrt{4}}{2}\\ x&=\frac{4\pm2}{2}\\ \end{align*}

In this case, we have two solutions, one given by separating \pm into plus and minus

    \begin{align*} x&=\frac{4\boxed{+}2}{2} &\text{or}&& x&=\frac{4\boxed{-}2}{2}\\ x&=\frac{6}{2} & \text{or}&& x&=\frac{2}{2}\\ \boxed{x=3}& & \text{or}&& &\boxed{x=1}\\ \end{align*}

If you notice this is the same solution we found in the previous section (As we solved the same equation). I hope you agree that by factoring we solved the equation faster.

Special Cases

When the equation is missing terms, the linear term bx or the constant term c. We can solve the equation easily.

Missing bx

When b=0 the equation looks like ax^2+c=0. Examples are 3x^2+2=0, x^2-2=0 or \frac{1}{2}x^2-\frac{1}{3}.

To solve this equation we need only need to isolate x (You could solve it by factoring or using the quadratic formula as well, but I think this is the easiest approach).

    \begin{align*} 2x^2-32&=0 &|&& ax^2+c&=0\\ 2x^2&=32 &|&& ax^2&=-c\\ x^2&=\frac{32}{2} &|&& x^2&=\frac{-c}{a}\\ x&=\pm\sqrt{\frac{32}{2}} &|&& x&=\pm\sqrt{\frac{-c}{a}}\\ x&=\pm4 &|&& \\ \boxed{x=4} \quad &\text{or} \quad \boxed{x=-4} \end{align*}

Missing c

When c=0 the equation looks like ax^2+bx=0. Examples are 2x^2+2x=0, x^2-3x=0 or \frac{1}{2}x^2-\frac{1}{3}x.

To solve this equation we need only need to take the Common Factor (Factoring again!).

    \begin{align*} 3x^2-6x&=0 &|&& ax^2+bx&=0\\ x(3x-6)&=0 &|&& x(ax+b)&=0\\ x=0 \quad &\text{or} \quad 3x-6=0 &|&& x=0 \quad &\text{or} \quad ax+b=0\\ x=0 \quad &\text{or} \quad x=\frac{6}{3} &|&& x=0 \quad &\text{or} \quad x=\frac{-b}{a}\\ \boxed{x=0} \quad &\text{or} \quad \boxed{x=2} \end{align*}

Notice that zero is always a solution for these types of equations.

No solutions for Quadratic Equations

Sometimes, a quadratic equation has no solutions in the real numbers. This happens whenever in one of our methods we encounter the square root of a negative number (i.e. \sqrt{-3}, \sqrt{-4}, \sqrt{-\frac{1}{2}}). If you are working in real numbers and you see one of these square roots your answer should be immediately:

 “There is no solution in the real numbers”

As a disclaimer, these equations have solutions in complex numbers, but that is a different topic that deserves its own post.

Solving Quadratic Equations Using Technology

Lastly, I want to show you how to solve a quadratic equation technology. You can use for example this widget from Wolfram Alpha.

If you need to write \sqrt{2} for example you can do so by writing “sqrt(2)” and exponents like 3^2 can be written as 3^2.

Step-by-step solutions are available in Symbolab. Wolfram Alpha also has this option but it is paid.

Calculators and Quadratic Equations

Some calculators can solve quadratic equations as well. Check yours! You would be surprised by the power of your calculator. This video from Blackpenredpen has an example of how to solve a quadratic equation using a Casio calculator.

This has been a brief introduction to the concepts of quadratic equations. For more information on this topic as well as assistance with homework and test preparation, feel free to reach out to an Academic Director toll-free at 1 (877) 545-7737 or via our Contact Us page.

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The post Solving Quadratic Equations first appeared on Free Homework Help.]]> Finding Asymptotes and Holes of Rational Functions https://schooltutoring.com/help/finding-asymptotes-and-holes-of-rational-functions/ Tue, 16 Jan 2024 20:30:57 +0000 https://schooltutoring.com/help/?p=67280 Here we will be looking at some defining features of rational functions: asymptotes and holes. We will begin by defining rational functions,. We will then proceed to uncover how to identify different types of asymptotes and how to distinguish them from holes. In each case, we will use example functions to visualize the effects and representations of these features.

The post Finding Asymptotes and Holes of Rational Functions first appeared on Free Homework Help.]]> Rational Functions

Rational functions are functions of the form:

    \[f(x)=\frac{P(x)}{Q(x)}\]

where P(x) and Q(x) are polynomials. An important feature of many rational functions is the existence of asymptotes and/or holes.

Asymptotes

An asymptote is a straight line that a rational function approaches, but never actually touches. There are three types of asymptotes:

  1. Vertical asymptotes
  2. Horizontal asymptotes
  3. Oblique asymptotes

Vertical Asymptotes

Vertical asymptotes occur at x values where Q(x) (the denominator in a rational function) is zero but P(x) (the numerator in a rational function) is non-zero. We can see this in the following example:

    \[f(x)=\frac{1}{x}\]

Here P(x)=1 and Q(x)=x. We see that at x=0, P(0)=1 and Q(0)=0. Thus, we have a vertical asymptote at x=0.

Horizontal Asymptotes

Horizontal asymptotes occur when a rational function approaches a specific y value as x becomes very big in the positive or negative directions. If we return to our example function:

    \[f(x)=\frac{1}{x}\]

we can see that if x is a very big negative number, then f(x) will get extremely close to 0.  The same is true if x is a very big positive number. This means that f(x) has a vertical asymptote at y=0. Thus, we have seen that f(x)=\frac{1}{x} is an interesting function in that it has both a horizontal asymptote and a vertical asymptote. These are graphed along with f(x) below.

Graph of function f(x)=1/x along with its asymptotes.

Note the vertical (green) and horizontal (blue) asymptotes. f(x) gets infinitely close to these asymptotes without ever touching them.

Oblique Asymptotes

Oblique asymptotes occur when the degree of P(x) is one greater than the degree of Q(x). Q(x) must also not be a factor of P(x). The degree of a polynomial is the highest power of x it contains. An oblique asymptote is a line that a rational function approaches for large values of x. It is different from a vertical or horizontal asymptote in that it is not vertical or horizontal (it will have non-zero, definite slope). A good example of this is the function below:

    \[f(x)=\frac{x^2+1}{x}\]

Here, the degree of P(x) is two and the degree of Q(x) is one. To find the equation of the line representing the oblique asymptote of f(x), we must divide the coefficient of the highest power of x in P(x) by the coefficient of the highest power of x in Q(x). In both cases, this coefficient is 1, so the equation of the oblique asymptote of f(x) is y=x. This is visualized in the graph below:

Graph of function f(x) = (x^2+1)/x with its associated vertical and oblique asymptotes.

Note the oblique asymptote (blue) which operates similarly to a vertical or horizontal asymptote, except that it is “tilted” due to having non-zero, definite slope.

Holes

Finally, we want to consider holes. A hole occurs at any x value where P(x) and Q(x) are both 0. Holes are aptly named, as they represent points where the function does not exist. They look like a hole in the graph of the function! A good example of this is the following function:

    \[f(x)=\frac{x^2}{x}\]

We can see that f(0)=\frac{0}{0}, so we have identified a hole at x=0. For all other x, f(x) can be simplified to f(x)=x. Thus, the graph of f(x) will be identical to the line y=x except that the function does not exist at x=0. This is shown in the graph below, with the whole represented as an unfilled circle.

Graph of function f(x) = x^2/x showing the hole at x=0.

Note the hole at x=0. Aside from the hole, f(x) is identical to the line y=x. It is also important to remember that there is no oblique asymptote, as although the degree of the numerator is 1 greater than the degree of the denominator, the denominator is a factor of the numerator.

This has been a brief introduction into the concepts of asymptotes and holes in rational functions. For more information on this topic as well as assistance with homework and test preparation, feel free to reach out to an Academic Director toll-free at 1 (877) 545-7737 or via our Contact Us page.

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