## Simplifying “3-stack” and “4-stack” fractions

I had a physics student a number of years ago who worked a complicated problem and ended up with the following (I’ve changed the actual numbers to make our work here easier to follow):

$\dfrac{(2)(3)}{4}=\dfrac{(5)(6)(7)B}{(8)(9)}$

Of course, his next step was to solve for B. It should be clear to you, as it was to him, that he needed to multiply both sides by 8 and 9 and divide by 5, 6 and 7. That is what he did. But this is how he wrote the solution:

$B= \dfrac{9 \cdot \left( \dfrac{ \left(\dfrac{ 8 \cdot \left( \dfrac{(2)(3)}{4} \right)}{5} \right)}{6} \right)}{7}$

His expression was totally correct, and he found the correct value of B, but he made the problem so much harder than he needed to. He created a fraction with five different stacks in it and he needed all those parentheses to keep track of which number was a numerator and which was a denominator.

This is an extreme case of what I see so many students do: they create “3-stack” and “4-stack” fractions all the time when they are simplifying problems. Let’s look at two examples. I’ll solve them with “bad” solutions and then with much smarter solutions (I don’t simplify the answers here because I want to focus on the first step of the solution):

Example 1. Solve:

$3x= \dfrac{2}{5}$

$x= \dfrac{\dfrac{2}{5}}{3}$

Smarter solution:

$x= \dfrac{2}{5 \cdot 3}$

Example 2. Solve:

$\dfrac{4}{7}x=\dfrac{2}{5}$

$x= \dfrac{\dfrac{2}{5}}{\dfrac{4}{7}}$

Smarter solution:

$x= \dfrac{2 \cdot 7}{5 \cdot 4}$

You can see that the smarter solutions are simpler to read and easier to simplify.

It is easy to train yourself to write answers as “2-stack” fractions if you remember one simple rule:

Respect the vinculum.

Um, respect the what?!?

The vinculum. When you write a fraction, the horizontal line that separates the numerator from the denominator is called the vinculum. No one ever talks about it, but it’s a very powerful symbol. It tells you to multiply by every number that is above it and divide by every number that is below it. And it’s as easy as that. So when you are simplifying expressions like the ones in the two examples above or the more complicated example at the beginning of this post, all you need to do is to put numbers that are multiplied above the vinculum and numbers that are divided below the vinculum. And what happens if you are multiplying or diving by a fraction? Then put the numerator on top and the denominator on the bottom if you are multiplying. Flip the fraction over first if you’re dividing (as I did in Example 2). When you do this, you will end up with a 2-stack fraction that can be easily evaluated. Let’s look at the original problem again:

$\dfrac{(2)(3)}{4}=\dfrac{(5)(6)(7)B}{(8)(9)}$

To solve for B, you will multiply by 8 and 9, so they go on top. And you will divide by 5, 6 and 7, so they go on the bottom. In one step, you’ve solved for B as follows:

$B= \dfrac{2 \cdot 3 \cdot 8 \cdot 9}{5 \cdot 6 \cdot 7}$

Compare that to the monstrosity at the beginning. So much easier!

P.S. You don’t really have to remember the name “vinculum”. Most people don’t know what the line is called and don’t care. Chances are good your math teacher doesn’t even know the term. To make things even more bizarre, when you write a fraction with a slash instead of a horizontal line like this—2/3 – the slash is called a “virgule”.  Most people don’t know that one and don’t care about it either. All you have to remember is that the horizontal line in a fraction tells you to multiply on top and divide on the bottom.

## Angles in circles and arc measures

When angles are formed by chords, secants and tangents in a circle, there are several formulas that help you determine the arc measures and angle measures that are formed. With a little bit of thought, you can often find values for all of the arcs and angles around the circle. For many students, the equations are difficult to remember. Your best approach is to categorize the angle by the location of the vertex. Then learn the rule for the angles with that particular vertex. There are four possibilities:

• Central angles: This is an angle whose vertex is at the center of the circle. For all central angles, the angle measure is equal to the arc measure of the subtended arc.
• Inscribed angles: These angles have their vertex on the circle. They can be chord-chord angles or chord-tangent angles. For these angles, the angle measure is one half the arc measure of the subtended arc.
• Internal angles: These are angles whose vertex is in the interior of the circle but not at the center. For these angles, the angle measure is one half the sum of the arc measures of the two subtended arcs.
• External angles: These are angles whose vertex is outside the circle. They can be secant-secant angles or secant-tangent angles. For these angles, the angle measure is one half the difference of the arc measures of the subtended arcs.

## A mnemonic for remembering values on the unit circle

In an earlier post, I told you that you must memorize the values on the unit circle if you want to be great at trigonometry. So let’s say that you have taken my advice, but on a test you have a big brain cramp and you forget the values. Here’s a trick that will help you fill in the values in the first quadrant. The five angles in the first quadrant are 0°, 30°, 45°, 60°, and 90°. And the sines of these angles just happen to be:

$\underline{ \qquad \theta \qquad \sin \theta \qquad \qquad \qquad}$

$0 \textdegree \qquad \; \; 0 \qquad =\dfrac{\sqrt{0}}{2}$

$30 \textdegree \qquad \dfrac{1}{2} \qquad = \dfrac{\sqrt{1}}{2}$

$45 \textdegree \qquad \dfrac{\sqrt{2}}{2} \quad = \dfrac{\sqrt{2}}{2}$

$60 \textdegree \qquad \dfrac{\sqrt{3}}{2} \quad = \dfrac{\sqrt{3}}{2}$

$90 \textdegree \qquad 1 \qquad = \dfrac{\sqrt{4}}{2}$

Do you see the pattern in the last column? The sines increase from

$\dfrac{\sqrt{0}}{2}$ to $\dfrac{\sqrt{4}}{2}$

That’s an easy pattern to remember even during the worst brain cramp. Why does it work? It’s just a coincidence. But if you use this pattern, you can fill in the sine values in the first quadrant. Then the cosine values are the same in reverse order. Once you’ve completed the first quadrant, you can fill in the rest of the unit circle by using reference angles. Now you have a completed unit circle!

## Using the reciprocal key on your calculator

Your graphing calculator has a reciprocal key on it, designated as $x^{-1}$. It’s the easiest way to find the reciprocal of a number on your screen. Just press the key, and then Enter.

The reciprocal key is also a great way to do calculations of resistors in parallel or capacitors in series. For example, find the equivalent resistance of a circuit that has a 10 Ω and a 15 Ω resistor in parallel. Simply enter it this way into the calculator to find the answer is 6 Ω.

## Factoring the difference of two squares

You will spend a lot of time in algebra (and courses beyond) factoring polynomials into linear and quadratic terms. There are some special polynomials that occur so frequently that you should recognize them on sight so that you know the method for factoring them. The most common is the binomial of the form $a^2-b^2$. This is called the difference of two squares because both terms are perfect squares and you are subtracting (finding the difference between) the two terms.

Some examples of this binomial are $x^2-4$, $9x^4-16y^2$ and $121y^2z^2-25w^4a^6b^{12}$. Note that in each case, both terms are perfect squares.

It is very easy to factor these expressions—you can do it by inspection. The rule is simple and you must memorize it: $a^2-b^2=(a+b)(a-b)$. Once you determine that an expression is the difference of two squares, you can write out its factors immediately. Let’s see how it works with the three examples given above:

$x^2-4=(x+2)(x-2)$

$9x^4-16y^2=(3x^2+4y)(3x^2-4y)$

$121y^2z^2-25w^4a^6b^{12}=(11yz+5w^2a^3b^6)(11yz-5w^2a^3b^6)$

It’s really that easy once you learn the format. Of course it doesn’t matter what order you write the two factors, so $x^2-4=(x-2)(x+2)$ is also correct.

One other important point: this factoring rule works in reverse, too! If you have to distribute (multiply out) two factors and you see they are in the form of $(a+b)(a-b)$, you should recognize that this gives the difference of two squares: $a^2-b^2$. You can do this immediately, without multiplying out every term. Learning this formula will save you lots of time.

## Graphing rational functions—4. Drawing the graph

In previous posts, I described the steps you follow to analyze a rational function. If you follow these steps, you can be a rational function superstar too. In this post, I put all the pieces together to show how you use the information you’ve obtained to plot the graph. [For details on how to execute the various steps, please see other posts on this website.]

Example: Graph the following rational function.

$f(x)=\dfrac{2x^3-16x^2+38x-24}{x^3-4x^2+x+6}$

Solution: We break this into many steps.

1. Find the $y$-intercept:

$f(0)=\dfrac{-24}{6}=-4$

2. Fully factor the function:

$f(x)=\dfrac{2x^3-16x^2+38x-24}{x^3-4x^2+x+6}=\dfrac{2(x-1)(x-3)(x-4)}{(x+1)(x-2)(x-3)}$

3. Find the vertical asymptotes, if any:

The vertical asymptotes are at $x=-1$ and $x=2$.

4.Find the $x$-intercept(s), if any:

$f(x)=0$ @ $x=1$ and $x=4$

5. Find the hole(s), if any:

There is a hole at the point $(3,-1)$

6. Find the horizontal or oblique asymptote, if any, or characterize the end behavior:

The horizontal asymptote is at $y=2$

7. Place all of this information on the graph:

8. Use all the information you have plotted to complete the graph of the function:

## Graphing rational functions—3. Finding horizontal or oblique asymptotes

In a previous post, I listed the steps you need to follow to analyze and graph a rational function. In this post, I show you how to find the horizontal or oblique asymptotes, if they exist. If there are no horizontal or oblique asymptotes, then you can determine the end behavior of the function. [Check other posts on this website for other steps in analyzing and graphing a rational function.]

The horizontal and oblique asymptotes, if they exist, tell you the end behavior of the function. That is, they describe what the function is doing as $x$ goes to ±∞. In order to determine the end behavior, examine the degrees of the polynomials in the numerator and denominator. There are four possible cases:

Case 1: The degree of the denominator is greater than the degree of the numerator.

When the denominator has the higher degree, the denominator grows much faster than the numerator as $x$ gets very large. Like all fractions, as the denominator gets very large, the fraction gets very small and approaches zero. So the asymptote is $y=0$.

Example 1: Find the end behavior for the following rational function:

$f(x)=\dfrac{x^2+3x+2}{x^3+4x^2+x-6}$

Solution: Because the degree of the numerator is $2$ and the degree of the denominator is $3$, this function has a horizontal asymptote $y=0$.

Case 2: The degree of the denominator is equal to the degree of the numerator.

When the degrees of the numerator and denominator are the same, the function approaches a finite non-zero number. This number is the ratio of the leading coefficients of the two polynomials and this is the value of the horizontal asymptote.

Example 2: Find the end behavior for the following rational function:

$f(x)=\dfrac{6x^2+7x-4}{5x^2-4x-6}$

Solution: Because the degree of the numerator and denominator are the same, this function has a horizontal asymptote. Divide the leading coefficients ($6$ and $5$) and the horizontal asymptote is $y=6/5$.

Case 3: The degree of the numerator is exactly one higher than the degree of the denominator.

In this case, there is an oblique asymptote (called a slant asymptote in some textbooks). To find the asymptote, divide the denominator into the numerator using long division. The quotient is the oblique asymptote. (Ignore the remainder; it has no effect on the asymptote.)

Example 3: Find the end behavior for the following rational function:

$f(x)=\dfrac{x^3+4x^2+x-6}{x^2+3x+2}$

Here, the numerator (degree 3) is exactly one higher than the denominator (degree 2). To find the asymptote, perform the following long division:

The quotient is the oblique asymptote:

$y=x+1$

Case 4: The degree of the numerator is more than one higher than the degree of the denominator.

In this case, there is no horizontal or oblique asymptote. Instead, the end behavior of the function is the same as the end behavior of a polynomial whose degree is the same as the difference between the degrees of the numerator and the denominator. For example, if the numerator is degree $5$ and the denominator is degree $2$, the rational function will look like a cubic function as $x$ approaches ±∞.

## Graphing rational functions—2. Finding holes and vertical asymptotes

In a previous post, I listed the steps you need to follow to analyze and graph a rational function. In this post, I show you how to find the hole(s) and vertical asymptote(s), if they exist. [Check other posts on this website for other steps in analyzing and graphing a rational function.]

In a rational function, holes and vertical asymptotes will only occur where the denominator is equal to zero. So fully factor the function, and look for the roots of the denominator. When a factor in the denominator cancels out with a factor in the numerator, the function will have a hole where the factor equals zero. If the factor in the denominator does not cancel out with a factor in the numerator, there will be a vertical asymptote where the factor equals zero.

Example: Find the holes and vertical asymptotes, if any, for the following rational function:

$f(x)=\dfrac{x^2+3x+2}{x^3+4x^2+x-6}$

First, fully factor the function:

$f(x)=\dfrac{x^2+3x+2}{x^3+4x^2+x-6}=\dfrac{(x+1)(x+2)}{(x-1)(x+2)(x+3)}=\dfrac{(x+1)}{(x-1)(x+3)}$

The roots of the denominator are $-3$, $-2$ and $1$. At $x=-2$, there is a hole. To find the “value” of the hole, evaluate the reduced function:

$f(-2)=\dfrac{(-2+1)}{(-2-1)(-2+3)}=\dfrac{-1}{-3}=\dfrac{1}{3}$

This means that when you graph the function, there will be a hole at the point $(-2, 1/3)$.

The vertical asymptotes will be $x=-3$ and $x=1$.

## Graphing rational functions—1. Finding x- and y- intercepts

In a previous post, I listed the steps you need to follow to analyze and graph a rational function. In this and subsequent posts, I will walk you through each of those steps. Usually the first step in plotting a rational function is finding the $x$– and $y$-intercepts. You have done this before for other functions and the process here is no different.

• To find the $y$-intercept, set $x=0$ and solve for $f(x)$.
• To find the $x$-intercept(s), set $f(x)=0$ and solve for $x$. Recall that a rational function is a fraction, and a fraction can only be equal to zero when its numerator is equal to zero (as long as the denominator is not also equal to zero!). So in practice, you find the $x$-intercept(s) by setting the numerator equal to zero and solving for $x$.

Example: Find the $x$– and $y$-intercepts, if any, for the following rational function:

$f(x) = \dfrac{x^3+4x^2+x-6}{x^2+3x+2}$

Finding the $y$-intercept is very easy if the polynomials are written in standard form as above. $f(0)$ is simply the ratio of the constant terms: $\frac{-6}{2} = -3$. If the rational function has already been factored, you need to plug zero in to each term and evaluate.

To find the $x$-intercept (and all the other parts of the rational function), the first thing you need to do is fully factor the numerator and denominator.

$f(x)=\dfrac{x^3+4x^2+x-6}{x^2+3x+2}=\dfrac{(x-1)(x+2)(x+3)}{(x+1)(x+2)}$

From here, it is easy to see that the roots of the numerator are $-3$, $-2$ and $1$. When $x=-2$, the denominator is also zero, so this is not an $x$-intercept; it is a hole. The $x$‑intercepts are at $-3$ and $1$.

Is it possible that a rational function does not have a $y$-intercept? Yes, if the function is not defined for $x=0$. In other words, if a rational function has either a hole or a vertical asymptote at $x=0$, there is no $y$-intercept.

[Check out my other posts in this section that describe how to perform the other steps when analyzing a rational function.]

## Graphing rational functions—overview

Graphing rational functions can be scary for a lot of students because there are so many details to manage. But you can make the process simpler by breaking down all the requirements into small steps. In this post I list all of the steps you should follow to analyze a rational function. Note that you can do these steps in any order. In subsequent posts, I will walk you through each of the steps and show you how to execute them.

Let’s start with the definition of a rational function. A rational function is a function that consists of a fraction, where both the numerator and denominator are polynomials. Or in symbols, if g(x) and h(x) are polynomials, then

$f(x)=\dfrac{g(x)}{h(x)}$

is called a rational function.

To analyze and graph a rational function, you need to do all of the following steps:

• Find the y-intercept (if it exists)
• Find the x-intercept(s) (if they exist)
• Determine the location of holes, if any
• Find the vertical asymptotes, if any
• Find the horizontal or oblique asymptotes, if any
• Determine the end behavior (if no horizontal or oblique asymptotes)

[Note that this is a list for algebra 2 and precalc students. When you get to calculus, there will be additional steps to completely analyze the function.]

This probably feels like a long and complicated list, but for the most part, each step by itself is not very difficult. Just perform each step in turn and you can become an expert at rational functions. Check out my other posts in this section that describe how to perform each of these steps.

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