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For the Precalc/Trig category

Calculating permutations and combinations

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When counting up the number of ways an event can occur, you use the formulas for permutations and combinations. You should be familiar with the nPr and nCr commands on your calculator, and this is the easiest way to evaluate these problems. But if your calculator doesn’t have these functions, there is a fairly simple way to set up these operations. This is the way we had to calculate permutations and combinations when calculators did not have these functions built in. Practice a couple of these examples and you’ll see that you can calculate permutations and combinations almost as quickly as your calculator can do it.

Calculating nPr

To calculate nPr, you will multiply together r consecutive numbers, starting with n and counting down. For example, 12P 3 is equal to 12*11*10 = 1320. We started with 12 (the value of n) and counted down to 10 so that we had 3 numbers (3 is the value of r). As another example, 7P5 = 7*6*5*4*3 = 2520.

Calculating nCr

To calculate nCr, create a fraction. The numerator is the same as above; that is, start with n and count down r consecutive numbers. The denominator is the smaller of r! and (n-r)!. For example,12P3 is

\dfrac{12*11*10}{1*2*3}

Before you calculate this fraction, simplify it. All of the terms in the denominator will always cancel out with terms in the numerator, leaving you with just numbers in the numerator to multiply together. For example,

\dfrac{12*11*10}{1*2*3} = 2 * 11 * 10 = 220

To calculate 7C5, note that 7C5 = 7C2 . Then,

\dfrac{7*6}{1*2} = 7 * 3= 21

Graphing sine and cosine functions like a pro

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When graphing a sine or cosine curve, the first thing you must do is determine the amplitude, period, phase shift and vertical shift. See my previous post (Graphing Sine and Cosine Functions – Intro) if you need help with this analysis. In this post, we will graph the function

\displaystyle f(x) = -3 \sin (2x + \frac{\pi}{2}) -1

We quickly determine the four values we need:

Amplitude = |-3| = 3

Period = 2 \pi /2 = \pi

Phase shift = -(\pi /2)/2 = - \pi/4  (that is, \pi /4 units to the left)

Vertical shift = -1

 

This is all the information we need in order to complete the graph. Just follow this procedure step-by-step.

 1. Put values on the coordinate axes. On the y-axis, you typically make each square equal to one unit, but you can change this if you want. To determine the scale on the x-axis, take the period and divide by 4. This will be the scale on the x-axis. In our example, the period is \pi , so each square will be \pi /4 . The vertical axis will be one unit per square. What do you do if your teacher gives you a grid with the numbers already in place? You should get a blank piece of graph paper and do your own grid!
 2. Use the vertical shift to draw a dashed line across the figure. This is the location of the midline of your graph. In our example, the vertical shift is -1, so we draw a dashed line at y= -1.
 3. Use the amplitude to draw two more dashed lines—one above the midline and one below. These represent the maximum and minimum values of your function. In our example, the amplitude is 3. Three units above -1 is 2—that’s our maximum dashed line. Three units below -1 is -4—that’s where our minimum is located.
 4. Plot the starting point of your graph, using the vertical shift and phase shift as a guide. Our function is a sine curve, which starts at the midline. The phase shift is \pi /4  to the left, so our initial point is \pi /4  units left of the y-axis. If our function had been a cosine curve, our initial point would be plotted on the maximum line instead of the midline (or on the minimum line if A is negative). It’s hard to see, but note that I’ve placed a green dot at the “start” point; the coordinates are ( - \pi /4, -1).
 5. Moving one square to the right at a time (because each square is one quarter of a period), plot points at the maximum, midline, minimum and midline. This is one period of your function. If you want to graph more than one period, continue the process. In our example, we’ve plotted points for two complete periods. Note that because A is a negative number (-3), our first point after the starting point is at the minimum instead of the maximum. Look closely, and you will see that I’ve placed a green dot every square to the right of our first point.

 6. Connect the dots with a nice smooth curve. You’ve graphed the sine curve like a pro!

Graphing sine and cosine functions– an Intro

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One of the most complicated skills you need to learn in your trig class is how to graph sine and cosine functions. This scares a lot of students, but you can tame this process if you make one simple observation: Every sine and cosine curve has exactly the same shape! No matter the amplitude or period or phase shift, the curve looks just like this:

 

 

 

 

 

You only need to place the graph in its proper position on the coordinate axes. This is (mostly) easy to accomplish if you can remember only two things about the sine and cosine parent curves:

1)     The sine curve y = sin x “starts” at the origin and goes up to its maximum, while the cosine curve y = cos x “starts” at its maximum.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2)     For either curve, you can break one period into four equal intervals. At each interval, the curve moves from its midline to the maximum to the midline to the minimum to the midline to the maximum to… over and over again. So all you need to do is find the starting point, and plot the points on the curve at each ¼-period interval.

We will always write our functions in standard form:

f(x) = A \; sin(Bx+C) + D \; or \; f(x)= A \; cos(Bx+C) + D

(Note that some textbooks prefer to write the formula in a slightly different form:

f(x) = A \; sin(B(x+C)) + D \; or \; f(x)= A \; cos(B(x+C)) + D

We will discuss how that affects your work below.)

Each of the constants A, B, C & D affects the position of the curve and you need to analyze this before you graph the curve. Let’s look at each of them in turn:

A: The absolute value of this number tells you the amplitude of your curve.

B: The period of your curve is determined by dividing 2\pi by B.

C: The phase shift is found by dividing -C by B. A positive value means the phase shift is to the right. A negative value means the phase shift is to the left. (If your class uses the version of the equation above with the B factored out, then the phase shift is equal to C.)

D: The vertical shift is equal to D.

Here’s an example to show how you would calculate all these values.

\displaystyle f(x) = -3 \sin (2x + \frac{\pi}{2}) -1

Here, A = -3; B = 2; C = \pi/2 ; and D = -1. Therefore,

Amplitude = |-3| = 3

Period = 2 \pi /2 = \pi

Phase shift = - (\pi /2)/2 = - \pi /4  (that is, \pi /4 units to the left)

Vertical shift = -1

When you need to graph a sine or cosine curve, always determine these four values first. Then you are ready to graph the function. We’ll do that in our next post.

Using multiplicity of factors to characterize graphs of rational functions

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Rational functions can be scary because there are so many details to manage. Check other posts on this website for information on how to graph rational functions. In this post, I look at one small clue that can help you figure out the behavior of a rational function as it approaches the vertical asymptotes. All you need to do is check the multiplicity of the factor in the denominator.

If the multiplicity of the factor is even, then the graph approaches +∞ from both sides of the asymptote, or it approaches -∞ from both sides of the asymptote.

If the multiplicity of the factor is odd, then the graph approaches +∞ on one side of the asymptote and approaches -∞ on the other side.

Here is an example that demonstrates this property:

\text{Graph } \dfrac {(x-2)(x+1)}{(x-1)(x+2)^2}

There are two vertical asymptotes for this function, at x=-2 and at x=1. The (x+2) factor is multiplicity 2 (even), so the graph approaches the same limit from both sides of the asymptote. The (x-1) factor is multiplicity 1 (odd), so the graph approaches opposite limits on either side of the asymptote. Here is the graph of the function, demonstrating this property:

Using multiplicity of factors to characterize graphs of polynomials

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When you are asked to sketch the graph of a polynomial, you do not want to make a tree to calculate the values of various points. You don’t know where the “turning points” are, so you won’t be able to connect the dots for the points you plot. Instead, you need to fully factor the polynomial and use the zeroes you find to draw the polynomial. In addition, the multiplicity of each factor tells you whether the polynomial crosses the x-axis at that zero or “bounces”. The rule is very simple: If the factor has an odd multiplicity, the graph crosses the x-axis. If the multiplicity is even, the graph bounces.

multiplicity behavior at x ‑axis
odd crosses
even bounces

 

Example: Sketch the graph of

f(x)=x^3(x+1)(x-1)^2

Solution: First of all, plot the zeroes. For this problem, the zeroes are at x=-1, x=0, \text{ and } x=1.

 

 

 

 

 

 

 

Next, determine the degree of the polynomial. In this case, it is degree 6. (Add the exponents of all the factors: 3+1+2=6.) The degree tells you the end behavior, and you can draw arrows to show that the function will go to positive infinity on the left and the right.

 

 

 

 

 

 

 

 

Now you can sketch the graph. At x=-1, the zero is multiplicity 1, so the graph crosses the x-axis. At x=0, the zero is multiplicity 3, so the graph also crosses the x-axis. Note that for multiplicity 3, the graph doesn’t cross straight through the axis, but flattens out as it goes through. At x=1, the zero is multiplicity 2, so the graph bounces at the x-axis. The final sketch is shown below:

Finding the inverse of a 3X3 matrix

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What’s the easiest way to find the inverse of a 3×3 matrix? Use your graphing calculator, of course! But if you need to find the inverse without a calculator, here’s a method that will give you the solution with the least amount of trouble. We’ll demonstrate with an example.

Find the inverse of the following matrix:

 \begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 10 \end{bmatrix}

The first step in finding the inverse is to calculate the determinant of the matrix. The easiest way to calculate a 3×3 determinant is to write the matrix out, and append the first two columns at the end:

\begin{bmatrix}  1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 10  \end{bmatrix} \begin{matrix}  1 & 4 \\ 2 & 5 \\ 3 & 6\\  \end{matrix}

From here, you need to find the six different products along each of the diagonals shown:

 

 

 

 

 

 

Add the blue products together and add the red products together, then subtract the red total from the blue total: (50 + 96 + 84) - (105 + 48 + 80) = -3. This is the determinant. By the way, if the determinant is 0, stop. Your matrix does not have an inverse.

Next, you need to find the elements of the inverse matrix. Here’s a clever trick that will help you do that. Start by writing the transpose of the original matrix. This is done by changing all the rows into columns:

\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{bmatrix}

Then write the first two columns over on the right and the first two rows over again on the bottom. Your array should look like this:

\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{bmatrix} \begin{matrix} 1 & 2 \\ 4 & 5 \\ 7 & 8 \end{matrix} \\ \begin{matrix} \text{ }1 & 2 & 3 & \,\, 1 & 2 \\ \text{ }4 & 5 & 6 & \,\, 4 & 5 \end{matrix}

The next step is a little tricky to explain, though once you’ve done it, it’s pretty easy to figure out. For each of the nine positions in the matrix, you find the value of the determinant of the 2×2 array that is just below it and to the right. [One way to think of this is that each position in the matrix is the upper-left number of a 3×3 array. If you mentally delete the first row and column of that array, you have a 2×2 array left. This is the array for which you find a determinant. I demonstrate this below with the 1 and the 6 from the array above.


 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Do this process for all nine elements of the matrix and you should end up with this matrix:

\begin{bmatrix} 2 & 2 & -3 \\ 4 & -11 & 6 \\ -3 & 6 & -3 \end{bmatrix}

Finally, divide this matrix by the determinant you found above. The result will be the inverse of the original matrix:

\dfrac{1}{-3} \begin{bmatrix} 2 & 2 & -3 \\ 4 & -11 & 6 \\ -3 & 6 & -3 \end{bmatrix}= \begin{bmatrix} ^{-2} \! /_3 & ^{-2} \! /_3 & 1 \\ ^{-4} \! /_3 & ^{11} \! /_3 & -2 \\ 1 & -2 & 1 \end{bmatrix}

I’ll leave it to you to verify that this is indeed the inverse of the original matrix. For any 3×3 matrix that has an inverse, this method will calculate it for you correctly every time.

Sum( and Seq( commands on your calculator

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Can you find the sum of the following series?

\displaystyle \sum_{i=1}^{15}\dfrac{2n+1}{3n-2}

 This is neither an arithmetic nor a geometric series, so you don’t have a formula for it. This would be a tedious problem to do by hand. Fortunately, your graphing calculator can do these problems quickly and efficiently.

There are two functions you need to use on your calculator. The seq( command creates a sequence of terms based on a rule that you give. The sum( command adds together the terms in a sequence. Both functions are found on the LIST menu on your calculator. The seq( command is on the OPS submenu and the sum( command is on the MATH submenu.

To sum a series, you combine the two commands. If you have the new operating system on your calculator, it will prompt you for the entries when you select the seq( command. If you have the old operating system, you need to know the syntax for the command. The syntax for the series above is:

 sum(seq((2x + 1)/(3x – 2),x,1,15))

Note that the seq( command has four parameters in the parentheses. From left to right, these are 1) the rule for the nth term of the sequence; 2) the variable name; 3) the first value of the variable; and 4) the final value of the variable. Now all you need to do is type this in to your calculator and let it do the crunching:

 

The “parent” functions

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Do you know your parent functions? I’m surprised every year when some of my students don’t know how to graph some of the elementary functions they’ve learned in class. Your chances for success in precalc and calculus are significantly better if you memorize the shapes of these ten functions. No excuses, no shortcuts. Just learn them.

 y = xn (Power functions)

 

 

 

 

 

 

 

 

 

 

y = |x| (absolute value function)

 

 

 

 

 

 

 

 

 

 

y = ⌊x⌋ (Greatest integer function)

 

 

 

 

 

 

 

 

 

 

y = √x

 

 

 

 

 

 

 

 

 

 

y = 1/x

 

 

 

 

 

 

 

 

 

y= bx (exponential function – growth and decay)

 

 

 

 

 

 

 

 

 

y = log x

 

 

 

 

 

 

 

 

y = sin x and y = cos x

 

 

 

 

 

 

 

 

y = tan x

 

Factoring sums and differences of fifth powers (and higher!) when the power is odd

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Occasionally, you are required to factor a polynomial that is in the form of a sum or difference of two power terms (i.e., x– yn). In my last post, I showed you a simple procedure you can learn to help you factor the binomial if the value of the power is 4 or greater and the power is even. [If you are reading this post, you should already be familiar with factoring the difference of two squares and the difference of two cubes. Check this website for posts on those procedures.] In this post, I show you a procedure (more complicated than the others, I admit) that you can use to factor these binomials when the power is odd.

First, let’s see how this works on some examples:

 

 

 

 

Do you see the pattern? When factoring a– bn or a+ bn (for n odd), there is always a linear term a ± b, where the plus or minus sign is the same as in the original binomial. The remaining factor is a little more complicated, but it does have a simple pattern. Each term consists of the first term in the binomial (with its exponent decreasing from n – 1 to 0) and the second term in the binomial with its exponent increasing from 0 to n – 1. In the first example above, the x term decreased from x4 to x0. The 2 term increased from 20 to 24. If you started with the difference of two powers, all of the signs in the second factor are plus signs. If you started with the sum of the two powers, the first sign in the second factor is a minus sign and the signs alternate after that.

Factoring differences of fourth powers (and higher!) when the power is even

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Occasionally, you are required to factor a polynomial that is in the form of a difference of two power terms (i.e., xn – yn). There is a simple procedure you can learn to help you factor the polynomial if the value of the power is 4 or greater and the power is even. [If you are reading this post, you should already be familiar with factoring the difference of two squares and the difference of two cubes. Check this website for posts on those procedures.]

First, let’s see how this works on some examples:

 

 

 

 

 

 

You can see that whenever the power is even, you treat the binomial as the difference of two squares and find its factors. If the factors are still the sum and difference of perfect squares, you repeat the process with the term that is the difference of two squares. If the factors are now the sum and difference of perfect cubes, you follow the method for factoring the sum or difference of two cubes. If the powers are sums and differences of odd powers higher than 3, there is another procedure to factor them. You can find another post on this website that shows that procedure.

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