Archive

For the Precalc/Trig category

Jan 08

Finding the roots of a quadratic in vertex form

By Tutor GuyNo Comments

 

If you have taken algebra 2, you know you can write quadratic functions in three forms:

  • Standard form: y = ax^2 + bx +c
  • Vertex form: y = a(x-h)^2 + k
  • Factored form: y = a(x-r_1)(x-r_2)

No matter which of the forms you have, you are often asked to find the roots (x-intercepts). The factored form is already done for you: the roots are the values of r_1 and r_2 . If your equation is in standard form, you use the quadratic formula. But what do you do if you have a quadratic in vertex form? Most students are taught to expand the equation into standard form (and then use the quadratic formula). But there’s a quick shortcut that is pretty easy to use. In fact, you can write the roots by inspection. If 0 = a(x-h)^2 + k , then the roots are

x = h \pm \sqrt{\dfrac{-k}{a}}

There are three cases:

  • k=0 : There is one (double) root;
  • a \text{ and } k have different signs: the two roots are real;
  • a \text{ and } k have the same sign: the two roots are complex.

In the complex case, you can express the roots as

x = h \pm \sqrt{\dfrac{k}{a}} \; i

Here are two quick examples:

Example 1: 0 = 2(x-3)^2 + 6

Using the formula above, the roots can be written immediately as

x = 3 \pm \sqrt{\dfrac{6}{2}} \; i =3 \pm \sqrt{3} \; i

Example 2: 0 = 2(x-3)^2 - 8

The roots are:

x = 3 \pm \sqrt{\dfrac{8}{2}} = 3 \pm 2 = 1 \text{ and } 5

Dec 25

graphing sec x, csc x functions

By Tutor GuyNo Comments

 

Graphing secant functions and cosecant functions can be daunting for students. But you can tame them if you recognize how they are related to graphs of sine and cosine functions. If you can graph the associated sine or cosine function, then secant and cosecant graphs will be pretty easy for you.

We will demonstrate this process with an example: Draw the graph of

-3 \csc{(2x+\dfrac{\pi}{2})} - 1

The first step for you is to graph a different function. If you are asked to graph a cosecant curve as in this example, change csc to sin and graph that instead. If you are graphing a secant function, change sec to cos and graph that instead. In our example, we will graph

-3 \sin{(2x+\dfrac{\pi}{2})} - 1

I showed you how to graph sine and cosine functions in a previous post, so check that out if you need help. In that post, we graphed the sine function listed above. Here, we will assume you have already graphed the sine function:


 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Next, put in the vertical asymptotes everywhere the sine curve crosses the midline:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Then use the asymptotes to place the branches of the cosecant function. Also, notice that the minimum points of the cosecant curve touch the maximum points of the sine curve and vice versa:

That’s all there is to it! You can erase the sine curve (in green) or you can leave it. I usually draw in the sine or cosine curve in light pencil or as a dashed line, so that it is obvious which curve is the secant or cosecant curve. Here, the red curve is the graph we desire.

Nov 06

Making sense of math induction

By Tutor GuyNo Comments

 

Do you know the sum of the first five positive odd numbers? How about the sum of the first 20 positive odd numbers? The first 100?  Wouldn’t it be great if there were a formula for the sum of the first n positive odd numbers so we could easily answer this question? Let’s see if we can find a pattern:

n Expression Sum
1 1 1
2 1 +3 4
3 1 + 3 + 5 9
4 1 + 3 + 5 + 7 16
5 1 + 3 + 5 + 7 + 9 25

 

It certainly appears that the sum of the first n positive odd numbers is n2. Or, if we write this symbolically:

\displaystyle \sum_{i=1}^n (2i - 1) = n^2

We’ve just shown that this formula is true for the first five values of n, but how do we know it is true for every value of n? We could test our theorem for n = 6, n = 7, n = 8, and so forth. But you can see that this process would take us a very long time to show it for every value of n. Like, forever! Fortunately, there is a better method for proving conjectures of this sort. It is called mathematical induction. Math induction is a very powerful tool in analysis. If you think about it, it allows us to prove an infinite number of theorems. And we can do this in only three steps!

How does math induction work? It depends on a subtle idea: if the theorem is true for one particular value of n (say n= k) and we can use this fact to show that it’s true for the next value of n (that is, n=k+1), then it will be true for every value of n ≥ k. That is, one proof turns into an infinite number of proofs! How is this possible? Because each value of n leads to the next value of n. If it’s true for n=3 then we’ve shown it’s true for n=4. But since it’s true for n=4, we’ve also shown it’s true for n=5. And therefore it’s also true for n=6, and then n=7, etc. Most textbooks that cover math induction use a dominoes analogy—if you knock down the first domino, it knocks down the second, the second knocks down the third and so on.

 

Here are the three steps in math induction, and you will see that they are quite simple to describe:

Step 1: prove the theorem is true for n=1. This is a really easy step. You just plug in n=1 into the theorem and show that it is true. (Note: on occasion, you will start at a value of n greater than 1. This does not change the validity of the process.)

Step 2: Assume that the theorem is true for n = k. (Note that you aren’t even proving anything in this step. You just write out what the theorem would look like if it were true for n=k.)

Step 3: Use the expression in Step 2 to show that the theorem is true for n=k+1. (You usually do this by applying a little bit of algebra.)

That’s all there is to it! Let’s see how it works on the theorem we proposed above. We restate the theorem here for convenience:

\displaystyle \sum_{i=1}^n (2i - 1) = n^2

Step 1: Prove for n = 1.

1 = 1^2     (See? This is a really easy step.)

Step 2: Assume for n=k

1+3+5+ \dots + (2k-1) = k^2    (Just write down the formula for n = k.)

Step 3: Prove for n = k+1

1+3+5+ \dots + (2k-1) = k^2  \; \; \text{[line 1]}

1+3+5+ \dots + (2k-1)+(2k+1) = k^2 + (2k+1)  \; \; \text{[line 2]}

k^2 + (2k+1) = (k+1)^2 \; \; \text{[line 3]}

[This step is a little harder to follow, so I’ve numbered the lines to explain the process.

  • Line 1 is simply a repeat of Step 2. I wrote it here for clarity; you don’t need to repeat your work.
  • In line 2, I’ve added (2k + 1) to both sides. That’s a valid algebra operation, so if Line 1 is true, Line 2 is also true. Why did I add (2k +1)? Because it’s the next term in the sum on the left side of the equation.
  • In Line 3, I’ve simplified the right side of the equation.]

But (k+1)2 is exactly what our theorem says the right side of the equation will be when n = k+1. So the assumption that the theorem is true for n=k led to the conclusion that it is also true for n=k+1. And our proof is complete.

Oct 30

Managing exponential equations

By Tutor GuyNo Comments

 

Many algebra 2 students get intimidated by exponential equations because the answers are rarely nice simple integers. But keep in mind that like all the equations you solved in algebra 1, you are merely trying to find the value of x that satisfies the equation. The process is a little more complicated because your variable is now in an exponent, but just follow these steps and you’ll soon be an exponential expert!

There are two types of exponential equations and they each have a preferred strategy for solving them. The first type has an exponential expression on both sides of the equal sign, and the two bases are both powers of the same number. The second type of equation has exponential expression with bases that aren’t powers of the same number (or has an exponential expression on only one side of the equals sign).

Type 1. Bases that are powers of the same number. Check out these two problems.

\text{a.  } 4^{x+4} = 8^{x-1}

Solution: note that the two bases (4 and 8) are both powers of 2. We rewrite both bases as powers of 2 and simplify:

\text{a.  } 4^{x+4} = 8^{x-1} \rightarrow (2^2)^{x+4} = (2^3)^{x-1} \rightarrow (2)^{2x+8} = (2)^{3x-3}

Now we take advantage of a simple property that says if a^x = a^y \text{ then } x = y.  Since both bases are the same, we just set the exponents equal to each other:

(2)^{2x+8} = (2)^{3x-3} \rightarrow 2x + 8 = 3x - 3. Solving this gives x = 11.

 

\text{b.  } 5^{4x} = 125^{x+1}

Solution: note that both bases are powers of 5. Proceed exactly as in the last problem:

\text{b.  } 5^{4x} = 125^{x+1} \rightarrow 5^{4x} = (5^3)^{x+1} \rightarrow 5^{4x} = 5^{3x+3} \rightarrow 4x = 3x + 3 \rightarrow x = 3

 

Type 2: Bases are not powers of the same number. To solve these types of problems, you will need to use logarithms. Here are two examples.

\text{a.  } 2^{x+3} = 12

Solution: The bases (2 and 12) are not powers of the same base. (Always check this first, because the method shown in the Type 1 examples above is almost always simpler.) So we solve by taking logs of each side. In the old days (before graphing calculators) you would need to use a common log (base 10) or a natural log (base e). I’ll show you that method first:

\text{a.  } 2^{x+3} = 12 \rightarrow \ln{(2^{x+3})} = \ln{12} \rightarrow (x+3) \ln{2} = \ln{12}

\therefore x = \dfrac{\ln{12}}{\ln{2}} - 3 \approx 0.5840

If your graphing calculator has the logbase command on it, you can solve this problem even more easily by taking a log base 2 of each side:

\text{a.  } 2^{x+3} = 12 \rightarrow \log_2{(2^{x+3})} = \log_2{12} \rightarrow x+3 = \log_2{12}

\therefore x = \log_2{12} - 3 \approx 0.5840

 

\text{b.  } 5^{x+3} = 7^{x-2}

Solution: This looks a lot uglier than the previous example, but the solution starts the same way. Take the log of both sides:

\text{b.  } 5^{x+3} = 7^{x-2} \rightarrow \log_5{(5^{x+3})} = \log_5{(7^{x-2})} \rightarrow (x+3) = (x-2) \log_5{7}

We need to collect the x terms to solve for x, so distribute on the right side and solve:

(x+3) = (x-2) \log_5{7} \rightarrow (x+3) = x \; \log_5{(7)} - 2 \; \log_5{(7)} \rightarrow

3 + 2 \log_5{(7)} = x \; \log_5{(7)} - x \rightarrow 3 + 2 \; \log_5{7} = x (\log_5{(7)} - 1) \rightarrow

x = \dfrac{3 + 2 \; \log_5{(7)}}{\log_5{(7)} - 1} \approx 25.92

Whew!

Sep 11

Calculating permutations and combinations

By Tutor GuyNo Comments

 

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

Aug 14

Graphing sine and cosine functions like a pro

By Tutor GuyNo Comments

 

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!
Aug 07

Graphing sine and cosine functions– an Intro

By Tutor GuyNo Comments

 

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.

Mar 23

Using multiplicity of factors to characterize graphs of rational functions

By Tutor GuyNo Comments

 

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:

Mar 16

Using multiplicity of factors to characterize graphs of polynomials

By Tutor GuyNo Comments

 

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:

Jan 26

Finding the inverse of a 3X3 matrix

By Tutor GuyNo Comments

 

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.

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