Sunday, June 3, 2007

Stretching Graphs and Compressing Graphs


If you understand how to shift a curve horizontally or vertically, stretching graphs or compressing them isn't much different. Once again, it's only a small modification to the equation that causes the stretch or compression.  (Please click the Facebook Like button at the top of this post, or hit the Google +1 button at the end if you find this post helpful!)

Stretching and compressing graphs vertically is determined by the coefficient in front of the x (or more specifically, in front of the other direct modifications to x).
Let's look at a basic example.... f(x) = x2, a standard parabola.


Now, to vertically compress this curve, you put a 'fraction coefficient' in front of the x component of the graph. i.e. f(x) = (1/2)*x2. This squashes the graph down by a factor of 2. Or, another way to look at it, every y value in this curve is 1/2 of the value in the starting curve. Plot your own points to convince yourself of this. Note that the curve crosses (-2,4) and (2,4) in the original curve, whereas the new one crosses at (-2,2) and (2,2).



Now, naturally, if you put a whole number coefficient in front of the x term, you will be stretching graphs. For example: f(x) = 2x2 looks like this:


You can see that this has caused the parabola to stretch upwards. Note that it now crosses (1,2), not (1,1). Or once again, to look at it from a different angle, every y value is now twice the value as in the original graph.  Compared to the original version, this is a vertical stretch graph.

The only other thing that you should keep in mind is that the coefficient to stretch or compress the graph vertically MUST be in front of any brackets that might be surrounding x, and the coefficient will act on any horizontal translation component and the exponent. Convince yourself of this by looking at graphs such as:

f(x) = (x-3)2....... and f(x)=2(x-3)2
f(x) = (x+1)3...... and f(x)=1/2(x+1)3
f(x) = x + 5...........and f(x) = 4x + 5
f(x) = (x-1)5 + 7...... and f(x) = 4(x-1)5 + 7

So far, all that I have talked about explains the concept of vertical compression and stretching.  But, let's now consider how we would go about showing a horizontal compression or stretch.  If we want to squash a graph together along the x-axis, what would we do?  Well, for any value of x, we would want the graph to be at some fractional value of that x.  So mathematically, this means that we have to modify the x value.  To perform a horizontal compression or stretch on a graph, instead of solving your equation for f(x), you solve it for f(c*x) for stretching or f(x/c) for compressing, where c is the stretch factor.  The simplest way to consider this is that for every x you want to put into your equation, you must modify x before actually doing the substitution.  Let me show you.

Consider again the parabola f(x) = 2x2.  If we want to start plotting this graph, we could start by building a table of values and solving for f(1), f(2), f(3), etc.  Doing the quick math, you can see that f(1) = 2.  Now let's say we want to horizontally compress this graph by a factor of 10.  In this case, we do the modification on x before subbing in, so we apply the f(x/c) for compressing and see that f(1) becomes f(1/10).  You then substitute 1/10 in for x, and solve.  You then come up with the ordered pair of (1, 1/50).  This is a ten times compression along the x-axis.  If you draw some of these out, you will easily see that squishing or stretching along x is different from squishing or stretching along y.

Continuing with this same example, say we want to graph a horizontal stretch of a factor of 4.  We then solve f(c*x), giving us f(4*x).  For x = 1, f(4*1) becomes f(4), which works out to y = 32.  Again, it is easy to see how much we have stretched the graph by this simple modification.

As you can see, stretching graphs (or compressing them), both vertically and horizontally, really isn't that difficult.  There are obviously a few things that you need to remember, especially the distinction between horizontal and vertical changes.  But these modifications are just an extension of what you already know, building on your knowledge of horizontal or vertical shifts. Keep practicing, and you'll get it in no time. :)  (Remember to please hit Facebook Like button back at the beginning, or the Google +1 button if you liked my post!)


Saturday, May 19, 2007

Manipulating graphs


In the next few days, I'm going to begin posting on graph manipulations. For any graphical expression, small changes to the expression can result in a very predictable change in the position of the graph. For instance, the graph can be shifted along the x-axis (horizontally), or along the y-axis (vertically). Similarly, the graph can be stretched or compressed along either axis as well. This may sound confusing, but the alterations to the equations are simple and easily recognizable. I will provide examples and go into detail in the coming days.


Friday, May 11, 2007

Trigonometry - Secant, Cosecant, Cotangent


In addition to the three basic trig functions we've already looked at (Sine, Cosine, Tangent), there are three other related functions. These are Secant, Cosecant, and Cotangent. These functions have similar meanings as the first three, in that they represent the ratios of various side lengths of a right angle triangle, and can be used to find angles or unknown side lengths. I will not go into extensive detail on these functions, as they are less commonly required, but I will show you what they mean.  Please remember to click on the Like button if this is helpful, and at the end, please hit the +1 button as well.

So far, with the help of SOHCAHTOA, we have seen that:

Sine = opposite / hypotenuse
Cosine = adjacent / hypotenuse
Tangent = opposite / adjacent

These new functions are related to the originals because they represent the inverse ratios.

Cosecant = hypotenuse / opposite... (compare to Sine)
Secant = hypotenuse / adjacent....... (compare to Cosine)
Cotangent = adjacent / opposite...... (compare to Tangent)

Also, these functions can be abbreviated: Cosecant = Csc, Secant = Sec, Cotangent = Cot.

At the middle school and high school math level, you will rarely have a need to use these functions, but it is good for you to know what they are. However, most trig problems at this stage can easily be solved with the original three functions.  Just in case, though, it's always good to know all the trig functions: sine, cosine, tangent, secant, cosecant, cotangent.

Please remember to Like and +1 this post if you found it helpful!  Thanks!


Monday, May 7, 2007

Area of a Triangle


With all this talk about triangles and their side lengths and angles, we shouldn't forget to discuss how to find the AREA of a triangle.

You are probably familiar with one formula for finding the area of a triangle:

Area = 1/2 (base)(height)

Compare this to finding the area of a rectangle:

The area of the rectangle is equal to the product of (base) x (height)..... (or length x width). However, by drawing a diagonal within the rectangle which joins two opposite corners, you can see that each newly-formed triangle is equal to half of the area of the original rectangle. Therefore, the area of a triangle is one-half the area of the rectangle, as shown by this triangle area formula. Even if you are looking at a triangle that doesn't immediately look like it is half of a rectangle, this formula still applies.

To prove it, you can draw a line in to represent the height, as I have shown here, thus creating two smaller triangles, and you can rearrange them to see that they indeed are equal to the area of half a rectangle:
That is one way to find the area of a triangle. However, if instead of base and height measurements, you are given lengths of sides or angles, this method won't work for you. In this case, you need to use a trig equation to solve for the area of a triangle.

Let's start with the first equation we had above, and modify it. By the standard trig identities, we can show that:

height = (a)(SinC)

So substituting that into our formula:

Area = 1/2(base)(height)
Area = 1/2(b)(a)(SinC)

And this is the trig formula for solving the area of a triangle!

Area = (1/2)abSinC

You can use this to find the area of a triangle where you know any two sides and the angle between them! It's that easy!


Tuesday, May 1, 2007

Special Angles in Trigonometry


Some angles in trigonometry are so common, they are known as special angles.  The values of trig functions of these specific angles can be represented by known ratios, and are good to commit to memory to help you work through problems faster. These angles can be remembered by examining two different special angle triangles. Specifically, the trig functions for angles of 0, 30, 45, 60, 90 degrees are the special ones.  (Please hit the Like button or +1 button at the end if you think this post helps you with special angles!)

Let's look at the first of the special angle triangles, and this will hopefully become clear. Take a right angle triangle with two 45 degree angles, and with sides of 1 unit length. By the Theorem of Pythagoras, the hypotenuse of this triangle is of length √2. This is what this triangle looks like:
So then, from these values and SOHCAHTOA, you can obtain the trig values for this special angle of 45 degrees. You can see that:

Sin(45) = 1/√2
Cos(45) = 1/√2
Tan(45) = 1


Don't worry if you can't remember these exact ratios... the simplest thing to remember is how to construct the special angle triangle... which is as easy as remembering a right angle triangle with a 45 degree angle and 2 sides of length 1... you can easily fill in the rest, and then work out the trig ratios yourself!

The second of the special angle triangles, which represents the rest of the special angles to remember, is slightly more complex, but still straightforward. Take a right angle triangle with angles of 30, 60, and 90 degrees. The simplest lengths of the sides of this triangle are 1, 2, √3 (with 2 being the longest side, the hypotenuse). This 30 60 90 triangle looks like this:
So then, using the these values, you can obtain the trig values for these special angles as well:

Sin(30) = 1/2
Cos(30) = √3/2
Tan(30) = 1/√3

Sin(60) = √3/2
Cos(60) = 1/2
Tan(60) = √3/1 = √3

Again, just remember how to construct the triangle, and the ratios are easy to come up with!

For 0 and 90 degrees, there isn't a simple triangle scheme to remember the values (although please feel free to correct me if I am wrong!). However, these aren't scary square root numbers or weird fractions:

Sin(0) = 0
Cos(0) = 1
Tan(0) = 0

Sin(90) = 1
Cos(90) = 0
Tan(90) = undefined

If you can familiarize yourself with all of these special angles, or at least understand how to derive them from the special angle triangles, then you will have a much easier time working on trigonometry questions.  Again, please remember to Like and/or +1 this post if it helped.  Thanks.


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