Help with graphing trigonometric functions?

Question

I need help in breaking apart this problem so that I can properly graph it on my own:

sketch the graph of y= 2 sin (-2x+pi) +1 in the interval 0≤ x ≤ 2pi

I think that I need to factor out the -2, making the equation y= 2 sin -2(x-pi/2) +1

Any help would be appreciated. I know you can't necessarily graph but explanations would help. Please explain as thoroughly as possible so that I can learn the concept and complete similar problesm on my own. Thank you so much and 10 points to the best answer!

Answer 1

To do something like this, you just need to know the simple rules of stretching and translation.

In your case, consider sin(x) to be f(x)

When you have f(x) + a, you simply move the whole graph up or down corresponding to the value of a. This is the case with the "+1" at the end of your graph.

When you have b * f(x), your graph is stretched vertically by a factor of b. This is the case with the "2" in front of "sin(-2x + pi)." Every one of the infinite points on the graph is simply 2 times as much as it used to be. If b is negative, this will flip the graph upside down over the x axis as well. Note that the "+1" in your graph is applied after the stretch because it is not included in your f(x) and is therefore not multiplied by 2.

For your x variable, things work the opposite way. If you have f(x+c), the whole graph will move left, in the negative direction, by the value of c. This is the case with "+ pi" in your function; the whole graph will be moved to the left by pi.

When you have f(d * x), your function is shrunk horizontally by the factor of d. That is, the period of f(d * x) is (the period of f(x))/d). You have "-2x," so d in your case would be -2. Because it is negative, you must flip the function horizontally, over the y axis. The 2 will make your period half as big.

When you combine all of these, you end up with a graph of sin(x) that has been flipped over the y axis, shrunk horizontally by a factor of 2, shifted left by pi, stretched vertically to twice the height, and finally shifted up by 1.

Answer 2

You graph the y values (vertically, usually) and the x values (horizontally).  Typically -∞ is left side for x and bottom for y.  But you know all this...

To graph the line you need to graph enough pairs of points (x,y) to make a smooth drawing.  You know this, too.

  So, all you need to do is compute enough (x,y) pairs to draw.  You seem a little shaky here.

  I fail to see what advantage (in computing the y value) there would be in changing sin(-2x + π) to sin( -2 ( x-π/2)).  I guess if you think its easier to add π/2 to your various x values and then multiply by -2, go right ahead.  Personally, I'd prefer to start with the x values, multiply by -2 then add π.  You say poe - tae - toe, I say pah-tah-toe.

Depending on whether you're doing the calculations by hand, or not, simplifying the expression might make sense: sin(X+π) = sin(-X)

so that would switch the domain to the negative side if you use substitute x' = -x for x.  

Or you could go with sin(2(π/2 - x)) = 2sin(π/2-x)cos(π/2-x)

= 2*cos(x)*sin(x)... but the only advantage in that approach is if you don't have a (scientific) calculator (or computer) and do have a table of sines and cosines and a lot of time on your hands to multiply sine by cosine.

So what I'd do is prepare 5 columns and about 16 rows, and starting the leftmost column I'd label it "X" and fill the rows with x = 0, 0.4, 0.8, ...,6.4 (6.4 is outside the domain for x so don't graph it) I'd then label the next column to the right -2x and fill it in, the next column would be -2x+π, the next sin(-2x+π) and finally twice that (which is y).

Then I'd plot the x, y points.

BTW, its lots easier with a spreadsheet...and 0.4 is about the biggest step you can take and still preserve the shape of the curve, you could graph more points at smaller steps between if you like.   You may want to add points at the maximum and minimum values of y to get the shape there just right... don't forget that  0 ≤ x means the end of the graph would be a filled-in point (0<x would be an open circle).  Same for other end, 'course.