I'm trying to graph a sine function and I have five points. I transform them using the parameters of the rule I'm given but I don't have enough coordinates in order to draw the curve properly. Does anybody know how I can create more coordinates by using the ones I already have (I have 5) so that I can graph it?

Any help would be greatly appreciated!

2. Originally Posted by s3a
I'm trying to graph a sine function and I have five points. I transform them using the parameters of the rule I'm given but I don't have enough coordinates in order to draw the curve properly. Does anybody know how I can create more coordinates by using the ones I already have (I have 5) so that I can graph it?

Any help would be greatly appreciated!
What is the function you're trying to plot? For example y = sin(x) goes through $(-\pi, 0) (-\frac{\pi}{2}, -1) (0,0) (\frac{\pi}{2}, 1) (\pi, 0)$

if you have y=sin(x+h) then all your points would move h to the left on the x axis

3. Originally Posted by e^(i*pi)
What is the function you're trying to plot? For example y = sin(x) goes through $(-\pi, 0) (-\frac{\pi}{2}, -1) (0,0) (\frac{\pi}{2}, 1) (\pi, 0)$

if you have y=sin(x+h) then all your points would move h to the left on the x axis
The rule is:
f(x) = 5 sin (-πx/4) - 3

I got the following points by doing (x/b + h, ay + k):

(0,-3)
(-2,2)
(-4,-3)
(-6,-8)
(-8,-3)

4. Just remembering this myself, so maybe I won't be of much help but I will try!

f(x) = 5 sin (-πx/4) - 3

I believe that 5 at the beginning can be used as the slope (or 5/1). I think this is true because of the form y = mx + b.

y = f(x)
m = 5/1
x = sin(-nx/4)
b = -3

m is a slope.

Using the slope you can take a point and add 5 to the y value, and 1 to the x value. This is because a slope is arranged in rise over run form.

5(rise) / 1 (run)

Well... this could be completely wrong but if you have an answer sheet that you can check your answer on it's a worth a shot to see if it is correct. However, if you don't have an answer sheet, you should probably wait for someone that knows for certain.