# LAst one i Promise.....:D

• Apr 25th 2007, 04:28 PM
MathNeedy18
LAst one i Promise.....:D
:o http://webwork1.math.utah.edu/math10...3-prob6-p6.gif
I can't figure out the formula for this graph. So far i have this:

2sin(3x+2)

I can find the amplitude just fine, i think i can find period, but please explain how to get it! Thanks!
• Apr 25th 2007, 05:01 PM
ecMathGeek
Quote:

Originally Posted by MathNeedy18
:o http://webwork1.math.utah.edu/math10...3-prob6-p6.gif
I can't figure out the formula for this graph. So far i have this:

2sin(3x+2)

I can find the amplitude just fine, i think i can find period, but please explain how to get it! Thanks!

The period is easy enough to find. Just measure the interval alone which the graph starts at one point till it gets back to that same point (also having the same slope).

For example, notice that the graph is at equilibrium (equals 0) and is downward sloping at x = .4, and that the graph is again at equilibrium and downward sloping at x = 2.4. According to this, the period is:

P = 2.4 - .4 = 2

The coefficient of x is b, where:

b = 2pi/P = 2pi/2 = pi.

So the equation should be:

2sin[pi(x - h)]

Where h is the shift in the graph. Recall that the normal unshifted sine function starts at equilibrium with an upward slope at x = 0. In this graph, we see that the function is at equilbrium with an upward slope at x = 1.4. Therefore h = 1.4. So the equation becomes:

2sin[pi(x - 1.4)] = 2sin(pi*x - 1.4pi)