: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!

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- April 25th 2007, 04:28 PMMathNeedy18LAst 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! - April 25th 2007, 05:01 PMecMathGeek
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)