Just to make sure we're using the same language, I'm going to use the function form of:
[] I would agree that k = 2, since the period is only half as long as a normal sine function. So, we so far, have y = A sin(2x) + h. We still need to find A and h.
[] The y-intercept is 3. Remember that the y-intercept happens when x = 0. So, plugging in x = 0 into our formula, we have: 3 = A sin(2*0) + h. In other words, 3 = A sin(0) + h = 0 + h = h. So, we now know that h = 3. The formula is now y = A sin(2x) + 3.
[] Finally, there is a single x-intercept. Picture what the sine function looks like right now, it is floating in the air around y = 3. We need to stretch it vertically until it just grazes the x-axis.
If A = 1, then our sine function bounces between 2 and 4 (+/- 1 around h = 3). But that doesn't touch 0, so no good.
If A = 2, then our sine function bounces between 1 and 5 (+/- 2 around h = 3). Again, not quite touching 0 yet.
The answer should be A = 3, then our sine function bounces between 0 and 6 (+/- 3 around h = 3).
The final formula is y = 3 sin(2x) + 3.
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