Posted by Shreya on Wednesday, January 23, 2013 at 10:41pm.
Graphing trigonometric functions
A sinusoidal function has a maximum at (2,10) and its next minimum (5,2) find an equation that represents this situation
A sinuosoidal function has a zero at (5,0) and its next minimum is (7,4). find an equation that represents this situation.
I really don't get how to come up with an equation, this chapter was really the hardest for me. I don't get how to find amplitude, vertical displacement, phase shift
Can someone please explain step by step

Pre Calculus 12  Reiny, Wednesday, January 23, 2013 at 11:04pm
Notice the max is 10 and the min is 2, for a range of 12
which makes the amplitude = 6
the axis would be 4 units high
so far we have something like
y = 6 sin (.......) + 4
the distance from the max to the min along the horizontal is 3 (from 2 to 5 is 3 units)
but the period would be twice that
so the period is 6
remember 2π/k = period
2π/k = 6
6k = 2π
k = 2π/6 = π/3
Ok, now also have the period
and our equation must look something like this:
y = 6 sin (π/3)(x + d) + 4
almost done....
we just have to sub in one of our points to find d , the phase shift
let's use (2,10)
10 = 6sin π/3(2  d) + 4
1 = sin π/3(2d)
I know that sin π/2 = 1
so π/3(2d) = π/2
divide both sides by π
(1/3)(2d) = 1/2
times 6
2(2d) = 3
2d = 3/2
d = 3/2  2 = 1/2
d = 1/2
y = 6 sin (π/3)(x + 1/2) + 4
check for the other point:
if x = 5, we should get y = 2
y = 6 sin (π/3)(5+1/2) + 4
= 6 sin ((π/3)(11/2)) + 4
= 6 sin (11π/2) + 4
= 6(1) + 4 = 2 , YEAHHHHH
I chose to use a sine curve for no specific reason
We could have used a cosine as well.
The period and amplitude and vertical shift would be the same
the change would be in the value of d
Try the second one, it is slightly different.
Make a sketch to see what you are doing. 
Pre Calculus 12  Shreya, Wednesday, January 23, 2013 at 11:16pm
so you added 10 and 2 to get 12, and amplitude is the middle of max and min?
how did you get the axis 4 units?
a period is always twice of something?
the phase shift is sort of confusing me, because isnt that just between the x values? 2, and 5? is there some other way to figure it out? 
Pre Calculus 12  Reiny, Thursday, January 24, 2013 at 12:03am
"how did you get the axis 4 units? "
where is the middle value between 2 and 10
sketch on a number line to see.
the amplitude, or the a value, is 1/2 the difference between the max and the min
difference = 10  (2) = 12
half of that is 6
I illustrated with detailed calculations, how I got the phase shift d