# Pre Calculus 12

posted by on .

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 - ,

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(2-d)
I know that sin π/2 = 1
so π/3(2-d) = π/2
divide both sides by π
(1/3)(2-d) = 1/2
times 6
2(2-d) = 3
2-d = 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 - ,

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 - ,

"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