Can someone help me with determining the equation for a trig function? I'm aware of how to find the amplitude and period, i'm just confused with this graph!

- anonymous

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- schrodinger

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- anonymous

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- MTALHAHASSAN2

wait how can we find the amplitute

- anonymous

it's usually the "a", I know how to determine it by looking at an equation

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## More answers

- jim_thompson5910

\[\large \text{amplitude} = \frac{|m-n|}{2}\]
m = largest y coordinate of a point on the curve
n = smallest y coordinate of a point on the curve

- jim_thompson5910

|dw:1444532222391:dw|

- jim_thompson5910

|dw:1444532254763:dw|

- anonymous

in this case it would be 2+4/2

- anonymous

which would be 3..

- jim_thompson5910

yes

- jim_thompson5910

the midline is found by averaging the two extreme values
\[\Large D = \frac{m+n}{2}\]
y = D is the equation of the midline
|dw:1444532397834:dw|

- anonymous

the midline is -1

- jim_thompson5910

|dw:1444532428898:dw|

- jim_thompson5910

`the midline is -1` yes

- anonymous

so for the period since the points aren't actually on the radians how would I calculate where it starts and where it ends?

- jim_thompson5910

look at the peak that lies on the y axis
this is when x = 0
the next peak over to the right is at x = pi
so the period is |pi - 0| = |pi| = pi units

- anonymous

I'm a little confused..

- anonymous

when x=0 y=2, is that the peak that you're talking about?

- jim_thompson5910

see the attached image

##### 1 Attachment

- anonymous

ok thanks! I got that. I'm just confused why we are subtracting the pi-0. I thought the formula for period was 2pi/k

- jim_thompson5910

you might be thinking of T = 2pi/B and that is useful when you have an equation of this form y = A*cos(Bx - C) + D

- jim_thompson5910

I subtracted the x values to find the distance between them

- jim_thompson5910

the absolute value is to ensure the distance is positive

- anonymous

so for this graph we don't need to use that form. We just find distance simply by finding out the x values from the y and subtract them from one another?

- jim_thompson5910

you can pick on any 2 neighboring peaks and do the same thing

- jim_thompson5910

we can't use that form because we don't have the equation

- anonymous

oh ok! That's where I was confused at :)

- jim_thompson5910

you might have learned how to pick apart y = A*cos(Bx - C) + D (eg find the amplitude to be |A|, etc)
but now we're going in reverse: finding the key elements and using them to construct the equation

- anonymous

so our equation would be y=3sin(x)-1?

- anonymous

y=sin(pi)-1

- jim_thompson5910

what is y when x = 0 ?

- anonymous

2

- jim_thompson5910

but does plugging in x = 0 give y = 2 when you use y = 3*sin(x) - 1 ?

- anonymous

no

- jim_thompson5910

so that equation doesn't fit

- jim_thompson5910

let's try cos instead of sin

- jim_thompson5910

if x = 0, does y = 2 for y = 3*cos(x) - 1 ?

- anonymous

it does fit

- jim_thompson5910

yep, ok how about if x = pi/2, does y = -4 like it should?

- anonymous

we know that cos(x)=1

- anonymous

I don't think so

- anonymous

cos of pi/2 is 1

- jim_thompson5910

cos(pi/2) = 0 actually

- anonymous

and 3-1=2, and 2-1=1

- jim_thompson5910

recall that the period is pi units
so T = pi
Solve the equation T = 2pi/B for B to get...
T = 2pi/B
pi = 2pi/B
B*pi = 2pi
B = 2pi/pi
B = 2
so the equation should be y = 3*cos(2x) - 1

- jim_thompson5910

now if we try x = pi/2, then
y = 3*cos(2x) - 1
y = 3*cos(2*pi/2) - 1
y = 3*cos(pi) - 1
y = 3*(-1) - 1
y = -3 - 1
y = -4
and it fits

- anonymous

I'm a little lost. Why are we using t=2pi/3?

- anonymous

B*

- jim_thompson5910

The period is T = pi
going from peak to peak is pi units across

- jim_thompson5910

plug T = pi into T = 2pi/B and solve for B to get B = 2

- jim_thompson5910

T = 2pi/B is the period formula
If you know B, you can find the period T (or vice versa)

- anonymous

so we just knew the T, but in order to find b we had to plug in pi for t

- jim_thompson5910

exactly

- anonymous

and how did you know that y=-4?

- anonymous

just the point on the graph?

- jim_thompson5910

##### 1 Attachment

- jim_thompson5910

Yes the point (pi/2, -4) is on the graph
Notice how in my work above I replaced x with pi/2 in the equation y = 3*cos(2x) - 1
Then I evaluated/simplified to get y = -4. So that confirms the equation works for x = pi/2

- anonymous

oh so you just replaced it to check to see if the equation that we wrote was correct.. makes sense. :) Thank you lots!

- jim_thompson5910

yes I recommend you check x = pi as well. Make sure that the output is y = 2

- anonymous

Thanks again :) You're a life saver! I honestly had a lot of trouble with this problem

- jim_thompson5910

I'm glad it's making sense now

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