Question

Write the trigonometric equation for the function with a period of 5, a low point of -3 at x=1 and an amplitude of 7

(Found f(x) = 4cos(2pie/5(x-1)) + 1/2 but the min wasn't at 1 when graphed)

Answer 1

Firstly, because amplitude is 7, it must be 7 cos 2pi/5 blah blah blah

secondly, it says "a low point of -3 at x=1" so is the lowest point at x=-3 or x=1

Answer 2

Oh my gosh. That is true. I just realized that. Been doing too much physics stuff before hand. Thank you!

Answer 3

\[Acos(b(x+c))+D\] A is amplitude, Period = 2pi/B, C = phase shift, D=vertical shift

So it should be just \[7cos\frac{2\pi}{5}+ something\]
but what is the lowest point tho, because that affects the value of D

Answer 4

The lowest point is y=-3, so basically (1,-3)

Answer 5

Oh

Answer 6

Well, you have \[7cos(\frac{2\pi}{5}(x+C))+4\]
and C is phase shift
you see that right now it's (2.5,-3) but since you want it at (1,-3) which means that you need to subtract 1.5 from 2.5 to get 1... So you add 1.5... if that makes sense .-.
\[7cos(\frac{2\pi}{5}(x+1.5))+4\]

Answer 7

Wait, I just got f(x) = 7cos(2pie/5(x+1.5)) + 4 but then the h isnt 1

Answer 8

Wdym 'h' what's h

Answer 9

Thank you for clearing that up

Answer 10

Sorry, h is c. Different variables

Answer 11

Thank you!

Answer 12

And why does C have to 1-?

Answer 13

I just realized it doesn't. My mind is so dead right now

Answer 14

lol, well, yw~ xd