Question

Help! I cant figure this out for the life of me! Trig graph attached....

Answer 1

not too bad
cosine starts at 1 and goes to -1, this one starts at -2 and goes to 2, so we know
a) the amplitude is 2 and
b) the coefficient is negative

Answer 2

so we know it looks like
\[y=-2\cos(bx)\] and we only need \(b\) is that part clear?

Answer 3

Yes, but why is it negative?

Answer 4

because it hits the y intercept at a negativ point?

Answer 5

because \(\cos(0)=1\) but here you have \(a\cos(0)=-2\)

Answer 6

so you know \(a\times 1=-2\) and therefore \(a=-2\)

Answer 7

yeah, what you said. it starts at -2 and not at 2

Answer 8

ok! i get that part..

Answer 9

so all you need now it \(b\) and to do that you need to see what the period of this function is

Answer 10

or figure that it goes through 3 complete periods as x goes from 0 to \(2\pi\)

Answer 11

looks like the period is \(\frac{2\pi}{3}\) and the period of \(-2\cos(bx)\) is \(\frac{2\pi}{b}\) so you can set
\[\frac{2\pi}{3}=\frac{\pi}{b}\] and in your head solve to get \(b=3\)

Answer 12

jeez something is wrong with me
i mean

set \[\frac{2\pi}{3}=\frac{2\pi}{b}\] and get \(b=3\)

Answer 13

http://www.wolframalpha.com/input/?i=-2cos%283x%29++domain+-1+to+5

Answer 14

ooo! ok i get it now!

Answer 15

thanks for your help :D

Answer 16

I wish i could print this and keep it in my notes