#1

Oj guys, I need some help with this math project. I need to find the phase shift and vertical shift of three trigonometric equations. I don't know if I'm doing these right so that's why I came on here. The pi doesnt look right on heree but i'll try it anyways (its going to be π for pi) ok here they are...

y= 3 sin (2x+π/4)

y= .5 cos (4x-π+1

y= -2 sec .5x

Please help!

y= 3 sin (2x+π/4)

y= .5 cos (4x-π+1

y= -2 sec .5x

Please help!

#2

shifting from what? the normal equations:

y=sin x

y=cos x

y= sec x

?

more specific please.

y=sin x

y=cos x

y= sec x

?

more specific please.

#3

I guess, all it says on the paper is what I said....but that's what I was thinking.

#4

y= 3 sin (2x+π/4)

ok, sin (2x) means there's going to be twice as many cycles, ie. sin repetes after pi, rather than 2 pi.

+ pi/4 in the bracket means you shift the graph left by pi/4 (i think)

3 of every thing means the graphs limits at 3 and -3, rather than 1 and -1.y= .5 cos (4x-π+1

see: above except now its (4x) rather than (2x) and .5 rather than 3. the -pi means shift the graph to the right by pi.

and the +1 at the end means the whole graph is shifted up 1, so it limits at 0 and 2, rather than 1 and -1.y= -2 sec .5x

simpler version of the above

enjoy.

#5

So for the first one, the vertical shift would be 3?

#6

it doesn't shift vertically. it stretches.

#7

So what should I put for vertical shift? 0?

#8

i honestly don't know what your questions mean either.