Question

trying to figure out phase shift for
y = -1/2 sin (pi^2- Pi*x/2) ? I am completely stumped? I know what the graph looks like but I can't figure out whats' "C" or "B" values? (i.e. C/B)

Answer 1

y = A sin[B (x+C) ] + D

has the period of 2pi/B

so your goal is to write the pi^2- Pi*x/2 in the form of B(x+C)

Answer 2

so I would need to factor out a PI?

Answer 3

or one can say \(\bf y=Asin(Bx+C)\qquad \textit{phase shift}\implies \cfrac{C}{B}\)

Answer 4

So the "b" value is pi/2?

Answer 5

but i'm still lost on how to put it into bx+c form?

Answer 6

\(\bf y=-\frac{1}{2}sin\left(\pi^2-\frac{\pi}{2}x\right)\implies y=-\frac{1}{2}sin\left({\color{blue}{ -\frac{\pi}{2}}}x+{\color{blue}{ \pi^2}}\right)
\\ \quad \\
\textit{phase shift}\implies \cfrac{{\color{blue}{ \pi^2}}}{{\color{blue}{ -\frac{\pi}{2}}}}\)

Answer 7

so -2pi is the phase shift? I think i'm grasping straws:/

Answer 8

yes

Answer 9

but when i calculated the period i got 4? and the graph doesn't look like that?...

Answer 10

I think i missed step somewhere?

Answer 11

hmm

Answer 12

\(\bf y=-\frac{1}{2}sin\left(\pi^2-\frac{\pi}{2}x\right)\implies y=-\frac{1}{2}sin\left({\color{blue}{ -\frac{\pi}{2}}}x+{\color{blue}{ \pi^2}}\right)
\\ \quad \\

period\implies \cfrac{2\pi}{{\color{blue}{ -\frac{\pi}{2}}}}\)

Answer 13

actually you got the period correct :)

Answer 14

I appreciate the help, I think maybe my graph is wrong, im not sure. I'm just going to try and keep re working. ty

Answer 15

https://www.wolframalpha.com/input/?i=%E2%88%921%2F2+sin%28pi%5E2%E2%88%92pi%2F2x%29&dataset=

Answer 16

see, yes thats the exact same graph I had, but i'm confused as why the sine function is not intersecting at 0,0 even though its reflective across the x-axis, and the amplitude is 1/2, but yet it's like it starts it cycle at .2?