Sketch (1 Viewer)

[Kutay]

Hey i was wondering what this graph would look like ??

y = 3 sin (x - theta / 2)
for -theta <(orequal) x <(orequal) theta

 

[Miikke]

yup..

its a normal sin curve except:

the amplitude (the height) is 3 so the tip and bott0m of the curve goes to 3 and -3.

[the domain:

-theta=<x-theta/2=<theta
-2theta=<2x-theta=<2theta
-theta=<2x=<3theta
-theta/2=<x=<3theta
] er...i dont think what i just typed for the domain is relavent..

umm for period its 2pi/(x-theta/2) i THINK..will some maths pro coorect me im the worse grapher in grammar..

 

[|Axis_]

For f(x) = M+Acos(Bx+C) = M+Acos[B(x+C/B)]

the graph of y=f(x) is like the cos graph you know, except that:

Amplitude = |A|
Period = |2pi/B|
The graph is shifted up by M (or down if M is negative)
The graph if shifted left by C/B (or right if C/B is negative)
If A is negative then remember that the shape of the graph is like upside-down cos.

You can replace "cos" by "sin" above.