anonymous
  • anonymous
Please help...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
http://screencast.com/t/N9H6p6CGYc
anonymous
  • anonymous
@satellite73 @Shadowys @RadEn
anonymous
  • anonymous
\(a\) is the amplitude. since this starts and 3 and goes up do 9, then comes back down to 3 and then to -3, the amplitude is 6

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anonymous
  • anonymous
that is, the range is of length 12 from -3 to 9, so the amplitude is half of that, therefore \(a=6\)
anonymous
  • anonymous
IS THERE ANY OTHER METHOD TO SOLVE IT WITH EQUATIONS.
anonymous
  • anonymous
from your eyes you see that the period is \(\pi\) the period of \(\sin(bx)\) is \(\frac{2\pi}{b}\) so set \(\frac{2\pi}{b}=\pi\) and solve for \(b\)
anonymous
  • anonymous
no there are no equations here, you have to visualize, since you are given a picture
anonymous
  • anonymous
well there is an equation to find \(b\) . it is \[\frac{2\pi}{b}=\pi\] but you only know that the period is \(\pi\) from looking at the graph
anonymous
  • anonymous
and i know c, now can i substitute it in the main equation and find a?
anonymous
  • anonymous
that is the entire point of this exercise, not to use equations, but to visualize the period, and amplitude from the picture
anonymous
  • anonymous
we know \(c=3\) because this is the graph of sine lifted up 3 units
anonymous
  • anonymous
can u explain why period of sin(bx) = 2pi/b
anonymous
  • anonymous
it is always the case that the period of \(\sin(bx)\) is \(\frac{2\pi}{b}\) we can think of it this way. since is periodic with period \(2\pi\) so it does everything on the interval \([0,2\pi)\) now if \(bx=0\) that means \(x=0\) and if \(bx=2\pi\) that means \(x=\frac{2\pi}{b}\) so that gives you the period
anonymous
  • anonymous
sub x = pi then x = 2pi when x = pi y =3 when x = 2pi y = 9 then solve the equations
anonymous
  • anonymous
@satellite73 period is for one complete wave right, how come its 2 pi, it has to be pi
anonymous
  • anonymous
when x = 0 y = 0 then solve the equations for a b and c
anonymous
  • anonymous
cn u show the working @allamiro
anonymous
  • anonymous
dsregard the 2 pi thing = 9 just x = pi and x = 0
anonymous
  • anonymous
what you mean show the work x = pi y = 3 3 = a sin ( b* pi ) + c x= 0 y = 0 0 = a sin ( b * 0 ) + c
anonymous
  • anonymous
yes?
anonymous
  • anonymous
sorry again y = 3 when x = 0 I didnt focus at the graph
anonymous
  • anonymous
so c = 3
anonymous
  • anonymous
ok how do u find b now.
anonymous
  • anonymous
yes so now lets say 9 = a sin ( bx) + 3 the highest value for sin when sinx b = 1 so bx = pi /2 so from there a =6
anonymous
  • anonymous
b?
anonymous
  • anonymous
y = 3 when x = pi /2 3 = 6 sin ( b 2 pi ) + 3 6 sin ( 2 b pi) = 0 sin ( 2 b pi ) = sin ( 2b pi ) = sin ( pi ) b = 1/2
anonymous
  • anonymous
y = 3 when x = 2pi * correctiion
anonymous
  • anonymous
:)

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