anonymous
  • anonymous
write the equation of the function represented by the following graph medal
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
jim_thompson5910
  • jim_thompson5910
See if you can determine * The amplitude * The period
anonymous
  • anonymous
3 and pi/4

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jim_thompson5910
  • jim_thompson5910
The period is actually pi/2
jim_thompson5910
  • jim_thompson5910
But you have the right amplitude
anonymous
  • anonymous
got it then what next
jim_thompson5910
  • jim_thompson5910
The y-intercept is not 0, and the relative extrema is at the y-intercept, so we have a cosine function here Since the y-intercept is negative, this means that the cosine function has a negative coefficient
jim_thompson5910
  • jim_thompson5910
General form y = A*cos(Bx-C)+D C = 0 and D = 0, so don't worry about those |A| = amplitude T = 2pi/B is the period
jim_thompson5910
  • jim_thompson5910
with T = 2pi/B you can solve for B to get B = 2pi/T plug in the period T=pi/2 and simplify
anonymous
  • anonymous
b=0 and t=0? @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
T = pi/2 is the period its not t = 0
jim_thompson5910
  • jim_thompson5910
\[\Large B = \frac{2\pi}{T}\] \[\Large B = \frac{2\pi}{\pi/2}\] \[\Large B = ??\]
anonymous
  • anonymous
b= 90 degrees
jim_thompson5910
  • jim_thompson5910
idk how you're getting that
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=+pi%2F2
anonymous
  • anonymous
0
jim_thompson5910
  • jim_thompson5910
can you simplify this expression? \[\Large \frac{2\pi}{\pi/2}\]
anonymous
  • anonymous
1
jim_thompson5910
  • jim_thompson5910
not quite
jim_thompson5910
  • jim_thompson5910
\[\Large \frac{2\pi}{\pi/2} = \frac{2\pi}{1} \times \frac{2}{\pi} = ???\]
anonymous
  • anonymous
0
anonymous
  • anonymous
jim_thompson5910
  • jim_thompson5910
no, forget about the pi think of the pi as x if you have to
jim_thompson5910
  • jim_thompson5910
2x/1 times 2/x simplifies to what?
anonymous
  • anonymous
4
jim_thompson5910
  • jim_thompson5910
so B = 4
jim_thompson5910
  • jim_thompson5910
what is A?
anonymous
  • anonymous
1
jim_thompson5910
  • jim_thompson5910
no, remember that above I said that |A| = amplitude and you said earlier that the amplitude was 3
anonymous
  • anonymous
sorry it 3
jim_thompson5910
  • jim_thompson5910
it's actually -3 to make sure that the y-intercept is -3
jim_thompson5910
  • jim_thompson5910
so A = -3 B = 4
jim_thompson5910
  • jim_thompson5910
y = A*cos(Bx-C)+D A = -3 B = 4 C = 0 D = 0
anonymous
  • anonymous
so its y=-3cos(4x):)
anonymous
  • anonymous
y=-3cos(4x) is the answer
jim_thompson5910
  • jim_thompson5910
yes correct
anonymous
  • anonymous
thanks so much !!!

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