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Mathematics 15 Online
OpenStudy (anonymous):

write the equation of the function represented by the following graph medal

OpenStudy (anonymous):

jimthompson5910 (jim_thompson5910):

See if you can determine * The amplitude * The period

OpenStudy (anonymous):

3 and pi/4

jimthompson5910 (jim_thompson5910):

The period is actually pi/2

jimthompson5910 (jim_thompson5910):

But you have the right amplitude

OpenStudy (anonymous):

got it then what next

jimthompson5910 (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

jimthompson5910 (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

jimthompson5910 (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

OpenStudy (anonymous):

b=0 and t=0? @jim_thompson5910

jimthompson5910 (jim_thompson5910):

T = pi/2 is the period its not t = 0

jimthompson5910 (jim_thompson5910):

\[\Large B = \frac{2\pi}{T}\] \[\Large B = \frac{2\pi}{\pi/2}\] \[\Large B = ??\]

OpenStudy (anonymous):

b= 90 degrees

jimthompson5910 (jim_thompson5910):

idk how you're getting that

OpenStudy (anonymous):

http://www.wolframalpha.com/input/?i=+pi%2F2

OpenStudy (anonymous):

0

jimthompson5910 (jim_thompson5910):

can you simplify this expression? \[\Large \frac{2\pi}{\pi/2}\]

OpenStudy (anonymous):

1

jimthompson5910 (jim_thompson5910):

not quite

jimthompson5910 (jim_thompson5910):

\[\Large \frac{2\pi}{\pi/2} = \frac{2\pi}{1} \times \frac{2}{\pi} = ???\]

OpenStudy (anonymous):

0

OpenStudy (anonymous):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

no, forget about the pi think of the pi as x if you have to

jimthompson5910 (jim_thompson5910):

2x/1 times 2/x simplifies to what?

OpenStudy (anonymous):

4

jimthompson5910 (jim_thompson5910):

so B = 4

jimthompson5910 (jim_thompson5910):

what is A?

OpenStudy (anonymous):

1

jimthompson5910 (jim_thompson5910):

no, remember that above I said that |A| = amplitude and you said earlier that the amplitude was 3

OpenStudy (anonymous):

sorry it 3

jimthompson5910 (jim_thompson5910):

it's actually -3 to make sure that the y-intercept is -3

jimthompson5910 (jim_thompson5910):

so A = -3 B = 4

jimthompson5910 (jim_thompson5910):

y = A*cos(Bx-C)+D A = -3 B = 4 C = 0 D = 0

OpenStudy (anonymous):

so its y=-3cos(4x):)

OpenStudy (anonymous):

y=-3cos(4x) is the answer

jimthompson5910 (jim_thompson5910):

yes correct

OpenStudy (anonymous):

thanks so much !!!

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