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postnew [5]
3 years ago
13

write the equation of the transformed graph of sine with period pi that has been shifted vertically up 3 unit and has an amplitu

de of 3/4
Mathematics
2 answers:
gregori [183]3 years ago
6 0
(3/4)sin(2x)+3 is the transformed equation
____ [38]3 years ago
3 0

Answer:

f(x)=\frac{3}{4}\sin (2x)+3

Step-by-step explanation:

We have the function, f(x)=\sin x

It is required to form a function with period \pi, shifted vertically 3 units upwards and having amplitude = \frac{3}{4}

Now, as we know, 'If a function f(x) has the period P, then f(bx) will have period \frac{P}{|b|}'.

Since, the new function need to have period \pi, that is the value \frac{P}{|b|}=\pi i.e. \frac{2\pi}{|b|}=\pi.

So, b= 2 implies the new function is f(x)=\sin (2x)

Further, as the function need to be vertically shifted 3 units upwards, we get the new function,  f(x)=\sin (2x)+3.

Finally, the amplitude of the function must be \frac{3}{4}, this means that the maximum and minimum value of the function is \frac{3}{4} and \frac{-3}{4}.

This gives us the transformed final function is f(x)=\frac{3}{4}\sin (2x)+3.

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