Question

write the equation of the transformed graph of sine with period pi that has been shifted vertically up 3 unit and has an amplitude of 3/4

Answer 1

(3/4)sin(2x)+3 is the transformed equation

Answer 2

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