Write a sine function that has a midline of y=2 an amplitude of 5, a period of 3pi and a horizontal shift of (3pi )/(2) to the right. Answer Attemptiout of 2 f(x)=

To write a sine function with the given parameters, we need to consider the following:<br /><br />1. Midline: The midline is the horizontal line that divides the graph of the function into two equal parts. In this case, the midline is $y=2$, which means the function will have a vertical shift of 2 units upwards.<br /><br />2. Amplitude: The amplitude is the distance between the midline and the maximum or minimum value of the function. In this case, the amplitude is 5, which means the function will have a vertical stretch of 5 units.<br /><br />3. Period: The period is the length of one complete cycle of the function. In this case, the period is $3\pi$, which means the function will repeat every $3\pi$ units.<br /><br />4. Horizontal shift: A horizontal shift moves the graph of the function left or right. In this case, the horizontal shift is $\frac{3\pi}{2}$ to the right, which means the function will be shifted $\frac{3\pi}{2}$ units to the right.<br /><br />Using these parameters, we can write the sine function as:<br /><br />$f(x) = 5\sin\left(\frac{2\pi}{3\pi}(x - \frac{3\pi}{2})\right) + 2$<br /><br />Simplifying the expression inside the sine function, we get:<br /><br />$f(x) = 5\sin\left(\frac{2\pi}{3\pi}x - \pi\right) + 2$<br /><br />Therefore, the sine function with the given parameters is:<br /><br />$f(x) = 5\sin\left(\frac{2\pi}{3\pi}x - \pi\right) + 2$