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$$f(x)=3\sin\left(x-\frac{5\pi}{6}\right)$$

Amplitude= $3$

Period= $2 \pi$

Midline= $y=0$

Phase Shift= $\frac{5 \pi}{6}$

I found the maximum to be $(\frac{9 \pi}{8}, 3)$, and the minimum to be $(\frac{17 \pi}{8}, -3)$. Neither were correct, and I don't know what else to do. I must have answered them each 7 times. Help please! Thank you!

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    Oh yes, I see. I really do apologize! It is a sine equation. So the full equation is f(x)= 3sin(x- 5pi/6) – 2017-02-16
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    What do you mean by maximum and minimum $(x,y)$ ? As you've written, the amplitude is 3, so the maximum and minimum y value you can attain is $3$ and $-3$, however this function is periodic for all possible x, so there are an infinite amount of x values which will provide you with 3 or -3 – 2017-02-16
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    $$\frac {9\pi}8 - \frac{5\pi}6 = \frac{7\pi}{24}$$ $\sin \frac{7\pi}{24} \ne 1$, so no, you have not found the correct maximum (and similarly for the minimum). Brush up on your fractions. Whether they are willing to settle for a maximum and minimum, or want formulas that provide them all as Hushus46 indicates is another matter. If they are demanding all maximums and minimums, recall that $\sin(\theta + 2k\pi) = \sin(\theta)$ for all integers $k$. – 2017-02-16
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    They would like x and y coordinates for the maximum and minimum closest to (0,0) on the graph, so x-values for the coordinates should reside between 0 and 2 pi on a graph. Obviously noted by the amplitude, the y-value will be 3 for the max or -3 for the min. – 2017-02-16

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