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The graph of a sinusoidal function has a maximum point at $(0,8)$ and a minimum point of $(5,2)$.

Write the formula, where $x$ is in radians.

I don't know where to start, and how to do the problem. Don't we have two possible choices:\begin{align*} & y=a\cos (bx)+c\\ & y=a\sin (bx)+c\end{align*}And I'm not sure which one to use. Since it seems like that both could work.

I know that the midline is $y=5$ so $c=5$ and this function has an ampitude of $3$. But I'm not sure what next...

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    $\begingroup$ "The formula" is clearly a mistake... $\endgroup$ Nov 19, 2016 at 17:14
  • $\begingroup$ @DanielW.Farlow I didn't quite understand, could you please elaborate? $\endgroup$
    – Frank
    Nov 19, 2016 at 17:16
  • $\begingroup$ What Daniel means is that there isn't any one formula to describe what the problem wants. In fact there are infinitely many formulas. And since cosine is just the sine graph transitioned over, you can use either $\endgroup$
    – WaveX
    Nov 19, 2016 at 18:06

2 Answers 2

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Since your maximum occurs at $(0,8)$ and your minimum at $(5,2)$, you know you have an amplitude of 3, as you stated, a period of 10, and phase shift...well, that depends on which function you want to use, $\sin$ or $\cos$. I argue for simplicity...indicating that we should use $\cos$. Try out the function $y=3\cos(\pi x/5)+5$.

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  • $\begingroup$ Why is the period of $10$? $\endgroup$
    – Frank
    Nov 19, 2016 at 17:22
  • $\begingroup$ Well, that assumes the maximum and minimum are as close as possible together. Of course, your period could be much smaller if those are just arbitrary max/min. The main problem with your question is lack of specificity. Whoever wrote it...you should make a complaint. $\endgroup$ Nov 19, 2016 at 17:24
  • $\begingroup$ Khan academy... $\endgroup$
    – Frank
    Nov 19, 2016 at 17:28
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Hint:

first note that you can use a $cos$ function or a $sin $ function because the two functions differ only by a phase, i.e. $\cos x= \sin(\pi/2 -x)$.

Sowe can write the function in the form $y=a\cos (\omega x+\varphi) +b$ (the other form $y=a \sin(\omega' x + \varphi')+b$ being equivalent ) and we have four unknowns parameters that we can found with the four conditions: $$ \begin{cases} y'(0)=0\\ y'(5)=0\\ y(0)=8\\ y(5)=2 \end{cases} $$

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