Respuesta :

Solution

- The general form of a sinusoidal function is:

[tex]\begin{gathered} f(x)=A\sin B(x-h)+k \\ where, \\ |A|=\text{ Amplitude} \\ Period(T)=\frac{2\pi}{|B|} \\ \\ y=k\text{ \lparen The Midline\rparen} \end{gathered}[/tex]

- First of all, let us find the value of B using the period formula given above.

[tex]\begin{gathered} \text{ We are told that the period is 6} \\ \therefore T=\frac{2\pi}{|B|} \\ \\ 6=\frac{2\pi}{|B|} \\ \text{ Make \mid B\mid the subject of the formula} \\ \\ \therefore|B|=\frac{2\pi}{6} \\ \\ |B|=\frac{\pi}{3} \\ \\ \text{ Thus,} \\ B=\pm\frac{\pi}{3} \\ \\ \text{ Since there is only the positive }\frac{\pi}{3}\text{ in the options,} \\ \\ B=\frac{\pi}{3} \end{gathered}[/tex]

- Next, we should apply the condition given to us that f(2.5) = 5

[tex]\begin{gathered} f(x)=A\sin Bx+H \\ B=\frac{\pi}{3} \\ \\ f(x)=A\sin\frac{\pi}{3}x+H \\ \\ f(2.5)=5\text{ implies that }x=2.5,f(x)=5 \\ \\ 5=A\sin\frac{\pi}{3}(2.5)+H\text{ \lparen Equation 1\rparen} \end{gathered}[/tex]

- We also know that the Amplitude and Valley are related to the midline as follows:

[tex]\begin{gathered} H-MinValue=A \\ MinValue=2 \\ \\ H-2=A \\ \therefore H-A=2\text{ \lparen Equation 2\rparen} \end{gathered}[/tex]

Solving equations 1 and 2 simultaneously, we have

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