Given:
The graph of a sine function.
To find:
The sine function.
Solution:
The general form of a sine function is
[tex]y=Asin(Bx+C)+D[/tex] ...(i)
Where, |A| is amplitude, [tex]\dfrac{2\pi}{B}[/tex] is period, [tex]-\dfrac{C}{B}[/tex] is phase shift and D is mid-line.
From the given graph it is clear that the minimum value of the function is -4 and the maximum value is 0.
[tex]|A|=\dfrac{Maximum-Minimum}{2}[/tex]
[tex]|A|=\dfrac{0-(-4)}{2}[/tex]
[tex]|A|=\dfrac{4}{2}[/tex]
[tex]|A|=2[/tex]
The graph is reflected across the mid-line, so A=-2.
And,
[tex]D=\dfrac{Maximum+Minimum}{2}[/tex]
[tex]D=\dfrac{0+(-4)}{2}[/tex]
[tex]D=\dfrac{-4}{2}[/tex]
[tex]D=-2[/tex]
Period of the function is 4π because it completer its one cycle in the interval of 4π.
[tex]4\pi=\dfrac{2\pi}{B}[/tex]
[tex]B=\dfrac{2\pi}{4\pi}[/tex]
[tex]B=\dfrac{1}{2}[/tex]
Phase shift is [tex]\dfrac{\pi}{4}[/tex].
[tex]-\dfrac{C}{B}=\dfrac{\pi}{4}[/tex]
[tex]-\dfrac{C}{\frac{1}{2}}=\dfrac{\pi}{4}[/tex]
[tex]C=-\dfrac{1}{2}\times \dfrac{\pi}{4}[/tex]
[tex]C=-\dfrac{\pi}{8}[/tex]
Putting [tex]A=-2,B=\dfrac{1}{2},C=-\dfrac{\pi}{8},D=-2[/tex] in (i), we get
[tex]y=-2\sin\left(\dfrac{1}{2}x-\dfrac{\pi}{8}\right)-2[/tex]
Therefore, the required equation is [tex]y=-2\sin\left(\dfrac{1}{2}x-\dfrac{\pi}{8}\right)-2[/tex].
