Draw a graph of the following sinusoidal situation. Then write an equation in the form Draw a graph of the following sinusoidal situation. Then write an equation in the form 
$\displaystyle y\,\, = \,\,a\cos {{2\pi \left( {t\,\, - \,\,c} \right)} \over p}\,\, + \,\,b$
    The high tide in a harbour occurs at 2:30 am and the depth at high tide is 16 metres. 
    8.4 hours later the low tide occurs. The depth of low tide is 4 metres.


 A: Since high tide is $16$ and low tide is $4$, the amplitude is half the difference, which gives us $a=\dfrac{16-4}{2}=6$. By taking the average, the "middle" tide corresponds to the vertical displacement of the graph, giving us $b=\dfrac{4+16}{2}=10$. Note that 1 period occurs after we go from high tide, to low tide, then back to high tide. Thus, by symmetry, we have $p=2(8.4)=16.8$.
Now assume that $t$ represent the number of hours that have passed after $12$am (midnight). Then since high tide occurs at $2$:$30$am, we know that the graph contains the point $(2.5, 16)$. Hence, since the graph of $y=\cos x$ contains the point $(0,1)$, the phase shift must be $c=2.5$.
This gives us the equation:

$$
y=6\cos{\left(\dfrac{2\pi}{16.8}(x-2.5) \right)}+10
$$

You can see how Wolfram|Alpha plots this here. The important points you need to plot are:

  
*
  
*$(2.5, 16)$
  
*$(6.7, 10)$
  
*$(10.9, 4)$
  
*$(15.1, 10)$
  
*$(19.3, 16)$
  

where the $x$-coordinates increase by a quarter of the period each time and the $y$-coordinates oscillate from its max to its min then back to its max.
