According to the description given, the graph of lies between the lines and . It crosses the y-axis at , rises up to 5 then goes down to 1 and finally back up to 2 at . Such a graph with the same basic shape as a sine graph is shown below.
To get an equation for this, of the form , we might start by noticing that the constant m is midway between the extremes – i.e. at , and that the amplitude R is half of the total range – i.e. . For the period to be 6, we need . So giving . Finally, to determine s, we use the fact that the y-intercept is 2. This gives , so .
From the picture on the right we see that there are two points corresponding to .
At Q , , and at P,
But, at Q, increasing t reduces y. So, since f(t) is increasing at , s must correspond to point P.
So will do, and .