<PRE><B>Question 1122862</B>
&lt;br&gt;
1. Clearly there can&#39;t be a local maximum at (-pi/3,1) and another at (2pi/3,-7).  The maximum values have to be the same.&lt;br&gt;
So I will assume that (2pi/3,-7) is the local minimum that follows the local maximum at (-pi/3,1).&lt;br&gt;
We want a function in the form&lt;br&gt;
{{{a*sin(b(x-c))+d}}}&lt;br&gt;
where a is the vertical stretch, b defines the period, c is the phase shift, and d is the vertical shift.&lt;br&gt;
From the local maximum to the next local minimum is half a period; in this example that is from -pi/3 to 2pi/3, a difference of pi.  That means the period of the function is 2pi; so there is no horizontal compression of the graph; b = 1.&lt;br&gt;
The difference between the maximum and minimum values is 8, so the vertical stretch is 4: a = 4.&lt;br&gt;
The center line is halfway between the maximum 1 and the minimum -7, at -3; d = -3.&lt;br&gt;
The basic cosine function has its maximum at x = 0; since this function has its maximum at x = -pi/3, the phase shift is -pi/3: c = -pi/3.&lt;br&gt;
We have all the pieces; the function with a=4, b=1, c=-pi/3, and d=-3 is&lt;br&gt;
{{{4*cos(x+pi/3)-3}}}&lt;br&gt;
A graph, showing a local maximum at (-pi/3,1) and the following local minimum at (2pi/3,-7)...&lt;br&gt;
{{{graph(400,200,-pi,2pi,-8,2,4cos(x+pi/3)-3)}}}&lt;br&gt;
2. We need a function in the form&lt;br&gt;
{{{atan(b(x-c))+d)}}}&lt;br&gt;
where a is the vertical stretch, b defines the period, c is the phase shift, and d is the vertical shift.&lt;br&gt;
The halfway points are (-pi/2,-2) and (pi/4,4).&lt;br&gt;
The difference in x values between the halfway points is 3pi/4, so the period is 3pi/2.  Since the period of the basic tangent function is pi, the graph is stretched horizontally by a factor of 3/2.  That means b = (pi)/(3(pi)/2)) = 2/3.&lt;br&gt;
The difference between the y values of the midpoints is 6.  Since the difference between the y values at the midpoints of the basic tangent function is 2, the vertical stretch is 3: a = 3.&lt;br&gt;
The center line is halfway between the y values at the midpoints, which is 1: d = 1.&lt;br&gt;
The center of the period is halfway between the two midpoints, at x = -pi/8.  Since the basic tangent function has the center of its period at x=0, the phase shift is -pi/8: b = -pi/8.&lt;br&gt;
Again we have all the pieces.  The tangent function with a=3, b=2/3, c=-pi/8, and d=1 is&lt;br&gt;
{{{3tan((2/3)(x+pi/8))+1}}}&lt;br&gt;
A graph, showing (-pi/2,-2) and (pi/4,4) as the halfway points of a period....
{{{graph(400,400,-pi,pi,-8,8,3tan((2/3)(x+pi/8))+1,-2,4)}}}&lt;br&gt;
3. Your answer can&#39;t be right; the &quot;2x&quot; still has to be part of the argument for the sine function.&lt;br&gt;
The given function is&lt;br&gt;
5cos(2x-pi/4)-3&lt;br&gt;
To find an equivalent sine function, you only need to know that cos(x) = sin(x+pi/2).  Note that is easy to see if you see that the maximum value of the basic cosine function is at x = 0 while the maximum value of the basic sine function is at x = pi/2.&lt;br&gt;
So all you need to do for this question is add pi/2 to the angle in the given formula:&lt;br&gt;
5sin(2x+pi/4)-3&lt;br&gt;
A graph of the two functions, with the sine function (green) shifted up 1 unit so you can see the two graphs...&lt;br&gt;
{{{graph(400,400,0,2pi,-12,6,5cos(2x-pi/4)-3,5sin(2x+pi/4)-2)}}}</PRE>