Question 1062508
*[illustration triggraph].
<pre>
That's a sine graph shifted left from the origin (0,0)
to the first x-intercept left of the x-axis.

{{{f(x)}}}{{{""=""}}}{{{a*sin(bx+c)}}}

Since the highest it goes in 2, and the lowest it goes is
-2, the amplitude, a, is 2.
So we have so far:

{{{f(x)}}}{{{""=""}}}{{{2*sin(bx+c)}}}

The period is given by the formula {{{2pi/b}}}

Each mark on the x-axis represents &#960; because there
are 6 of them and the 6&#960; at the right end of the graph
lets us know that each mark counts as &#960;.

We can tell the period because it's the distance along
the x-axis from 1 peak to the next.

The peak marked (-&#960;/2,2) has x-coordinate -&#960;/2 and the
next peak is halfway between 3&#960; and 4&#960; which is {{{3&1/2pi}}}, 
or {{{expr(7/2)pi}}} or {{{(7pi)/2}}}. So the period is the 
difference of these two x-values, which is 
{{{(7pi)/2-(-pi/2)}}}{{{""=""}}}{{{8pi/2}}}{{{""=""}}}{{{4pi}}}

Since the period is given by the formula {{{2pi/b}}}, we have
the equation:  

{{{2pi/b}}}{{{""=""}}}{{{4pi}}}

Multiply both sides by b

{{{2pi}}}{{{""=""}}}{{{4pi*b}}}

Divide both sides by 4&#960;

{{{2pi/(4pi)}}}{{{""=""}}}{{{b}}}

Cancel the &#960;'s:

{{{2/4}}}{{{""=""}}}{{{b}}}

{{{1/2}}}{{{""=""}}}{{{b}}}

So we have so far:

{{{f(x)}}}{{{""=""}}}{{{2*sin(expr(1/2)*x+c)}}}

Finally we need to know the phase shift which is {{{c/b}}},
right if it's negative and left if it's positive.

The first x-intercept left of the origin is halfway
between -&#960; and -2&#960; which is {{{-1&1/2pi}}}, 
or {{{expr(-3/2)pi}}} or {{{-(3pi)/2}}}. 

So c = that with a positive sign since it shifts left from (0,0,)

{{{c/b}}}{{{""=""}}}{{{((3pi)/2)/2}}}{{{""=""}}}{{{((3pi)/2)*(1/2)}}}{{{""=""}}}{{{3pi/4}}}

So the final answer is

{{{f(x)}}}{{{""=""}}}{{{2*sin(expr(1/2)*x+3pi/4)}}}

Edwin</pre>