1. The graph has 4 turning points, so the lowest degree it can have
   is degree which is 1 more than the number of turning points 5.  
   So it has degree 5.

2. The graph touches and "bounces off" the x-axis at (-6,0) and (5,0),
   so x=-6 and x=5 are zeros of even multiplicity.  The least possible even
   multiplicity is 2.  Therefore f(x) has factors (x+6)² and (x-5)² 

3. The graph cuts through the x-axis at (2,0), So x=2 is a zero of odd
   multiplicity.  The least possible odd multiplicity is 1.  Therefore,
   f(x) has factor (x-2).

4. f(x) contains the factors (x+6)²(x-5)²(x-2). That would multiply out
   to be a fifth degree polynomial but it may also have a constant factor
   other than 1 as well.  We will let the contant factor be k.  

So we know that f(x) has this form:

         f(x) = k(x+6)²(x-5)²(x-2). 
   
We only need to determine the value of k.  We do that by observing that 
the graph has y-intercept (0,4).  Therefore f(0) = 4.  So we substitute
0 for x in f(x) and set it equal to 4:
 
         f(0) = k(0+6)²(0-5)²(0-2) = 4 =

                   k(-6)²(-5)²(-2) = 4 

                     k(36)(25)(-2) = 4
                        
                            -1800k = 4
   
                                 k = 

                                 k = 
Therefore:

           f(x) = k(x+6)²(x-5)²(x-2)

becomes    
           f(x) = (x+6)²(x-5)²(x-2).

Here is a more accurate graph than the one above:



Edwin