SOLUTION: Find the lowest degree polynomial f(x) that match the graph below. Leave your answer in factored form. Graph: <img src="http://i.imgur.com/2CrqoMi.jpg" >

Algebra ->  Equations -> SOLUTION: Find the lowest degree polynomial f(x) that match the graph below. Leave your answer in factored form. Graph: <img src="http://i.imgur.com/2CrqoMi.jpg" >      Log On


   

Question 762287: Find the lowest degree polynomial f(x) that match the graph below. Leave your answer in factored form.
Graph:

Found 2 solutions by jim_thompson5910, Edwin McCravy:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
There are 3 roots, and the roots on the very left and very right are double roots (since they only touch the x axis and does not pass through)

So there are really 2+1+2 = 5 roots total (2 of which are repeating, only one is a non-repeating root)

So the lowest possible degree is a 5th degree polynomial.

The roots are: -6, 2, 5

The factors would then be: x - (-6), x-2, x-5 which turn into: x-6, x-2, x-5

But remember -6 and 5 are double roots, so the factors are really (x-6)^2, (x-2), (x-5)^2


Put this all together to get (x-6)^2 * (x-2) * (x-5)^2

Then stick a constant k out front to get: k*(x-6)^2 * (x-2) * (x-5)^2

Because f(0) = 4, we know that

f(x) = k*(x-6)^2 * (x-2) * (x-5)^2

f(0) = k*(0-6)^2 * (0-2) * (0-5)^2

4 = k*(0-6)^2 * (0-2) * (0-5)^2

Now solve for k

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!


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 = 4%2F%28-1800%29

                                 k = -1%2F450
Therefore:

           f(x) = k(x+6)²(x-5)²(x-2)

becomes    
           f(x) = -1%2F450(x+6)²(x-5)²(x-2).

Here is a more accurate graph than the one above:



Edwin