First we try a first degree polynomial with the first two
points, amd see if it works with the other numbers:

Try  

Substitute the first two points (x,y) = (0,1) and (x,y) = (1,2)

     
     

That system is




So B = 1 and substituting in 

 and 

So  becomes

   

     

Now we check by substituting the other three given points:

Substituting (x,y)=(2,4)

   
   
     

Oh, oh.  It doesn't check, so there is no use to 
check the other two.  We must try the next higher 
degree polynomial, so we try

     

Substitute the first three points (x,y) = (0,1), (x,y) = (1,2),
and (2,4)

     
     
     

That system is





Write the system as:





Write as a matrix:



Swap row 1 and row 2 to get a 1 in the upper
left:




Need a 0 where the 4 on the left bottom is.
So multiply row 1 by -4 and add to row 3



Swap rows 2 and 3 to get a 0 where the -2 is



This is in triangular form, 3 0's at the bottom
left.  Convert back to a system of equations:





or


  
    

Substitute  into  

    
  
     
       
           
 
Substitute  and  into 

  


Multiply through by 2 to clear of fractions:






So  becomes:

      

Now we check by substituting the other three given points:

Substituting (x,y)=(2,4)

    
   
     
   
     

That checks.

Substituting (x,y)=(3,7)

    
   
     
   
   
   
   

That also checks.

Finally, substitute (x,y)=(4,11)

    
   
     
   
   

So that also checks, so the answer is

    

Edwin