What's this about?
This is, in some way, the opposite of curve sketching. Curve sketching means you got a function and are looking for roots, turning and inflection points. What we do here is the opposite: Your got some roots, inflection points, turning points etc. and are looking for a function having those.
How to reconstruct a function?
Primarily, you have to find equations and solve them. This gives you the coefficients of your function. Here is an example: Let's assume we are looking for a function of degree
having a minimum turning point at (14) and a maximum turning point at (13).
You are looking for a function with: quadratic function Maximum turning point at (13) Minimum turning point at (14)
Mathepower found the following function:
This is the graph of your function.

 Roots at 0.386; 3.886
 yaxis intercept at (01.5)
 Maximum and minimum turning points at (1.754.563)
 Inflection points
This is how Mathepower calculated:
The point at (13) gives the equation :
simplified: : 1a1b+1c=3
The point at (14) gives the equation :
simplified: : 1a+1b+1c=4
So, we got the following system of equations: :
a  1b  +c  =  3  a  +b  +c  =  4 
This is how to solve this system of equations:
a  1b  +c  =  3  a  +b  +c  =  4 
   ( 1 times line 1 was added to line 2 )
  ( The 2 line was divided by 2 ) 
2 line:  b+0c = 3,5  c can be chosen freely  Solve for b : :  b = 0c 3,5 
1 line:   Substitute variables already known:   Solve for a :  a = 1c 0,5  Set a equal to This means that c is equal to 1,5
Inserting shows that the function equals to ist.

How to find a function through given points?
The general rule is that for any n given points there is a function of degree
whose graph goes through them. So e.g. you find by solving equations a function of degree
through the four points (13), (02), (11) und (24):
You are looking for a function with: Function of degree 3 Point at (13) Point at (02) Point at (11) Point at (24)
Mathepower found the following function:
This is the graph of your function.

 Roots at 2
 yaxis intercept at (02)
 Maximum and minimum turning points at (0.9133.014); (0.9130.986)
 Inflection points at (02)
This is how Mathepower calculated:
The point at (13) gives the equation :
simplified: : 1a+1b1c+1d=3
The point at (02) gives the equation :
simplified: : 0a+0b+0c+1d=2
The point at (11) gives the equation :
simplified: : 1a+1b+1c+1d=1
The point at (24) gives the equation :
simplified: : 8a+4b+2c+1d=4
So, we got the following system of equations: :
1a  +b  1c  +d  =  3     d  =  2  a  +b  +c  +d  =  1  8a  +4b  +2c  +d  =  4 
This is how to solve this system of equations:
1a  +b  1c  +d  =  3     d  =  2  a  +b  +c  +d  =  1  8a  +4b  +2c  +d  =  4 
  1a  +b  1c  +d  =  3     d  =  2  a  +b  +c  +d  =  1   4b  6c  7d  =  4 
 ( 8 times line 3 was added to line 4 )
 1a  +b  1c  +d  =  3     d  =  2   2b   +2d  =  4   4b  6c  7d  =  4 
 ( 1 times line 1 was added to line 3 )
 a  1b  +c  1d  =  3     d  =  2   2b   +2d  =  4   4b  6c  7d  =  4 
 ( The 1 line was divided by 1 )  a  1b  +c  1d  =  3     d  =  2   2b   +2d  =  4    6c  3d  =  4 
 ( 2 times line 3 was added to line 4 )
 a  1b  +c  1d  =  3   2b   +2d  =  4     d  =  2    6c  3d  =  4 
 ( the 3 line was interchanged with the 2 line )
 a  1b  +c  1d  =  3   b   +d  =  2     d  =  2    6c  3d  =  4 
 ( The 2 line was divided by 2 )  a  1b  +c  1d  =  3   b   +d  =  2    6c  3d  =  4     d  =  2 
 ( the 4 line was interchanged with the 3 line )
 a  1b  +c  1d  =  3   b   +d  =  2    c  +0,5d  =  0,667     d  =  2 
 ( The 3 line was divided by 6 ) 
3 line:   Substitute variables already known:   Solve for c :  c = 1,667 
2 line:   Substitute variables already known:   Solve for b :  b = 0 
1 line:   Substitute variables already known:  a  1⋅0  +⋅(1,667)  1⋅2  =  3 
 Solve for a :  a = 0,667 
Inserting shows that the function equals to ist.

How to find a function with a given inflection point?
An inflection point gives multiple equations: On the one hand, you got the yvalue. On the other hand, you know that the second derivative is
at an inflection point. Let's take a look at an example for a function of degree
having an inflection point at (13):
You are looking for a function with: Function of degree 3 root at 2 root at 4 Inflection point at (13)
Mathepower found the following function:
This is the graph of your function.

 Roots
 yaxis intercept at (00)
 Maximum and minimum turning points
 Inflection points
This is how Mathepower calculated:
The point at (13) gives the equation :
simplified: : 1a+1b+1c+1d=3
So, we got the following system of equations: :
This is how to solve this system of equations:
1 line:  c+1d = 3  d can be chosen freely  Solve for c : :  c = 1d +3 
Inserting shows that the function equals to ist.

And how to use that in my example?
Just enter your exercise above. Mathepower shows how it works by doing a free stepbystep calculation. Or just make up any interesting exercise and check out what Mathepower does.