## 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 (1|-4) and a maximum turning point at (-1|3). You are looking for a function with: quadratic function Maximum turning point at (-1|3) Minimum turning point at (1|-4) Mathepower found the following function: This is the graph of your function.
- Roots at -0.386; 3.886
- y-axis intercept at (0|-1.5)
- Maximum and minimum turning points at (1.75|-4.563)
- Inflection points
This is how Mathepower calculated: The point at (-1|3) gives the equation :simplified: : 1a-1b+1c=3 The point at (1|-4) gives the equation :simplified: : 1a+1b+1c=-4 So, we got the following system of equations: :
This is how to solve this system of equations:
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 (-1|3), (0|2), (1|1) und (2|4): You are looking for a function with: Function of degree 3 Point at (-1|3) Point at (0|2) Point at (1|1) Point at (2|4) Mathepower found the following function: This is the graph of your function.
- Roots at -2
- y-axis intercept at (0|2)
- Maximum and minimum turning points at (-0.913|3.014); (0.913|0.986)
- Inflection points at (0|2)
This is how Mathepower calculated: The point at (-1|3) gives the equation :simplified: : -1a+1b-1c+1d=3 The point at (0|2) gives the equation :simplified: : 0a+0b+0c+1d=2 The point at (1|1) gives the equation :simplified: : 1a+1b+1c+1d=1 The point at (2|4) gives the equation :simplified: : 8a+4b+2c+1d=4 So, we got the following system of equations: :
This is how to solve this system of equations:
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 y-value. 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 (1|3): You are looking for a function with: Function of degree 3 root at 2 root at 4 Inflection point at (1|3) Mathepower found the following function: This is the graph of your function.
- Roots
- y-axis intercept at (0|0)
- Maximum and minimum turning points
- Inflection points
This is how Mathepower calculated: The point at (1|3) 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:
Inserting shows that the function equals to ist. |