# Finding functions

Find a function

Degree of the function:
 1 2 3 4 5

( The degree is the highest power of an x. )

Symmetries:
axis symmetric to the y-axis
point symmetric to the origin

y-axis intercept

Roots / Maxima / Minima /Inflection points:
at x=
at x=
at x=
at x=
at x=

Characteristic points:
at |)
at |)
at |)
at (|)
at (|)

Slope at given x-coordinates:
Slope at x=
Slope at x=
Slope at

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:
Maximum turning point at (-1|3)
Minimum turning point at (1|-4)

Mathepower found the following function:

This is the graph of your function.
 Dein Browser unterstützt den HTML-Canvas-Tag nicht. Hol dir einen neuen. :P
• 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: :
 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
 a -1b +c = 3 2b = -7
( -1 times line 1 was added to line 2 )
 a -1b +c = 3 b = -3,5
( 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:
 a -1b +c = 3
 a -1⋅(0c -3,5) +⋅1c = 3
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 (-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.
 Dein Browser unterstützt den HTML-Canvas-Tag nicht. Hol dir einen neuen. :P
• 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: :
 -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 )

4 line:
 d = 2

3 line:
 c +0,5d = -0,667
 c +0,5⋅2 = -0,667
Solve for c : c = -1,667

2 line:
 b +d = 2
 b +⋅2 = 2
Solve for b : b = 0

1 line:
 a -1b +c -1d = -3
 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 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.
 Dein Browser unterstützt den HTML-Canvas-Tag nicht. Hol dir einen neuen. :P
• 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: :
 a +b +c +d = 3

This is how to solve this system of equations:
 a +b +c +d = 3

 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 step-by-step calculation. Or just make up any interesting exercise and check out what Mathepower does.