How do you find the equation of a Cubic function given 5 points? (no zeros)?

In summary, to find the equation of a polynomial with given points, one way is to create 5 equations and solve for ax^3+bx^2+cx+d using the elimination/substitution method. Another method is the finite difference method, which involves creating a table of differences and filling in the values to find the coefficients of the polynomial. This method can be faster and more efficient than the elimination/substitution method. The similarity of the values in the Δ3 column of the table for f(x) = ax^3+bx^2+cx+d shows that the x3 coefficient is 6, while the table for g(x) = x3 has all 6's in the Δ3 column, indicating that the x3
  • #1
srizen
14
0

Homework Statement



what the questions asks is that i need to find the equation of a polynomial with these given points:
1,1
2,-3
3,5
4,37
5,105i know that one way to solve is by creating 5 equations then solve for ax^3+bx^2+cx+d using the elimination/substitution method.
however is there another, much easier way of doing this question?

Homework Equations



ax^3+bx^2+cx+d

The Attempt at a Solution



1= a+b+c+d
-3=8a+4b+2c+d
5= 27a+9b+3c+d
37=64a+16b+4c+d
108=125a+25b+5c+d

fixed, yes, my mistake
 
Last edited:
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  • #2
srizen said:
1= a+b+c+d
-3=8a+4b+3c+d
5= 64a+16b+5c+d
37=64a+16b+5c+d
108=125a+25b+5c+d
Three of the equations are wrong. They should be
1= a+b+c+d
-3=8a+4b+2c+d
5= 27a+9b+3c+d
37=64a+16b+4c+d
108=125a+25b+5c+d

Also, you don't need the last equation, because there are 4 unknowns.

As for other methods, there's the finite difference method, but I don't think it will help for this particular problem (because you already told us that this is a cubic).
 
Last edited:
  • #3
eumyang said:
Three of the equations are wrong. They should be
1= a+b+c+d
-3=8a+4b+2c+d
5= 27a+9b+3c+d
37=64a+16b+4c+d
108=125a+25b+5c+d

fixed, i typed the equations too fast
 
  • #4
eumyang said:
...

As for other methods, there's the finite difference method, but I don't think it will help for this particular problem (because you already told us that this is a cubic).
True, but I got the coefficients fairly quickly using a difference method.

Actually, after playing around a bit with this, I got the result with two different difference methods.
 
  • #5
SammyS said:
True, but I got the coefficients fairly quickly using a difference method.

Actually, after playing around a bit with this, I got the result with two different difference methods.

what exactly is the difference method?
i ask because I've tried this question, and i kept getting it wrong.
 
  • #7
srizen said:
what exactly is the difference method?
i ask because I've tried this question, and i kept getting it wrong.

Make a table of differences

[tex]\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline
\quad i \quad & \quad x_i \quad & \quad f(x_i) \quad &\quad (\Delta^1)_i \quad &\quad (\Delta^2)_i \quad & \quad (\Delta^3)_i \quad & \quad (\Delta^4)_i \quad \\
\hline
& & & & & & \\
1 & 1 & 1 & -4 & 12 & 12 & 0 \\
& & & & & & \\
2 & 2 & -3 & 8 & 24 & & -- \\
& & & & & & \\
3 & 3 & 5 & & & -- & -- \\
& & & & & & \\
4 & 4 & 37 & & -- & -- & -- \\
& & & & & & \\
5 &5 & 105 & -- & -- & -- & -- \\
& & & & & & \\
\hline \end{array}[/tex]

Where: [itex](\Delta^1)_i=f(x_{i+1})-f(x_{i})\,,[/itex]

[itex](\Delta^2)_i=(\Delta^1)_{i+1}-(\Delta^1)_i[/itex]

etc.

See if you can fill in the rest.

If f(x) is truly a cubic function then the Δ3 column will all be the same.

Fill out a similar Table for g(x) = x3 . The Δ3 column will all be 6's.

What do you suppose that means about the x3 coefficient of f(x) ?
 
  • #8
SammyS said:
Make a table of differences

[tex]\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline
\quad i \quad & \quad x_i \quad & \quad f(x_i) \quad &\quad (\Delta^1)_i \quad &\quad (\Delta^2)_i \quad & \quad (\Delta^3)_i \quad & \quad (\Delta^4)_i \quad \\
\hline
& & & & & & \\
1 & 1 & 1 & -4 & 12 & 12 & 0 \\
& & & & & & \\
2 & 2 & -3 & 8 & 24 & & -- \\
& & & & & & \\
3 & 3 & 5 & & & -- & -- \\
& & & & & & \\
4 & 4 & 37 & & -- & -- & -- \\
& & & & & & \\
5 &5 & 105 & -- & -- & -- & -- \\
& & & & & & \\
\hline \end{array}[/tex]

Where: [itex](\Delta^1)_i=f(x_{i+1})-f(x_{i})\,,[/itex]

[itex](\Delta^2)_i=(\Delta^1)_{i+1}-(\Delta^1)_i[/itex]

etc.

See if you can fill in the rest.

If f(x) is truly a cubic function then the Δ3 column will all be the same.

Fill out a similar Table for g(x) = x3 . The Δ3 column will all be 6's.

What do you suppose that means about the x3 coefficient of f(x) ?

OMG! i love you forever, i had no idea this method existed, i already solved through almost an hour of writing matrices, with this i solved it in 3 minutes. thank you!
 

1. What is a cubic function?

A cubic function is a type of polynomial function with the highest degree of three. It can be written in the form f(x) = ax^3 + bx^2 + cx + d, where a, b, c, and d are coefficients.

2. Why is it important to find the equation of a cubic function given 5 points?

Finding the equation of a cubic function allows us to understand and analyze the relationship between the independent variable (x) and the dependent variable (y). This can be useful in various fields such as physics, engineering, and economics.

3. What are the steps to find the equation of a cubic function given 5 points?

The steps include:
1. Write the general cubic function equation: f(x) = ax^3 + bx^2 + cx + d
2. Plug in the given points to create a system of equations
3. Solve the system of equations using elimination or substitution
4. Once you have the values for a, b, c, and d, write the equation in standard form
5. Simplify the equation if needed and check if it satisfies all 5 points.

4. Can a cubic function have no zeros?

Yes, a cubic function can have no zeros. This means that the graph of the function will not intersect the x-axis at any point. This can occur when the function has a positive or negative leading coefficient and all the roots are complex numbers.

5. Are there any shortcuts or formulas to find the equation of a cubic function given 5 points?

There is no shortcut or formula to find the equation of a cubic function given 5 points. The process involves solving a system of equations, which can be done using various methods such as elimination or substitution. However, there are online calculators or software that can help with the calculations.

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