Solve 4th Degree Polynomial w/ Slope Constraints

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Homework Help Overview

The problem involves finding the coefficients of a fourth degree polynomial given specific points and slope constraints. The polynomial is defined as p(x) = ax^4 + bx^3 + cx^2 + dx + e, with conditions based on its values and derivatives at certain points.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss setting up systems of equations based on the polynomial's values at given points and its slopes at specific x-values. There is uncertainty about how to integrate slope information into the existing equations.

Discussion Status

Some participants have made progress in setting up matrices to solve for the coefficients, while others are clarifying the connection between different sets of equations. There is acknowledgment of the need to solve a system of equations, and some guidance has been provided regarding the setup of these equations.

Contextual Notes

Participants note the challenge of having more variables than equations in their initial attempts, leading to confusion about how to proceed with the problem. The discussion reflects a learning process as they navigate the requirements of the problem.

crombiecutie105
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Homework Statement



the question states: Find the coefficients of the fourth degree polynomial:
p(x) = ax^4 + bx^3 + cx^2 + dx + e whose graph goes through the points (0,0), (1,1), (-1,3) and whose slope at x=-1 is 20 and x=1 is 9.

Homework Equations





The Attempt at a Solution



i started by putting it into an augmented matrix, and solving...
my matrix was:

0 0 0 0 1 | 0
1 1 1 1 1 | 1
1 -1 1 -1 1 | 3

when i solved for the coefficients, i got a = 0, b = -1, c = 2, d = 0, and e = 0.
i'm not sure how to incorporate the information given by the slopes.

i know that you can take the derivative to find slope.. and i took the derivative of the polynomial and got:

4ax^3 + 3bx^2 + 2cx + d

but I'm not really sure where to go from there. do i plug in -1 for x and set it equal to 20 and 1 for x and set it equal to 9? if i do that i get the equations:

-4a + 3b - 2c + d = 20 and
4a + 3b + 2c + d = 9

by now i think I'm starting to make things up though! :wink:
can anyone offer any hints? thank you!
 
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crombiecutie105 said:
but I'm not really sure where to go from there. do i plug in -1 for x and set it equal to 20 and 1 for x and set it equal to 9? if i do that i get the equations:

Yes, you do. I didn't check your calculations, but the next step is obviously to solve a system of equations.
 
i set up a system of equations for those two equations and with only 2 equations and 4 variables i was unsure of how to proceed. i ended up with the matrix:

-4 3 -2 1 | 20
0 6 0 2 | 29

i'm just confused as to how the two steps are connected... like the first time i solved for the coefficients and this time. can i combine my results somehow?
 
I don't follow. Did you set up a system of 5 equations with 5 unknowns? Did you try to solve it?
 
ok, i think i got it now. i didnt think i could set up all 5 equations before because the first set of equations had 5 variables and the second set of equations had 4. i put a zero as a place holder though, and solved this matrix:

0 0 0 0 1 | 0
1 1 1 1 1 | 1
1 -1 1 -1 1 | 3
-4 3 -2 1 0 |20
4 3 2 1 0 |9

i think I'm good from here. thanks for your help!
 
The matrix looks okay.
 

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