Solve 4th Degree Polynomial w/ Slope Constraints

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SUMMARY

The discussion focuses on solving a fourth-degree polynomial, specifically of the form p(x) = ax^4 + bx^3 + cx^2 + dx + e, with given points and slope constraints. The points (0,0), (1,1), and (-1,3) are used alongside slope conditions at x=-1 (20) and x=1 (9) to establish a system of equations. The user initially derived coefficients a=0, b=-1, c=2, d=0, and e=0 but struggled to incorporate slope information. Ultimately, they successfully set up a system of five equations to solve for the coefficients.

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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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