Find equation of cubic polynomial which passes through...

Hi, I thought I knew how to solve this problem, but I don't really understand the answer I got.

The question is as followed:

Find the equation of a cubic polynomial which passes through (1,1), with slope -5 at this point, and through (-1,7), with slope -5 at this point.

So, consider some things...

p(t) = a0 + a1t + a2t^2 + a3t^3

p`(t) = a1 + a2t + a3t^2

p`(1) = -5, p(1) = 1, p(-1) = 7 and p`(-1) = -5

So, I have four linear equations:

a0 + a1 + a2 + a3 = 1, a0 - a1 + a2 - a3 = 7, a1 + a2 + a3 = -5, a1 -a2 + a3 = -5

So, I decided to plug these equations into my ti-89 and rref it.

1, 1, 1, 1, 1

1, -1, 1, -1, 7

1, 1, 1, 0, -5

1, -1, 1, 0, -5

but I get

1 0 1 0 0

0 1 0 0 0

0 0 0 1 0

0 0 0 0 1

I could use some help,

Thank you

Re: Find equation of cubic polynomial which passes through...

Re: Find equation of cubic polynomial which passes through...

Quote:

Originally Posted by

**zodiacbrave** So, I decided to plug these equations into my ti-89 and rref it.

1, 1, 1, 1, 1

1, -1, 1, -1, 7

1, 1, 1, 0, -5

1, -1, 1, 0, -5

Matrix is incorrect the matrix has to be

$\displaystyle \left( \begin{matrix} 1& 1& 1& 1\\ 1& -1& 1& -1\\ 0& 1& 1& 1\\ 0& 1& -1& 1\end{matrix} \right)$ $\displaystyle \left( \begin{matrix} a_0\\ a_1\\ a_2\\ a_3 \end{matrix} \right)$ = $\displaystyle \left( \begin{matrix} 1\\ 7\\ -5\\ -5 \end{matrix} \right) $