# Modelling a quadratic - Systems of linear equations

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• May 11th 2010, 07:15 PM
Ari
Modelling a quadratic - Systems of linear equations
Hi, i cant figure out how to set up this question:
ax^2 + bx + c passes through (-1,5) and has a horizontal tangent at (1,7) find the coefficents a b and c

Thanks.
• May 11th 2010, 07:23 PM
dwsmith
Quote:

Originally Posted by Ari
Hi, i cant figure out how to set up this question:
ax^2 + bx + c passes through (-1,5) and has a horizontal tangent at (1,7) find the coefficents a b and c

Thanks.

$a(-1)^2-1b+c=5\rightarrow a-b+c=5$

derivative

$2xa+b=0$ when x=-1
• May 11th 2010, 07:26 PM
Ari
k im confused, how does that help me find the the values for a,b,c?

edit: and can you please explain how you got that? i feel very confused atm lol.
• May 11th 2010, 07:29 PM
dwsmith
Setup the coefficient matrix for the two equations and solve simultaneously.
• May 11th 2010, 07:34 PM
Ari
Quote:

Originally Posted by dwsmith
Setup the coefficient matrix for the two equations and solve simultaneously.

i just started this course... which two equations are you talking about?
• May 11th 2010, 07:35 PM
dwsmith
Quote:

Originally Posted by Ari
i just started this course... which two equations are you talking about?

1: $a-b+c=5$
2: $-2a+b=0$
• May 11th 2010, 07:36 PM
Ari
Quote:

Originally Posted by dwsmith
1: $a-b+c=5$
2: $2a+b=0$

oh okay, how did you get that second equations from the

$a-b+c=5$\

edit: nevermind i see it now. its from the originial one.

now how exactly do i solve them simultaneously? i can get the parametric solutions, but how do i get the number values?
• May 11th 2010, 07:38 PM
dwsmith
Quote:

Originally Posted by dwsmith
$a(-1)^2-1b+c=5\rightarrow a-b+c=5$

derivative

$2xa+b=0\rightarrow -2a+b=0$ when x=-1

The 2nd equation came from the original equation.
• May 11th 2010, 07:41 PM
Ari
is the answer just going to be a,b,c in parametric form or will there be coefficent values, if so how do i get values for them?
• May 11th 2010, 07:43 PM
dwsmith
Quote:

Originally Posted by Ari
is the answer just going to be a,b,c in parametric form or will there be coefficent values, if so how do i get values for them?

Reduced row echelon form of the coefficient matrix.

$\begin{bmatrix}
1 & -1 & 1 & 5\\
-2 & 1 & 0 & 0
\end{bmatrix}$
• May 11th 2010, 07:46 PM
Ari
Quote:

Originally Posted by dwsmith
Reduced row echelon form of the coefficient matrix.

$\begin{bmatrix}
1 & -1 & 1 & 5\\
2 & 1 & 0 & 0
\end{bmatrix}$

$\begin{bmatrix}
1 & -1 & 1 & 5\\
0 & 3 & -2 & -10
\end{bmatrix}$

is this it or am i wayyyy off?
• May 11th 2010, 07:48 PM
dwsmith
$\begin{bmatrix}
1 & -1 & 1 & 5\\
0 & 1 & 2 & 10
\end{bmatrix}\rightarrow \begin{bmatrix}
1 & 0 & 3 & 15\\
0 & 1 & 2 & 10
\end{bmatrix}$
• May 11th 2010, 07:49 PM
Ari
Quote:

Originally Posted by dwsmith
$\begin{bmatrix}
1 & -1 & 1 & 5\\
0 & 3 & -2 & -10
\end{bmatrix}$

Close but off a little.

haha i quickly changed it but i wasnt fast enough. anyway there are 2 equations and 3 unknowns so there are infinietly many solutions?
• May 11th 2010, 07:52 PM
dwsmith
Yes you can right your solution as you would the nullity with a free variable.

Note: Not all under determined linear systems are consistent.
• May 11th 2010, 07:54 PM
Ari
Quote:

Originally Posted by dwsmith
Yes you can right your solution as you would the nullity with a free variable.

Note: Not all under determined linear systems are consistent.

okay so

a = 5 + b - c

b = -10 + 2

c = ?
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