# Math Help - Modelling a quadratic - Systems of linear equations

1. ## 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.

2. 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

3. 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.

4. Setup the coefficient matrix for the two equations and solve simultaneously.

5. 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?

6. Originally Posted by Ari
i just started this course... which two equations are you talking about?
1: $a-b+c=5$
2: $-2a+b=0$

7. 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?

8. 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.

9. 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?

10. 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.

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

11. 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?

12. $\begin{bmatrix}
1 & -1 & 1 & 5\\
0 & 1 & 2 & 10
\end{bmatrix}\rightarrow \begin{bmatrix}
1 & 0 & 3 & 15\\
0 & 1 & 2 & 10
\end{bmatrix}$

13. 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?

14. Yes you can right your solution as you would the nullity with a free variable.

Note: Not all under determined linear systems are consistent.

15. 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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