ok this is my lasst question of the night i promise!

Find the unique quadratic function that passes through the points (-2, 5), (1,-1), and (6, 29). use matrices to solve

2. Originally Posted by mamajen
ok this is my lasst question of the night i promise!

Find the unique quadratic function that passes through the points (-2, 5), (1,-1), and (6, 29). use matrices to solve
You need to create a sytem of equations given these 3 points.

Use $y =ax^2+bx+c$

Here's the first one $(-2,5) \implies 5 = a(-2)^2+b(-2)+c = 4a-2b+c$

You do the next 2, I will then show you the next step.

3. Originally Posted by pickslides
You need to create a sytem of equations given these 3 points.

Use $y =ax^2+bx+c$

Here's the first one $(-2,5) \implies 5 = a(-2)^2+b(-2)+c = 4a-2b+c$

You do the next 2, I will then show you the next step.
ok here is what i have.....
(1, -1) -1=a+b+c
(6, 29) 29=36a+6b+c

4. Originally Posted by mamajen
ok here is what i have.....
(1, -1) -1=a+b+c
(6, 29) 29=36a+6b+c
so we have,

$-1=a+b+c$
$29=36a+6b+c$
$5 = 4a-2b+c$

Which can be written in matrix form as

$
\left[ \begin{array}{c}
a\\
b\\
c\end{array}\right]=\left[ \begin{array}{ccc}
1 & 1 & 1 \\
36 & 6 & 1 \\
4 & -2 & 1 \end{array} \right]^{-1} \left[ \begin{array}{c}
-1\\
29\\
5\end{array}\right]
$

Now you need to find the inverse of a $3\times 3$ matrix, will be tricky

5. Originally Posted by pickslides
so we have,

$-1=a+b+c$
$29=36a+6b+c$
$5 = 4a-2b+c$

Which can be written in matrix form as

$
\left[ \begin{array}{c}
a\\
b\\
c\end{array}\right]=\left[ \begin{array}{ccc}
1 & 1 & 1 \\
36 & 6 & 1 \\
4 & -2 & 1 \end{array} \right]^{-1} \left[ \begin{array}{c}
-1\\
29\\
5\end{array}\right]
$

Now you need to find the inverse of a $3\times 3$ matrix, will be tricky
1, -1, 1

thanks so much for your help!