Equation of a Parabola

**bearej50** · May 10th 2009, 11:33 AM

Find the equation of the parabola y = ax² + bx + c that contains the points (1, -2), (2, 0), and (4, 22).

**Mush** · May 10th 2009, 11:36 AM

Originally Posted by bearej50

Find the equation of the parabola y = ax² + bx + c that contains the points (1, -2), (2, 0), and (4, 22).

Plug in the points and you'll get three equations.

$a+b+c = -2$

$4a+2b+c = 0$

$16a+4b + c = 22$

You can solve this system of equations in a number of ways. I recommend setting up a matrix.

$\bigg(\begin{array}{ccc} 1 & 1 & 1 \\ 4 & 2 & 1 \\ 16 & 4 & 1 \end{array}\bigg) \bigg( \begin{array}{ccc} a \\ b \\ c \end{array}\bigg) = \bigg(\begin{array}{ccc} -2 \\ 0 \\ 22 \end{array} \bigg)$

So let's write that as $AX = B$ . We want to find x, hence: $X = BA^{-1}$

And you know how to calculate the inverse of a 3x3 matrix hopefully! $A^{-1} = \frac{adj(A)}{det(A)}$

**bearej50** · May 10th 2009, 12:18 PM

I was taught a different way. Subtract the equations from one another.
I got a = 13/5, b = -23/5, and c = 0. Is this right?

**Mush** · May 10th 2009, 12:24 PM

Originally Posted by bearej50

I was taught a different way. Subtract the equations from one another.
I got a = 13/5, b = -23/5, and c = 0. Is this right?

Check yourself if it is right. If it satisfies the original equations it's right.

Unfortunately yours are not. I get a = 3 ,b = -7, c = 2, and that is correct.

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