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Given $y = a + bx + cx^2$ fits three given points, find and solve the matrix equation for the unknowns $a$, $b$, and $c$.

the equation fits the points $(1,0), (-1, -4),$ and $(2, 11)$

I really don't know what I'm supposed to do here. I tried setting it up like a normal matrix like

$$\begin{pmatrix} 1&-1&2\\ 0&4&11 \end{pmatrix} (x\quad y)=??$$

but this doesn't make any sense because the original equation has multiple x's in it and I feel like im not taking care of that.

Can you help me get started with this? I just don't get it and I'm so frustrated.

Thanks!

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The point $(x,y)$ is on the graph of a function $f$ if and only if $y = f(x)$. In other words, plug the three $x$-values into the function, and you get three equations with three unknowns. –  JavaMan Oct 7 '11 at 18:10

2 Answers 2

up vote 6 down vote accepted

First note that $y = a + bx + cx^2$ can also be written as

$$(1 \quad x \quad x^2) \left(\begin{array}{c} a \\ b \\ c \end{array}\right) = y$$

You are given three points, $(x_1,y_1) = (1,0)$, $(x_2,y_2) = (-1,-4)$ and $(x_3,y_3) = (2,11)$. So you can set up three of these equations:

$$\begin{align} (1 \quad x_1 \quad x_1^2) \left(\begin{array}{c} a \\ b \\ c \end{array}\right) &= y_1 \\ (1 \quad x_2 \quad x_2^2) \left(\begin{array}{c} a \\ b \\ c \end{array}\right) &= y_2 \\ (1 \quad x_3 \quad x_3^2) \left(\begin{array}{c} a \\ b \\ c \end{array}\right) &= y_3 \end{align}$$

Or, in one matrix form:

$$\left(\begin{array}{ccc} 1 & x_1 & x_1^2 \\ 1 & x_2 & x_2^2 \\ 1 & x_3 & x_3^2 \end{array}\right) \left(\begin{array}{c} a \\ b \\ c \end{array}\right) = \left(\begin{array}{c} y_1 \\ y_2 \\ y_3 \end{array}\right)$$

You know the values of $x_1, x_2, x_3, y_1, y_2, y_3$, so filling them in you will be left with an equation of the form $Ax = b$, with $A$ and $b$ known and $x$ unknown.

Edit: For completeness, filling in those values we get

$$\left(\begin{array}{ccc} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 2 & 4 \end{array}\right) \left(\begin{array}{c} a \\ b \\ c \end{array}\right) = \left(\begin{array}{c} 0 \\ -4 \\ 11 \end{array}\right)$$

You could do a Gaussian elimination, but this example has a simple solution which we can just calculate by hand. Subtracting the second row from the first we get

$$(0 \quad 2 \quad 0) \left(\begin{array}{c} a \\ b \\ c \end{array}\right) = 4 \quad \Longrightarrow \quad \fbox{b = 2}$$

Subtracting the first from the last we get

$$(0 \quad 1 \quad 3) \left(\begin{array}{c} a \\ b \\ c \end{array}\right) = 11 \quad \Longrightarrow \quad b + 3c = 11 \quad \Longrightarrow \quad \fbox{c = 3}$$

Then finally you could use the first row to get

$$(1 \quad 1 \quad 1) \left(\begin{array}{c} a \\ b \\ c \end{array}\right) = 0 \quad \Longrightarrow \quad a + b + c = 0 \quad \Longrightarrow \quad \fbox{a = -5}$$

(For the enthusiast, note that the matrix $A$ is a Vandermonde matrix.)

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If the equation fits three points, then if we assign the x-axis value and $y$-axis value to the equation, we can have a row in the matrix. Example, assign $(1,0)$ to the equation, we have $a + b+ c = 0$. The same for the other two points. Then you have a matrix that can be solved using a calculator or by Gaussian calculation.

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