1. ## Algebra

fit a quadratic polynomial to the points (1,4),(3,8),(-2,8) by writing the general equation, setting up and solving of 3 linear equations.

how to solve using row reduce method?

2. Hello, trythis!

Fit a quadratic polynomial to the points (1,4), (3,8), (-2,8)
by writing the general equation, setting up and solving of 3 linear equations.

how to solve using row-reduce method?

The general quadatic is: .$\displaystyle f(x) \:=\:ax^2+bx+c$

Those three points must satisfy the equation, so we have:

. . $\displaystyle \begin{array}{ccccc}f(1) = 4 & & a + b + c &=& 4 \\ f(3)=8 & & 9a + 3b + c &=& 8 \\ f(\text{-}2) = 8 & & 4a - 2b + c &=&8 \end{array}$

Then we have the augmented matrix: .$\displaystyle \left[\begin{array}{ccc|c}1 & 1 & 1 & 4 \\ 9 & 3 & 1 & 8 \\ 4 & \text{-}2 & 1 & 8 \end{array}\right]$

. . and I assume you know how to row-reduce it . . .

3. ## how u got this equation,plse explain?

4. Originally Posted by mathseek
This is how he did it

For the point (1,4). if x=1 and y=4 then $\displaystyle f(x)=ax^{2}+bx+c \Rightarrow f(1)=a(1)^{2}+b(1)+c \Rightarrow a+b+c=4$

For (3,8) $\displaystyle f(x)=ax^{2}+bx+c \Rightarrow f(3)=a(3)^{2}+b(3)+c \Rightarrow 9a+3b+c=8$

For (-2,8) $\displaystyle f(x)=ax^{2}+bx+c \Rightarrow f(-2)=a(-2)^{2}+b(-2)+c \Rightarrow 4a-2b+c=8$

Just plug in the values that you were given for each quadratic equation.