Question

A vertical parabola goes through (1, -2), (-3, 10), and (4, 31). Use matrices to solve for D, E, and F, given the general form (since it is a vertical parabola there is not a term)

Answer 1

The general equation of the vertical parabola is

$y= ax^2+bx+c$, where a,b and c are constant.

The parabola passes through the points D(1,-2), E(-3,10), and F(4,31).

For point D:

$-2=a(1)^2+b(1)+c$

$\Rightarrow -2=a+b+c\cdots(i)$

For point E:

$10=a(-3)^2+b(-3)+c$

$\Rightarrow 10=9a-3b+c\cdots(ii)$

For point F:

$31=a(4)^2+b(4)+c$

$\Rightarrow 31=16a+4b+c\cdots(iii)$

All the three equations can be written in the matrix for as:

$\[ \left[ {\begin{array}{cc} -2 \\ 10 \\31 \end{array} } \right]\]= \left[ {\begin{array}{ccc} 1 & 1 & 1 \\ 9 & -3 & 1 \\ 16 & 4 &1\end{array} } \right]\] \left[ {\begin{array}{cc} a \\ b \\c \end{array} } \right]\]$

The value of all the three coefficients can be determined by solving the above equation as

$\left[ {\begin{array}{ccc} a \\ b \\c \end{array} } \right]= \left[ {\begin{array}{ccc} 1 & 1 & 1 \\ 9 & -3 & 1 \\ 16 & 4 &1\end{array}\right]^{-1} \left[ {\begin{array} -2 \\ 10 \\31 \end{array} } \right]$

On solving this, we have

$\Rightarrow \[ \left[ {\begin{array}{ccc} a \\ b \\c \end{array} } \right]\]= \left[ {\begin{array}{ccc} 2 \\ 1 \\-5 \end{array} } \right]$

So, values of the constants,

a=2, b=1, and c=-5

Hence, the required equation of parabola is

$y=2x^2+x-5$