Question

Suppose a parabola passes through the points (1,4), (2, 8), and (4, 34). Determine theequation for the parabola in the form y = ax? + bx + c. I suggest you plug the threepoints into the parabola equation to get three equations with a, b, and c as unknowns.Then use matrices to solve with your calculator.

Answer 1

The given parabola passes through the points (1,4), (2, 8), and (4, 34)

The equation of the parabola is

`y=ax^2+bx+c`

Substitute x=1 and y=4, we get

`4=a(1)^2+b(1)+c`
`4=a+b+c`

`a+b+c=4\text{ take it as equation (1).}`

Substitute x=2 and y=8 in the parabola equation, we get

`8=a(2)^2+b(2)+c`

`8=4a+2b+c`

`4a+2b+c=8\text{ take it as equation (2).}`

Substitute x=4 and y=34 in the parabola equation, we get

`34=a(4)^2+b(4)+c`

`34=16a+4b+c`

`16a+4b+c=34\text{ take it as equation (3)}`

Hence we get the system of equations

`a+b+c=4\text{ }`

`4a+2b+c=8\text{ }`

`16a+4b+c=34\text{ }`

This can be written in the form AX=B, where

`A=\begin{bmatrix}{1} & {1} & {1} \\ {4} & {2} & {1} \\ {16} & {4} & {1}\end{bmatrix}\text{ , X=}\begin{bmatrix}{} & {a} & {} \\ {} & {b} & {} \\ {} & {c} & {}\end{bmatrix}\text{ and }B=\begin{bmatrix}{} & {4} & {} \\ {} & {8} & {} \\ {} & {34} & {}\end{bmatrix}`

The solution is

`X=BA^(-1)`

We need to find the inverse of the matrix A by using a calculator.

The inverse of the matrix A is

`A^(-1)=\begin{bmatrix}{(1)/(3)} & {-(1)/(2)} & {(1)/(6)} \\ {-2} & {(5)/(2)} & {-(1)/(2)} \\ {(8)/(3)} & {-2} & {(1)/(3)}\end{bmatrix}`

Multiplying inverse of A and B to find X.

`\begin{bmatrix}{} & {a} & {} \\ {} & {b} & {} \\ {} & {c} & {}\end{bmatrix}=\begin{bmatrix}{} & {4} & {} \\ {} & {8} & {} \\ {} & {34} & {}\end{bmatrix}\begin{bmatrix}{(1)/(3)} & {-(1)/(2)} & {(1)/(6)} \\ {-2} & {(5)/(2)} & {-(1)/(2)} \\ {(8)/(3)} & {-2} & {(1)/(3)}\end{bmatrix}`

`=\begin{bmatrix}{} & {4((1)/(3))+8(-2)+34((8)/(3))} & {} \\ {} & {4(-(1)/(2))+8((5)/(2))+34(-2)} & {} \\ {} & {4((1)/(6))+8(-(1)/(2))+34((1)/(3))} & {}\end{bmatrix}`

`=\begin{bmatrix}{} & {(4)/(3)-16+(272)/(3)} & {} \\ {} & {-2+20-68} & {} \\ {} & {(2)/(3)-4+(34)/(3)} & {}\end{bmatrix}`

`=\begin{bmatrix}{} & {(4-48+272)/(3)} & {} \\ {} & {-50} & {} \\ {} & {(2-12+34)/(3)} & {}\end{bmatrix}`

`=\begin{bmatrix}{} & {76} & {} \\ {} & {-50} & {} \\ {} & {8} & {}\end{bmatrix}`

Hence the solution is

`a=76,\text{ b=-50 and c=8}`

Substitute a=76, b=-46 and c=8 in the parabola equation, we get

`y=76x^2-50x+8`

Hence the required parabola equation is

`y=76x^2-50x+8`