# Thread: find equation of THIS PARABOLA

1. ## find equation of THIS PARABOLA

Axis of symmetry parallel to x-axis, contains points (0,1) (3,2) and (1,3)

2. ## Re: find equation of THIS PARABOLA

Hello, sluggerbroth!

Axis of symmetry parallel to x-axis, contains points (0,1) (3,2) and (1,3)
Find the equation of the parabola.

This is a "horizontal" parabola: .$\displaystyle f(y) \:=\:ay^2 + by + c$

Use the three points to set up a system of equations.

. . $\displaystyle \begin{array}{cccccccccc}f(1) = 0: & a + b + c &=& 0 & [1] \\ f(2) = 3: & 4a + 2b + c &=& 3 & [2]\\ f(3) = 1: & 9a + 3b + c &=& 1 & [3] \end{array}$

$\displaystyle \begin{array}{ccccccc}\text{Subtract [2] - [1]:} & 3a + b &=& 3 & [4] \\ \text{Subtract [3] - [2]:} & 5a + b &=& \text{-}2 & [5] \end{array}$

$\displaystyle \text{Subtract [5] - [4]: }\;2a \:=\:\text{-}5 \quad\Rightarrow\quad a \:=\:\text{-}\tfrac{5}{2}$

$\displaystyle \text{Substitute into [4]: }\;3(\text{-}\tfrac{5}{2}) + b \:=\:3 \quad\Rightarrow\quad b \:=\:\tfrac{21}{2}$

$\displaystyle \text{Substitute int [1]: }\;\text{-}\tfrac{5}{2} + \tfrac{21}{2} + c \:=\:0 \quad\Rightarrow\quad c \:=\:\text{-}8$

$\displaystyle \text{Therefore: }\:x \;=\;\text{-}\tfrac{5}{2}y^2 + \tfrac{21}{2}y - 8$