How to find the equation of a Parabola

**22upon7** · June 1st 2009, 07:38 PM

Hi,

Could someone explain step-by-step, how to work through this type of sums:

"Find the equation of the quadratic which passes through the points with coordinates:

(-2,-1), (1,2), (3,-16)"

I substituted the values in to the equation $ax^2+bx+c$

$-1=4a-2b+c$
$2 =a+b+c$
$-16=9a+3b+c$

I'm not sure how you work out the simultaneous equations though, Any help would be much appreciated,

Thanks,

Dru

**Soroban** · June 1st 2009, 08:05 PM

Hello, 22upon7!

You did the hard part . . . and you forgot your algrbra?

Find the equation of the parabola which passes through the points
. . with coordinates: .(-2,-1), (1,2), (3,-16)

I substituted the values in to the equation $y \:=\:ax^2+bx+c$

$-1=4a-2b+c$
$2 =a+b+c$
$-16=9a+3b+c$

We have: . $\begin{array}{cccc}a + b + c &=& 2 & {\color{blue}[1]} \\ 4a - 2b + c &=& \text{-}1 & {\color{blue}[2]}\\ 9a + 3b + c &=& \text{-}16 & {\color{blue}[3]} \end{array}$

Subtract ${\color{blue}[2] - [1]}\!:\;\;3a - 3b \:=\:\text{-}3 \;\;\quad\Rightarrow\quad a - b \:=\:\text{-}1\;\;{\color{blue}[4]}$

Subtract ${\color{blue}[3] - [2]}\!:\;\;5a + 5b \:=\:\text{-}15 \quad\Rightarrow\quad a + b \:=\:\text{-}3\;\;{\color{blue}[5]}$

Add ${\color{blue}[4] + [5]}\!:\;\;2a \:=\:-4 \quad\Rightarrow\quad\boxed{ a \:=\:-2}$

Substitute into ${\color{blue}[5]}\!:\;\;-2 + b \:=\:-3 \quad\Rightarrow\quad \boxed{b\:=\:-1}$

Substitute into ${\color{blue}[1]}\!:\;\;-2 -1 + c \:=\:2 \quad\Rightarrow\quad\boxed{ c \:=\:5}$

Therefore: . $y \;=\;-2x^2 - x + 5$

**22upon7** · June 1st 2009, 08:27 PM

Originally Posted by Soroban

Hello, 22upon7!

You did the hard part . . . and you forgot your algrbra?

We have: . $\begin{array}{cccc}a + b + c &=& 2 & {\color{blue}[1]} \\ 4a - 2b + c &=& \text{-}1 & {\color{blue}[2]}\\ 9a + 3b + c &=& \text{-}16 & {\color{blue}[3]} \end{array}$

Subtract ${\color{blue}[2] - [1]}\!:\;\;3a - 3b \:=\:\text{-}3 \;\;\quad\Rightarrow\quad a - b \:=\:\text{-}1\;\;{\color{blue}[4]}$

Subtract ${\color{blue}[3] - [2]}\!:\;\;5a + 5b \:=\:\text{-}15 \quad\Rightarrow\quad a + b \:=\:\text{-}3\;\;{\color{blue}[5]}$

Add ${\color{blue}[4] + [5]}\!:\;\;2a \:=\:-4 \quad\Rightarrow\quad\boxed{ a \:=\:-2}$

Substitute into ${\color{blue}[5]}\!:\;\;-2 + b \:=\:-3 \quad\Rightarrow\quad \boxed{b\:=\:-1}$

Substitute into ${\color{blue}[1]}\!:\;\;-2 -1 + c \:=\:2 \quad\Rightarrow\quad\boxed{ c \:=\:5}$

Therefore: . $y \;=\;-2x^2 - x + 5$

Thankyou!

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