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A(the?) equation of parabola is $y = ax^2 + bx + c$. That gives the equations below:

\begin{align*} 6 & = a - b + c\\ 4 & = a + b + c\\ 9 & = 4a + 2b + c \end{align*}

Then I simply solve for $(a, b, c)$ and substitute it into $y = ax^2 + bx + c$, correct? I guess it can be considered a quick question :)

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    $\begingroup$ Yes, this is correct. $\endgroup$ – Nikhil Jan 22 '16 at 21:51
  • $\begingroup$ @ Nikhil, It didn't generate a whole buncha discussion, but hopefully it will help someone out in the future. Thank you :) $\endgroup$ – user307277 Jan 22 '16 at 21:53
  • $\begingroup$ Your approach is correct. Are you able to solve these? Adding the first two is a good start. $\endgroup$ – Ross Millikan Jan 22 '16 at 21:58
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    $\begingroup$ Subtract the first two even better ;) to find -2b = 2 instead of 2a+ 2c = 10 $\endgroup$ – Pieter21 Jan 22 '16 at 22:18
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The graphical solution gives (Spoiler):

$(a,b,c) = (2,-1,3)$

graphical solution coefficent system $\quad\quad\quad$ result

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