Question: Find the equation of the parabola whose axis is parallel to the y-axis and which passes through the points (0,4) (1,9) and (-2,6)

Well as the parabola has its axis parallel to the y-axis and it is obviously opening upwards, the general form can be written as: $$(x-h)^2 = 4a(y-k)$$

Then I would have to go through the painstaking task of solving three variables from three equations. Obvious method to do it. I was wondering if there's any easier way of doing it?


The general formula for a parabola is $f(x)=ax^2+bx+c$.

  • Plugging in $(0,4)$, we get: $c=4$.
  • Plugging in $(1,9)$, we get: $a+b+c=9$.
  • Plugging in $(-2,6)$, we get: $4a-2b+c=6$.

The system of equations to be solved becomes:

$\begin{cases}a+b=5 \\ 4a-2b=2 \end{cases}$

This gives $a=2$ and $b=3$.

The formula for your parabola becomes: $f(x)=2x^2+3x+4$.

  • $\begingroup$ I see. But any easier way to do it? $\endgroup$ – Gummy bears Sep 10 '14 at 18:59
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    $\begingroup$ Even easier? No, I don't think so. $\endgroup$ – rae306 Sep 10 '14 at 19:00
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    $\begingroup$ What is easier than posting on Mathematics and having someone else solve it for you? $\endgroup$ – ja72 Sep 10 '14 at 19:02
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    $\begingroup$ @ja72 The point isn't to get the answer. Rather to get a method to use in future problems......... $\endgroup$ – Gummy bears Sep 10 '14 at 19:03
  • $\begingroup$ 48 seconds left lol $\endgroup$ – Gummy bears Sep 10 '14 at 19:05

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