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I was given the points (2, -1) and (10,-1) and also a max of 4. How would I go about finding the equation of the parabola given this info?

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You should be able to deduce first that the vertex of your parabola is at $(6,4)$ (why?). You now have three points; substitute the appropriate values in $y=ax^2+bx+c$, and you now have three equations in three unknowns. –  J. M. Dec 15 '11 at 5:06

3 Answers 3

Notice that the $y$ values of those points are equal, which implies they are equidistant from the axis of symmetry. Therefore the vertex is at $(6,4)$. Since $x^2$ increases from $0$ to $16$ as $x$ goes from $0$ to $4$, but your parabola decreases by $5$ when you go $4$ units to the right (or left) of the vertex, there must be a scaling factor of $-\frac{5}{16}$. So I would conclude that $y=-\frac{5}{16}(x-6)^2+4$ is the equation.

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The above answers are using the following form for the equation of a parabola:

If the vertex of the parabola is located at $(h,k)$, then the equation of the parabola is $$\tag{1}y=a(x-h)^2+k$$ for some constant $a$.

That this holds can be seen by taking the graph of the parabola $y=ax^2$ and translating it $h$ units to the right and $k$ units up.

Some notes:

1) The line $x=h$ is the line of symmetry of the parabola.

2) $a$ is the "scaling factor". The larger $a$ is in absolute value, the "narrower" the parabola is.

3) If $a>0$, the parabola opens up and $k$ is the minimum value of the $y$ coordinates on the parabola

4) If $a<0$, the parabola opens down and $k$ is the maximum value of the $y$ coordinates on the parabola

An example is shown below.

enter image description here


As made clear by Ross and Jonas, you should find the vertex first, then write the equation in the form (1) above. There will still be the unknown $a$ in the equation at this point. To find it, substitute the information given by one of the given points into the equation and solve for $a$.

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Such a great answer and graph. How did you drawn and took that graph, David? Which programme did you used for example? –  Kerim Atasoy Apr 27 '12 at 22:22
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@KerimAtasoy Thanks. It was done with the Javascript library JSXgraph. –  David Mitra Apr 27 '12 at 22:24

Presumably your max is the maximum $y$ coordinate as $x$ gets larger in absolute value. As the $y$ components are the same in the points you have, the max must come symmetrically between them, at $x=6$. So our parabola is $y=a(x-6)^2+b$. If you put these two points into this equation, you have two simultaneous equations in two unknowns.

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