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Finding the equation of a parabola

by Flatland
Tags: equation, parabola
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Oct24-05, 12:03 AM
P: 153
How do you find the equation of a parabola if you are given it's vertex and 1 point? For example, find the quadratic equation of a parabola that has a vertex of (-2,-2) and goes through the point (-1,0)
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Oct24-05, 01:33 AM
P: 8
General quadratic equation: y=ax^2 + bx + c

Derivative: y' = 2ax + b

At the vertex, the derivate equals to zero. Use this fact and simultaneous equations to arrive at the equation.
Oct24-05, 07:26 AM
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P: 39,568
More simply, since you are given the vertex of the parabola, you can write the equation y= a(x-x0)2+ y0 where x0 and y0 are the coordinates of the vertex. Choose a to force the parabola to go through the second point.

y= a(x-(-2))2- 2= a(x+2)2- 2. Setting x= -1, y= 0,
0= a(-1+2)2-2= a- 2 so a= 2.

By the way, this is assuming the parabola has a vertical line of symmetry. Otherwise there are an infinite number of parabolas satisfying these conditions.

Oct24-05, 11:40 PM
P: 37
Finding the equation of a parabola

Quote Quote by HallsofIvy
... assuming the parabola has a vertical line of symmetry...
Graph first, and you may find a shortcut for a given specific data.
If a point on a parabola is 1 to the right and 2 up from its vertex, it must be parabola
[tex]y = 2x^{2}[/tex]
shifted horizontally and vertically, so its vertex (0,0) moves into (-2,-2), i.e. 2 to the left and 2 down:
[tex]y = 2(x+2)^{2} - 2[/tex]

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