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Finding the equation of a parabola 
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#1
Oct2405, 12:03 AM

P: 150

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)



#2
Oct2405, 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. 


#3
Oct2405, 07:26 AM

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PF Gold
P: 39,363

More simply, since you are given the vertex of the parabola, you can write the equation y= a(xx_{0})^{2}+ y_{0} where x_{0} and y_{0} 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. 


#4
Oct2405, 11:40 PM

P: 37

Finding the equation of a parabola
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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