Finding the equation of a parabola
- Thread starter cosmic
- Start date
Oh good. Perhaps, you figured that either of the following forms work with two points.
y = Ax^2 + Bx .. :!: as long as neither of the two known points have x-coordinate zero
y = Ax^2 + C
EGs: given coordinates (1,0) and (-4,-4)
Solving the two systems of equations, the former gives y = -1/5*x^2 + 1/5*x
and the latter gives y = -4/15*x^2 + 4/15
Ciao
D
Deleted member 4993
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If you provide me with the location of the vertex (not on x-axis) and one of the x-intercepts for vertical parabola - I can see graphically "a general parabola" is defined.
Say if the parabola is vertical then the x-intercepts are symmetric about the vertex - and we will have 3 points defined (the property of the vertex hides one condition) . Similar with horizontal parabola. The slanted parabola will need four parameters.
Say if the parabola is vertical then the x-intercepts are symmetric about the vertex - and we will have 3 points defined (the property of the vertex hides one condition) . Similar with horizontal parabola. The slanted parabola will need four parameters.
Oh good. Perhaps, you figured that either of the following forms work with two points.
y = Ax^2 + Bx .. :!: as long as neither of the two known points have x-coordinate zero
y = Ax^2 + C
EGs: given coordinates (1,0) and (-4,-4)
Solving the two systems of equations, the former gives y = -1/5*x^2 + 1/5*x
and the latter gives y = -4/15*x^2 + 4/15
Ciao
Thank you for your reply. Yes I used a similar method.