Re: Conics - Parabola
- From: lrudolph@xxxxxxxxx (Lee Rudolph)
- Date: 8 Jun 2006 20:11:59 -0400
Shaynelle@xxxxxxxxx writes:
I have these three points of a parabola: (-1,5) (0,-1) (2,-1)
or, more precisely and correctly, three points of a parabola
that is the graph y = p(x) of a quadratic polynomial p(x) =
ax^2 + bx + c, or alternatively p(x) = a(x-d)^2 + e
and I am
trying to figure out where the origin is
that is, the value of d if p is written in the second form
(or of -b/2a if p is written in the first form).
. I know it must open up
that is, the "leading coefficient" a must be postive
as
the
y-coordinates of the
second and third point are both at -1,
from which you can conclude without further ado that d is
the average of the x-coordinates of those two points (why?
think about it), that is, (0+2)/2 = 1, so that p(x) = a(x-1)^2 + e
for some a and e...but now, knowing that p(-1) = 5 and p(0) = -1,
we deduce that a(-1-1)^2 + e = 5 and a(0-1)^2 + e = -1,
i.e., 4a + e = 5 and a + e = -1, i.e. (why?) a = 2 and e = -3,
i.e., p(x) = 2(x-1)^2 - 3 = 2x^2 - 4x -1 (check that this does
fit the three given points!)
but I can't for the life of
me figure out how it would open up and only one over (to the
left/negative) on the x axis for the third point. I'm assuming there
must be a compression but I can't easily tell what it is.
Does anyone have any suggestions for me? Ultimately the question is
asking me for the equation written in standard form for this parabola,
but I would appreciate only a hint to get me in the right direction if
anyone is kind enough to please help me out.
Lee Rudolph
.
- Follow-Ups:
- Re: Conics - Parabola
- From: Shaynelle
- Re: Conics - Parabola
- References:
- Conics - Parabola
- From: Shaynelle
- Conics - Parabola
- Prev by Date: Re: Conics - Parabola
- Next by Date: Looking for a surjection or R^2
- Previous by thread: Re: Conics - Parabola
- Next by thread: Re: Conics - Parabola
- Index(es):
Relevant Pages
|
