Start with the equation of a quadratic, that is
So is the xcoordinate of the vertex.
Therefore the vertex of any parabola is the point

Start with the equation of a quadratic, that is
So is the xcoordinate of the vertex.
Therefore the vertex of any parabola is the point
Last edited by ellatha; January 19th, 2013 at 08:29 PM.
how did you get there? i know its 1/2 (x,y) between (,x ,y). graph please a or b could be anything. the problem here is resolving the height of the parabola. You know the y intercept which is really all you have that you know for certain. how much can you be certain about? you have x x and c
Last edited by fiveworlds; January 18th, 2013 at 08:51 PM.
I already provided the proof in the above post; and yes, the formula works for any quadratic.
try this too.
take x and x take your y intercept draw a tangent from the y intercept to the line x x
get the distance from intercept to each x point.
find ratio to total width and half
multiply h by this number
half x between x and minus x too
place awnsers as (x,y)
and tell me it isnt a vertex
Except not all parabolas have an xintercept.
Last edited by ellatha; January 18th, 2013 at 10:11 PM.
they all have an x intercept you place it in position if you are smart about it. if you have no y intercept im sorry but you cannot solve it because you only have three pieces of info a parabola without a y intercept is not a parabola it is a line through x
You can still get the vertex for a parabola with no xintercept. Consider the parabola x^2 + 2x + 2, it has no xintercept but its vertex is the point (1, 1).
the x intercept is x=2 + x=1.
ps please tell me where y came from in your eqn
ax^2+bx+c=0 not =y
y was never part of your eqn
you dont have enough info from a parabola to solve for y
because c isnt the y intercept.
tell me what shape you know with a constant diameter.
basically a parabola is when somebody thinks its funny
to take the y cords from a circle equation. so basically
its a circle of unknown height which means its a curved
surface on both sides of the x axis passing an infinite number
of points
c could also be called r^2
that clarify?
a parabola slope must contain
fixed quantised packets of energy
called energy levels. now do you have
an idea of what parabolae are atoms
what i said for pi is that atoms consist
of a proton a neutron and an electron
so multiples of 3.3333333 however
taking height into account this becomes
4.44444444 or h
Last edited by fiveworlds; January 19th, 2013 at 08:41 AM.
Your text is incomprehensible.
i cant make it much simpler your eqn has no frame of reference on the y axis to construct your list of y coordinates.
it is also impossible to find your np list of points to construct an xaxis
you do not have x or y you have two points floating in infinte space and know the distance between them
No it isn't. Plot a graph of the equation and you can see it doesn't intercept the x axis:
plot y=x^2 + 2x + 2 from 2.5 to 0.5  WolframAlpha
Obviously,ps please tell me where y came from in your eqn
<meaningless drivel deleted>
well by that i mean the slopes themselves take two electrons separated by a distance. You have no idea at all about where the electron is going to be or how it is going to be moving all you know is that a slope must exsist corresponding to how the position of the electron is changing within an atom. but you do know that it must have a certain amount of energy corresponding to its orbital.is that better?
also the 4 bit is basically that on average an atom should be a sphere approx
its not of topic strange by claiming that she can find the vertex of any parabola she is saying she can find the position and velocity of any electron in an atom it amounts to the same thing really.
and its obvious that y=ax^2+bx+chowever if your say @y =0then your numbers for a b and cthen the equation no longer has aninverse and it is impossible to know that10x^2+13bx+c = 9 lies on the original parabolayou have no reference of how y changes
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