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vertex x is half way between intercepts
so its at least got vertex on the x=-2 axis
how would i set it up then
2+-6 = -4/2 = -2
then how would I find the y
f(x) = a(x+2)^2 + h
err, i think its +k for standard usage
yeah but how would i figure out k and the slope?
without any other information; your pretty much at a stand still; there is a whole family of equations to fit this
if k is negative; make a + and f(x) = a(x+2)^2 - k or the other way around since the vertex needs to be on the oterside of the opening end f(x) -a(x+2)^2 + k
anything but zeros for k and a should work, and make sure they gots opposite signs
one sec let me test it
but whats the slope then?
hmmm, i just tested a few on the wolf, my thoughts need some refinement :)
ok, try this; make k equal to half the distance between intercepts; in this case -4 should work for i trial run
.... recalculating lol we are sitting at 4; so if x=4 x^2 = 16 .... lets try -16 hope this aint to confusing yet
that one hit, i knew one of my ideas had to stick ;)
so the slope is 1?
a simpler way would have been to construct an equation form the zeros: a(x+6)(x-2)=0 but it wouldnt have been in vertex form
Thats ok, at least we found it :)
:) good luck