what is the vertex form of the equation? y=x^2-6x+8?
@jdoe0001 help?
http://www.mathwarehouse.com/geometry/parabola/images/standard-vertex-forms.png
now, you just need to factor it to reflect the "vertex form", then from that use (h,k) :)
wait so I basically just plud in the problem? im confused ha
one sec
ok
first off, you need to get a squared binomial, for the vertex form, the way to do it, is to get a perfect trinomial from it firstly
how do i do that,,
\[x^2-6x+8\ = x^2-6x+8+(1-1) \] so basically, we're using "0", (1-1), to try to make "8" to 9, since \[\sqrt{9}=3\]
remember that a perfect trinomial is \[A^2+2(AB)+B^2\]
okay.. i think i get it a little i guess... i was just skipped a class and put into this one so its really difficult
so in this case, we have A^2=(x^2), 2(AB)=(-6x), and B=(8), but 8 doesn't have a rational root, thus the padding of it using +1 to make it 9 and 9 does have a rational root, "3"
\[x^2-6x+8\ =\ x^2-6x+8+(1-1)\] \[=x^2-6x+9-1\] \[=(x^2-6x+9)-1\]
\[=(x^2-6x+9)\ = \ ?\]
hmmm would it be 6x^2+9
remember that a perfect trinomial is \[A^2+2(AB)+B^2\]
and from there you can get a binomial to the 2nd power
which is what you need for the vertex form
oops...so haha i still dont really get it :/ im trying i swear!
holy smokes!, that example is identical to yours :)
haha. I was gonna say! you went through all that trouble to make me something i would understand ;) but thank you!
anyhow, once you group those 3 terms into the binomail, you'd be left with the (-1), put them together and that's your vertex form, from there you get the (h,k) :)
okay okay so i plug it into the y=a(x-h)^2+k?
well, group those 3, to make squared binomial, what does the equation look like after that?
still confused?
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