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Mathematics 8 Online
OpenStudy (anonymous):

what is the vertex form of the equation? y=x^2-6x+8?

OpenStudy (anonymous):

@jdoe0001 help?

OpenStudy (jdoe0001):

now, you just need to factor it to reflect the "vertex form", then from that use (h,k) :)

OpenStudy (anonymous):

wait so I basically just plud in the problem? im confused ha

OpenStudy (jdoe0001):

one sec

OpenStudy (anonymous):

ok

OpenStudy (jdoe0001):

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

OpenStudy (anonymous):

how do i do that,,

OpenStudy (jdoe0001):

\[x^2-6x+8\ = x^2-6x+8+(1-1) \] so basically, we're using "0", (1-1), to try to make "8" to 9, since \[\sqrt{9}=3\]

OpenStudy (jdoe0001):

remember that a perfect trinomial is \[A^2+2(AB)+B^2\]

OpenStudy (anonymous):

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

OpenStudy (jdoe0001):

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"

OpenStudy (jdoe0001):

\[x^2-6x+8\ =\ x^2-6x+8+(1-1)\] \[=x^2-6x+9-1\] \[=(x^2-6x+9)-1\]

OpenStudy (jdoe0001):

\[=(x^2-6x+9)\ = \ ?\]

OpenStudy (anonymous):

hmmm would it be 6x^2+9

OpenStudy (jdoe0001):

remember that a perfect trinomial is \[A^2+2(AB)+B^2\]

OpenStudy (jdoe0001):

and from there you can get a binomial to the 2nd power

OpenStudy (jdoe0001):

which is what you need for the vertex form

OpenStudy (anonymous):

oops...so haha i still dont really get it :/ im trying i swear!

OpenStudy (jdoe0001):

holy smokes!, that example is identical to yours :)

OpenStudy (anonymous):

haha. I was gonna say! you went through all that trouble to make me something i would understand ;) but thank you!

OpenStudy (jdoe0001):

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) :)

OpenStudy (anonymous):

okay okay so i plug it into the y=a(x-h)^2+k?

OpenStudy (jdoe0001):

well, group those 3, to make squared binomial, what does the equation look like after that?

OpenStudy (jdoe0001):

still confused?

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