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The question is Using the completing-the-square method, rewrite f(x) = x2 − 8x + 3 in vertex form.
do you have to show your work or can you just do it?
you don't have to show your work
ok then lets to it quick you have \[f(x)=x^2-8x+3\] right?
what is half of 8?
so go right to \[f(x)=(x-4)^2+k\]
to find \(k\) plug in 4 in to \[f(x)=x^2-8x+3\]
so I solve f(x)= (x-4)^2 +k and for k, I plug in 4?
yes, let me know what you get and i will check it
okay I'll solve it real quick :)
I squared the -4 in the parenthesis and got f(x)= (x-16)+4
is that all that I have to do?
whoa hold on
lets back up a second
we got half of 8 is 4, so it s going to look like \[f(x)=(x-4)^2+k\]
now we need the number \(k\) right?
yes and you said to plug 4 in for k right?
to find \(k\) start with \[f(x)=x^2-8x+3\] and find \(f(4)\) \[f(4)=4^2-8\times 4+3\]
let me know what you get now
oh! okay. I totally switched up what you said :o I'll re-solve it
I got -13 on my calculator, but when I solved it on paper I got 35. Sorry if I'm annoying you, I just don't understand algebra 2 at all :/
it is ok dear, not problem the correct one is \(-13\) lets do it with pencil and paper
\[f(4)=4^2-8\times 4+3\] first square \[16-8\times 4+3\] then multiply \[16-32+3\] then subtract \[-16+3\] then add \[-13\]
final answer to your question "vertex form" is \[f(x)=(x-4)^2-13\]
Oh! I solved the exponent first and then subtracted. But thank you so much for helping me! :)
sure happy to!!
It's this - Rewrite f(x) = –2(x − 3)2 + 2 from vertex form to standard form.
ok this one is going the other way around from vertex to standard how is your algebra?
because that is what you need \[-2(x-3)^2+2=-2(x-3)(x-3)+2\] you have to multiply out, distribute, combine like terms
Should I solve that and then tell you what I got?
sure don't forgot to multiply out first \[(x-3)(x-3)\] requires 4 multiplications then multiply each term by \(-2\)
So for the first part of the equation, I got -2x+6^2 +2. Is that right so far?
\(-3\times (-3)=+9\) lets go slow
\[(x-3)(x-3)\]first we get \[x^2-3x-3x+9\] let me know when that is clear, or if not, ask
Yes that makes sense
you see where each term came from?
some times math teachers call it "foil"
Yeah. The x's get multiplied to each other and that makes x^2, and then you multiply the 3's to the x's and get -3x-3x. Last, you multiply the -3's to each other and you get 9. But how would we put that into standard form?
My answers are f(x) = –2x2 − 18 f(x) = –2x2 + 12x + 20 f(x) = –2x2 + 12x − 16 f(x) = 4x2 − 24x + 38
ok so we are at \[x^2-3x-3x+9\] which is the same as \[x^2-6x+9\]
then add the \(2\) at the end, get \[x^2+12x-16\]
So the answer would be f(x)= -2x^2+12x-16?
yes. Do you do FLVS?
lol no dear, i don't
good luck with the algebra \[\color\magenta\heartsuit\]
Thank you so much! And thanks for the help again :)