anonymous
  • anonymous
please help me with this question! Using the completing-the-square method, rewrite f(x) = x2 − 8x + 3 in vertex form. f(x) = (x − 8)2 f(x) = (x − 4)2 − 13 f(x) = (x − 4)2 + 3 f(x) = (x − 4)2 + 16
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
The question is Using the completing-the-square method, rewrite f(x) = x2 − 8x + 3 in vertex form.
misty1212
  • misty1212
HI!!
anonymous
  • anonymous
hi!

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misty1212
  • misty1212
do you have to show your work or can you just do it?
anonymous
  • anonymous
you don't have to show your work
misty1212
  • misty1212
ok then lets to it quick you have \[f(x)=x^2-8x+3\] right?
anonymous
  • anonymous
yes
misty1212
  • misty1212
what is half of 8?
anonymous
  • anonymous
4
misty1212
  • misty1212
so go right to \[f(x)=(x-4)^2+k\]
misty1212
  • misty1212
to find \(k\) plug in 4 in to \[f(x)=x^2-8x+3\]
anonymous
  • anonymous
so I solve f(x)= (x-4)^2 +k and for k, I plug in 4?
misty1212
  • misty1212
yes, let me know what you get and i will check it
anonymous
  • anonymous
okay I'll solve it real quick :)
anonymous
  • anonymous
I squared the -4 in the parenthesis and got f(x)= (x-16)+4
anonymous
  • anonymous
is that all that I have to do?
misty1212
  • misty1212
whoa hold on
misty1212
  • misty1212
lets back up a second
misty1212
  • misty1212
we got half of 8 is 4, so it s going to look like \[f(x)=(x-4)^2+k\]
misty1212
  • misty1212
now we need the number \(k\) right?
anonymous
  • anonymous
yes and you said to plug 4 in for k right?
misty1212
  • misty1212
no
misty1212
  • misty1212
to find \(k\) start with \[f(x)=x^2-8x+3\] and find \(f(4)\) \[f(4)=4^2-8\times 4+3\]
misty1212
  • misty1212
let me know what you get now
anonymous
  • anonymous
oh! okay. I totally switched up what you said :o I'll re-solve it
anonymous
  • anonymous
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 :/
misty1212
  • misty1212
it is ok dear, not problem the correct one is \(-13\) lets do it with pencil and paper
misty1212
  • misty1212
\[f(4)=4^2-8\times 4+3\] first square \[16-8\times 4+3\] then multiply \[16-32+3\] then subtract \[-16+3\] then add \[-13\]
misty1212
  • misty1212
final answer to your question "vertex form" is \[f(x)=(x-4)^2-13\]
anonymous
  • anonymous
Oh! I solved the exponent first and then subtracted. But thank you so much for helping me! :)
misty1212
  • misty1212
\[\color\magenta\heartsuit\]
anonymous
  • anonymous
Hey @misty1212 . Can you help me with one more question? It also involves the vertex form and standard form
misty1212
  • misty1212
sure happy to!!
anonymous
  • anonymous
It's this - Rewrite f(x) = –2(x − 3)2 + 2 from vertex form to standard form.
misty1212
  • misty1212
ok this one is going the other way around from vertex to standard how is your algebra?
misty1212
  • misty1212
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
anonymous
  • anonymous
Should I solve that and then tell you what I got?
misty1212
  • misty1212
sure don't forgot to multiply out first \[(x-3)(x-3)\] requires 4 multiplications then multiply each term by \(-2\)
anonymous
  • anonymous
So for the first part of the equation, I got -2x+6^2 +2. Is that right so far?
misty1212
  • misty1212
hmm no
misty1212
  • misty1212
\(-3\times (-3)=+9\) lets go slow
misty1212
  • misty1212
\[(x-3)(x-3)\]first we get \[x^2-3x-3x+9\] let me know when that is clear, or if not, ask
anonymous
  • anonymous
Yes that makes sense
misty1212
  • misty1212
you sure?
misty1212
  • misty1212
you see where each term came from?
misty1212
  • misty1212
some times math teachers call it "foil"
anonymous
  • anonymous
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?
anonymous
  • anonymous
My answers are f(x) = –2x2 − 18 f(x) = –2x2 + 12x + 20 f(x) = –2x2 + 12x − 16 f(x) = 4x2 − 24x + 38
misty1212
  • misty1212
ok so we are at \[x^2-3x-3x+9\] which is the same as \[x^2-6x+9\]
misty1212
  • misty1212
then \[-2(x^2-6x+9)=-2x^2+12x-18\]
misty1212
  • misty1212
then add the \(2\) at the end, get \[x^2+12x-16\]
anonymous
  • anonymous
So the answer would be f(x)= -2x^2+12x-16?
misty1212
  • misty1212
yes FLVS?
anonymous
  • anonymous
yes. Do you do FLVS?
misty1212
  • misty1212
lol no dear, i don't
anonymous
  • anonymous
oh haha
misty1212
  • misty1212
good luck with the algebra \[\color\magenta\heartsuit\]
anonymous
  • anonymous
Thank you so much! And thanks for the help again :)

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