- 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

- jamiebookeater

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- anonymous

The question is
Using the completing-the-square method, rewrite f(x) = x2 − 8x + 3 in vertex form.

- misty1212

HI!!

- anonymous

hi!

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## More answers

- misty1212

do you have to show your work or can you just do it?

- anonymous

you don't have to show your work

- misty1212

ok then lets to it quick
you have \[f(x)=x^2-8x+3\] right?

- anonymous

yes

- misty1212

what is half of 8?

- anonymous

4

- misty1212

so go right to \[f(x)=(x-4)^2+k\]

- misty1212

to find \(k\) plug in 4 in to \[f(x)=x^2-8x+3\]

- anonymous

so I solve f(x)= (x-4)^2 +k and for k, I plug in 4?

- misty1212

yes, let me know what you get and i will check it

- anonymous

okay I'll solve it real quick :)

- anonymous

I squared the -4 in the parenthesis and got
f(x)= (x-16)+4

- anonymous

is that all that I have to do?

- misty1212

whoa hold on

- misty1212

lets back up a second

- misty1212

we got half of 8 is 4, so it s going to look like \[f(x)=(x-4)^2+k\]

- misty1212

now we need the number \(k\) right?

- anonymous

yes and you said to plug 4 in for k right?

- misty1212

no

- 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

let me know what you get now

- anonymous

oh! okay. I totally switched up what you said :o I'll re-solve it

- 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

it is ok dear, not problem
the correct one is \(-13\) lets do it with pencil and paper

- 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

final answer to your question "vertex form" is \[f(x)=(x-4)^2-13\]

- anonymous

Oh! I solved the exponent first and then subtracted. But thank you so much for helping me! :)

- misty1212

\[\color\magenta\heartsuit\]

- anonymous

Hey @misty1212 . Can you help me with one more question? It also involves the vertex form and standard form

- misty1212

sure happy to!!

- anonymous

It's this -
Rewrite f(x) = –2(x − 3)2 + 2 from vertex form to standard form.

- misty1212

ok this one is going the other way around from vertex to standard
how is your algebra?

- 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

Should I solve that and then tell you what I got?

- misty1212

sure
don't forgot to multiply out first \[(x-3)(x-3)\] requires 4 multiplications
then multiply each term by \(-2\)

- anonymous

So for the first part of the equation, I got -2x+6^2 +2. Is that right so far?

- misty1212

hmm no

- misty1212

\(-3\times (-3)=+9\) lets go slow

- 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

Yes that makes sense

- misty1212

you sure?

- misty1212

you see where each term came from?

- misty1212

some times math teachers call it "foil"

- 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

My answers are
f(x) = –2x2 − 18
f(x) = –2x2 + 12x + 20
f(x) = –2x2 + 12x − 16
f(x) = 4x2 − 24x + 38

- misty1212

ok so we are at \[x^2-3x-3x+9\] which is the same as \[x^2-6x+9\]

- misty1212

then \[-2(x^2-6x+9)=-2x^2+12x-18\]

- misty1212

then add the \(2\) at the end, get \[x^2+12x-16\]

- anonymous

So the answer would be f(x)= -2x^2+12x-16?

- misty1212

yes
FLVS?

- anonymous

yes. Do you do FLVS?

- misty1212

lol no dear, i don't

- anonymous

oh haha

- misty1212

good luck with the algebra \[\color\magenta\heartsuit\]

- anonymous

Thank you so much! And thanks for the help again :)

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