Question

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

Answer 1

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

Answer 2

HI!!

Answer 3

hi!

Answer 4

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

Answer 5

you don't have to show your work

Answer 6

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

Answer 7

yes

Answer 8

what is half of 8?

Answer 9

4

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

okay I'll solve it real quick :)

Answer 15

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

Answer 16

is that all that I have to do?

Answer 17

whoa hold on

Answer 18

lets back up a second

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

no

Answer 23

to find \(k\) start with \[f(x)=x^2-8x+3\] and find \(f(4)\)
\[f(4)=4^2-8\times 4+3\]

Answer 24

let me know what you get now

Answer 25

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

Answer 26

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 :/

Answer 27

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

Answer 28

\[f(4)=4^2-8\times 4+3\] first square \[16-8\times 4+3\] then multiply \[16-32+3\] then subtract \[-16+3\] then add \[-13\]

Answer 29

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

Answer 30

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

Answer 31

\[\color\magenta\heartsuit\]

Answer 32

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

Answer 33

sure happy to!!

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

hmm no

Answer 41

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

Answer 42

\[(x-3)(x-3)\]first we get \[x^2-3x-3x+9\] let me know when that is clear, or if not, ask

Answer 43

Yes that makes sense

Answer 44

you sure?

Answer 45

you see where each term came from?

Answer 46

some times math teachers call it "foil"

Answer 47

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?

Answer 48

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

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

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

Answer 53

yes
FLVS?

Answer 54

yes. Do you do FLVS?

Answer 55

lol no dear, i don't

Answer 56

oh haha

Answer 57

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

Answer 58

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