## anonymous one year ago 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

1. anonymous

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

2. misty1212

HI!!

3. anonymous

hi!

4. misty1212

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

5. anonymous

you don't have to show your work

6. misty1212

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

7. anonymous

yes

8. misty1212

what is half of 8?

9. anonymous

4

10. misty1212

so go right to $f(x)=(x-4)^2+k$

11. misty1212

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

12. anonymous

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

13. misty1212

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

14. anonymous

okay I'll solve it real quick :)

15. anonymous

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

16. anonymous

is that all that I have to do?

17. misty1212

whoa hold on

18. misty1212

lets back up a second

19. misty1212

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

20. misty1212

now we need the number $$k$$ right?

21. anonymous

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

22. misty1212

no

23. misty1212

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

24. misty1212

let me know what you get now

25. anonymous

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

26. 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 :/

27. misty1212

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

28. 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 $-13$

29. misty1212

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

30. anonymous

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

31. misty1212

$\color\magenta\heartsuit$

32. anonymous

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

33. misty1212

sure happy to!!

34. anonymous

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

35. misty1212

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

36. 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

37. anonymous

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

38. misty1212

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

39. anonymous

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

40. misty1212

hmm no

41. misty1212

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

42. misty1212

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

43. anonymous

Yes that makes sense

44. misty1212

you sure?

45. misty1212

you see where each term came from?

46. misty1212

some times math teachers call it "foil"

47. 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?

48. anonymous

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

49. misty1212

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

50. misty1212

then $-2(x^2-6x+9)=-2x^2+12x-18$

51. misty1212

then add the $$2$$ at the end, get $x^2+12x-16$

52. anonymous

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

53. misty1212

yes FLVS?

54. anonymous

yes. Do you do FLVS?

55. misty1212

lol no dear, i don't

56. anonymous

oh haha

57. misty1212

good luck with the algebra $\color\magenta\heartsuit$

58. anonymous

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