## anonymous 3 years ago find the equation of this graph (picture included) i have started it off but don't know how to finish it

1. anonymous

2. anonymous

$h(x)=k(x+3)(x-2)(x-3)$

3. anonymous

3=k(2+3)(2-2)(2-3)

4. anonymous

3=k

5. anonymous

Looks alright so far.

6. anonymous

is there more to or is it all ? :S

7. anonymous

Er., wait, no

8. anonymous

In your solving for k, you need to put in 0 for x if you're using the y-intercept.

9. anonymous

would that make k zero ?

10. anonymous

No, what you did made 3=0.

11. anonymous

ohh i see

12. anonymous

"3=k(2+3)(2-2)(2-3)" 3=k(5)(0)(-1) -> 3=0

13. anonymous

i cancelled out k too :/

14. anonymous

(2,3) is not a solution to the equation, but (0,3) is.

15. anonymous

is dont see where you're getting (2,3) from

16. anonymous

You plugged in x=2 and set y=3 in your attempt to solve for k.

17. anonymous

So, I don't see where *you* got (2,3) from. ;-)

18. anonymous

ohhhh ! i'm supposed to set x as 0 right ?

19. anonymous

Yes, because you know the y-intercept is 3.

20. anonymous

ohh okay i see it now :\$

21. anonymous

You could use any other known point, but that one is the most convenient.

22. anonymous

so k would be 1/6

23. anonymous

and h(x) = 1/6 (x+3)(x-2)(x-3)

24. anonymous

never mind h(x) = k(x+3)(x-2)(x-3) h(0)=3 3=k(3)(-2)(-3) 3=18k

25. anonymous

Yep, it's all good.

26. anonymous

Thank-you once again :D

27. anonymous

if you're not busy, can you help me with one last question

28. anonymous

I think you know what you're doing, just be careful and don't rush through it.

29. anonymous

Oh, sure. I think I can handle one more.

30. anonymous

thanks :)

31. anonymous

Find the equation of a cubic polynomial that gives a remainder of 3, when divided by (x+2)

32. anonymous

how would i approach this problem ?

33. anonymous

Do you remember the remainder theorem?

34. anonymous

yes, the factor is equal to zero

35. anonymous

Sort of..

36. anonymous

It kind of goes like this. If (x+2) is a factor of the polynomial, then dividing by (x+2) would have a remainder of 0. And, if (x+2) is a factor, then x = -2 is a root of the polynomial. If (x+2) is not a factor, then f(-2) = the remainder.

37. anonymous

but how would i use this info to make an equation, this is like working backwards from the norm :/

38. anonymous

Yes, a lot of what you're doing here is using concepts to 'unsolve' equations.

39. anonymous

If dividing by (x+2) gives a remainder of 3, then you know (-2,3) is a solution to the equation.

40. anonymous

yes that's true

41. anonymous

It's a pretty open-ended question; there will be many functions (infinite, really) that you could develop that have that point on it.

42. anonymous

i still don't really get it, like i understand what it is asking for but i don't knw how to get to it :/

43. anonymous

It wants a cubic function, so what does an equation for one of those look like?

44. anonymous

like x^3-3x^2-4x-5

45. anonymous

Sure, something like that, so to be more general, start with $y=ax^3+bx^2+cx+d$

46. anonymous

You know what x and y are, and three out of those four variables are free parameters, so choose whatever you want for those, then solve for the fourth one.

47. anonymous

You could even set three of them =0 if you wanted to be really lazy about it. (General math tip: be lazy whenever possible)

48. anonymous

so something like $y=x^3+x^2+x+d$ ?

49. anonymous

I don't see why not.

50. anonymous

hmm but "d" has to be a specific number right

51. anonymous

Yep, that's what you're solving for. Me, I went with y=ax^3, then solved for a.

52. anonymous

(but I'm lazy like that)

53. anonymous

ohh that way a would be -3/8 right

54. anonymous

Uh huh.

55. anonymous

And if you want, you can verify that $\large\frac{-3}{8}x^3 \div (x+2)$ will give a remainder of 3.

56. anonymous

$:. y=\frac{ -3 }{ 8 }x^3$

57. anonymous

Looks beautiful.

58. anonymous

thank you so much for your helppp, you're amazing :D

59. anonymous

My pleasure.