## burhan101 find the equation of this graph (picture included) i have started it off but don't know how to finish it one year ago one year ago

1. burhan101

2. burhan101

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

3. burhan101

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

4. burhan101

3=k

5. CliffSedge

Looks alright so far.

6. burhan101

is there more to or is it all ? :S

7. CliffSedge

Er., wait, no

8. CliffSedge

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

9. burhan101

would that make k zero ?

10. CliffSedge

No, what you did made 3=0.

11. burhan101

ohh i see

12. CliffSedge

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

13. burhan101

i cancelled out k too :/

14. CliffSedge

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

15. burhan101

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

16. CliffSedge

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

17. CliffSedge

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

18. burhan101

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

19. CliffSedge

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

20. burhan101

ohh okay i see it now :\$

21. CliffSedge

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

22. burhan101

so k would be 1/6

23. burhan101

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

24. CliffSedge

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

25. CliffSedge

Yep, it's all good.

26. burhan101

Thank-you once again :D

27. burhan101

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

28. CliffSedge

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

29. CliffSedge

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

30. burhan101

thanks :)

31. burhan101

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

32. burhan101

how would i approach this problem ?

33. CliffSedge

Do you remember the remainder theorem?

34. burhan101

yes, the factor is equal to zero

35. CliffSedge

Sort of..

36. CliffSedge

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. burhan101

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

38. CliffSedge

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

39. CliffSedge

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

40. burhan101

yes that's true

41. CliffSedge

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. burhan101

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. CliffSedge

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

44. burhan101

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

45. CliffSedge

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

46. CliffSedge

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. CliffSedge

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. burhan101

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

49. CliffSedge

I don't see why not.

50. burhan101

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

51. CliffSedge

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

52. CliffSedge

(but I'm lazy like that)

53. burhan101

ohh that way a would be -3/8 right

54. CliffSedge

Uh huh.

55. CliffSedge

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

56. burhan101

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

57. CliffSedge

Looks beautiful.

58. burhan101

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

59. CliffSedge

My pleasure.