## burhan101 Group Title 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 Group Title

2. burhan101 Group Title

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

3. burhan101 Group Title

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

4. burhan101 Group Title

3=k

5. CliffSedge Group Title

Looks alright so far.

6. burhan101 Group Title

is there more to or is it all ? :S

7. CliffSedge Group Title

Er., wait, no

8. CliffSedge Group Title

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

9. burhan101 Group Title

would that make k zero ?

10. CliffSedge Group Title

No, what you did made 3=0.

11. burhan101 Group Title

ohh i see

12. CliffSedge Group Title

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

13. burhan101 Group Title

i cancelled out k too :/

14. CliffSedge Group Title

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

15. burhan101 Group Title

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

16. CliffSedge Group Title

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

17. CliffSedge Group Title

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

18. burhan101 Group Title

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

19. CliffSedge Group Title

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

20. burhan101 Group Title

ohh okay i see it now :\$

21. CliffSedge Group Title

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

22. burhan101 Group Title

so k would be 1/6

23. burhan101 Group Title

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

24. CliffSedge Group Title

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

25. CliffSedge Group Title

Yep, it's all good.

26. burhan101 Group Title

Thank-you once again :D

27. burhan101 Group Title

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

28. CliffSedge Group Title

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

29. CliffSedge Group Title

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

30. burhan101 Group Title

thanks :)

31. burhan101 Group Title

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

32. burhan101 Group Title

how would i approach this problem ?

33. CliffSedge Group Title

Do you remember the remainder theorem?

34. burhan101 Group Title

yes, the factor is equal to zero

35. CliffSedge Group Title

Sort of..

36. CliffSedge Group Title

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 Group Title

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

38. CliffSedge Group Title

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

39. CliffSedge Group Title

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

40. burhan101 Group Title

yes that's true

41. CliffSedge Group Title

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 Group Title

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 Group Title

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

44. burhan101 Group Title

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

45. CliffSedge Group Title

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

46. CliffSedge Group Title

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 Group Title

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 Group Title

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

49. CliffSedge Group Title

I don't see why not.

50. burhan101 Group Title

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

51. CliffSedge Group Title

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

52. CliffSedge Group Title

(but I'm lazy like that)

53. burhan101 Group Title

ohh that way a would be -3/8 right

54. CliffSedge Group Title

Uh huh.

55. CliffSedge Group Title

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 Group Title

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

57. CliffSedge Group Title

Looks beautiful.

58. burhan101 Group Title

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

59. CliffSedge Group Title

My pleasure.