## mathcalculus Group Title PLEASE HELP :( Please help: sketch the graph of a function. label the intercepts, relative extrema, points of inflection, and asymptotes! then state the domain of the function! y= (5-3x)/(x-2) 9 months ago 9 months ago

1. mathcalculus Group Title

so i found the vertical asymptote: x=2 and horizontal asymptote: y=3 and the intercepts i found that it is (0, -2.5) and (1.67,0) so the domain is: (-inf,2)(2, inf) now I'm stuck on how to do the relative extrema and the points of inflections. :( please help.

2. mathcalculus Group Title

i tried to also find the derivative and i got 1/ (x-2)^2

3. mathcalculus Group Title

4. zepdrix Group Title

So did you apply the Quotient Rule? One sec, I'll check it :O

5. mathcalculus Group Title

yeah i used the quotient rule

6. zepdrix Group Title

7. mathcalculus Group Title

yeah the first answer was correct.

8. mathcalculus Group Title

i mean I'm trying to find the relative extrema but i forgot how to. i know i have to find the derivative first. then I'm not sure where to go . :(

9. zepdrix Group Title

So we have our first derivative:$\Large y'\quad=\quad \frac{1}{(x-2)^2}$ To find extrema/critical points, we set the first derivative equal to zero and solve for x.

10. mathcalculus Group Title

right but wouldn't it be false?

11. mathcalculus Group Title

1=0? doesn't work.

12. zepdrix Group Title

Oh there is another little tid bit that I often forget. critical points also exist when the first derivative is undefined. Do you see any points where that might be true?

13. mathcalculus Group Title

by undefined what do you mean?

14. zepdrix Group Title

Numbers that would cause a problem :D Like dividing by zero.

15. zepdrix Group Title

Any x values that would cause a problem for our derivative?

16. mathcalculus Group Title

like this one?

17. mathcalculus Group Title

oh ya 2!

18. zepdrix Group Title

Yes! Good. x=2 is something we might consider to be a critical point. Before we can say that for certain though, is x=2 in the domain of our function?

19. zepdrix Group Title

You stated the domain earlier, so go back and look at it! :)

20. mathcalculus Group Title

no.

21. mathcalculus Group Title

x cannot equal to 2.

22. zepdrix Group Title

Good good. Since it's not in the domain of our function, it can't be a critical point.

23. mathcalculus Group Title

all real numbers except 2.

24. zepdrix Group Title

So there are no critical points. Should we look for inflection points next? :o

25. mathcalculus Group Title

so theres no relative extrema?

26. mathcalculus Group Title

yes last but not least, inflection points. :)

27. zepdrix Group Title

So we'll need to find the second derivative for that. Have you tried that step yet?

28. mathcalculus Group Title

no but let me try real quick

29. mathcalculus Group Title

so inflection points is always SECOND DERIVATIVE right?

30. zepdrix Group Title

y=0 gives us roots of the function. y'=0 gives us critical points of the function. y''=0 gives us inflection points of the function. yes.

31. mathcalculus Group Title

i got (-3x^3+17x^2-32x+20)/(x^2-4x+4)

32. mathcalculus Group Title

@zepdrix

33. zepdrix Group Title

Dude dude dude -_- what..? Don't expand out squares, especially if they're in the denominator.

34. zepdrix Group Title

Did you apply the quotient rule? We don't want to do that.

35. mathcalculus Group Title

yeah i did:/

36. mathcalculus Group Title

i used the rule book.

37. zepdrix Group Title

$\Large y'\quad=\quad \frac{1}{(x-2)^2}\quad=\quad (x-2)^{-2}$Power rule from there :O

38. mathcalculus Group Title

ohhhhh yes

39. mathcalculus Group Title

crap.

40. zepdrix Group Title

XD

41. mathcalculus Group Title

-2/(x-2)^3

42. mathcalculus Group Title

ok :D

43. zepdrix Group Title

Ok cool. Setting our second derivative equal to zero:$\Large 0\quad=\quad \frac{-2}{(x-2)^3}$

44. zepdrix Group Title

Hmm I think inflection points work the same way as critical points ~ they exist where the second derivative is undefined.. mmmmm

45. mathcalculus Group Title

huh?

46. mathcalculus Group Title

don't we plug in severals x's between the intervals or something?

47. mathcalculus Group Title

so threres no inflection point?

48. zepdrix Group Title

between what intervals though? We would need to find inflection points to determine which intervals to check. I guess we just check where the function has a discontinuity since we have no inflection points. So ya, check on the left side of 2 and also on the right. We should get different concavity on each side for this problem.

49. zepdrix Group Title

Oh the problem didn't ask for concavity did it? -_- oh.. hmm

50. mathcalculus Group Title

no it didn't...

51. mathcalculus Group Title

so help me with inflection point, do we have to include other x's between the intervals?im not sure

52. zepdrix Group Title

include other x's... whu? +_+

53. mathcalculus Group Title

so this is the answer: it doesn't say which is for which but i got the idea of what is what. my question is how'd they get the other curve on the right side

54. zepdrix Group Title

So they labeled the intercepts (as you found), the asymptotes x=2, y=-3, and nothing else... So I think we did everything correctly, seeing as how we found no extrema for this one and they didn't label any on their graph.. :\

55. zepdrix Group Title

Oh the other curve, umm

56. zepdrix Group Title

bah i dunno >:O my head hurts.. i need a maf break!