anonymous one year ago Graph

1. anonymous

$f(x)= \frac{ x^{2}+x-2 }{ x ^{2-}3x-4 }$

2. karatechopper

Lets factor!

3. karatechopper

Do you know how to factor?

4. anonymous

Yes $\frac{ (x+1)(x-2) }{ (x+1)(x-4) }$

5. karatechopper

Cool! Okay now, cross out the common factors and what are you left with?

6. anonymous

$\frac{ (x-2) }{ (x-4) }$

7. karatechopper

Right, okay. Now do you have the slightest idea of what we do now?

8. anonymous

Not really haha

9. karatechopper

Alright haha. Well the most simple way of going about this problem is to make an X,Y chart. Just some numbers for your x value, plug into the equation, pop out a Y value. - fill that into the chart. Once you get a few good points, plot. You are set!

10. karatechopper

11. anonymous

12. anonymous

@karatechopper

13. karatechopper

Alright. Let's start.

14. anonymous

Thank you :D

15. karatechopper

|dw:1434953607858:dw| Pick 5 x-values for me. Keep them a good range. :)

16. anonymous

-2,-1,0,1,2

17. anonymous

Isn't that what people usually do

18. karatechopper

|dw:1434953723896:dw| Now, go through each X value and plug each of them into the above equation we simplified to: (x-2)/(x-4), to find each y-value. I will check them after you finish.

19. karatechopper

Also yes, those are common points, sometimes people pick different points because they know what the shape of the graph will look like, but there is not a requirement about which points you pick haha.

20. anonymous

$\frac{ 2 }{ 3 }, -\frac{ 3 }{ 5 }, \frac{ 1 }{ 2 },-\frac{ 1 }{ 3 }, \frac{ 0 }{ -2 }$

21. anonymous

I'm pretty sure some of those are wrong haha

22. karatechopper

Could you show me your work for x-values -1 and 1 please?

23. campbell_st

can I just say... ignore the table of values.... look at the asymptotes the vertical asymptote is when x - 4 = 0 so x = 4 is a vertical asymptote

24. anonymous

(-1-2)=-3 (-1-4)=-5

25. anonymous

Is that correct?

26. karatechopper

Negatives cancel, leaving you with positive 3/5. Same with x-value you 1, you get a positive 1/3. Also Campbell is right.

27. campbell_st

then the numerator and denominator are both degree 1 polynomials so the horizontal asymptote is at y = x/x or y = 1 not look at the numerator to see the where the cruve cuts the x-axis x - 2 = 0 so the x-intercept is x = 2 lastly the y- intercept let x = 0 and you get y = 1/2 so the graph|dw:1434954017902:dw|

28. anonymous

Its two separate lines?

29. campbell_st

a table of values is a waste of time... you need to recognise the curve is a hyperbola... find asymptotes and intercepts then sketch the curve

30. campbell_st

there are 2 parts to the curve

31. campbell_st

and there would be a point of discontinuity at x = -1... in my humble opinion

32. UsukiDoll

asymptotes is a calculus topic though... what if the original poster is in an elementary algebra course? Then a chart / table of values is needed.

33. anonymous

pre cal

34. campbell_st

but Uki will probably clean up my solution and meake it more readable

35. UsukiDoll

ah...so.. ihateschool18 is getting there to asymptotes..

36. UsukiDoll

ha ha I need to eat dinner... no seriously...can't do math without food

37. karatechopper

Asymptotes is an Algebra 2 Subject for me. But that's interesting to see how its in Pre-calc too. I remember it being recovered there too.

38. campbell_st

the great is complex... and would need a very large table to get an idea of the whole graph

39. UsukiDoll

@karatechopper wow that's early...I was first exposed to it in Calculus I O_O

40. karatechopper

I need sleep. Lol yeah Campbell I thought a table of value would get some of the point across as to how such points can't cross a certain line lol.

41. campbell_st

and I tend to use quick and dirty methods rather than the long and boring ones

42. anonymous

Thank you guys so much btw

43. karatechopper

Oye, old school is cool sometimes when getting the point across. xD

44. karatechopper

Welp, I tried haha. Have a good day guys.

45. karatechopper

:)

46. anonymous

Thanks and you too

47. campbell_st

the method is sound... just tedious... and in a question of this type.. I would expect the asker has some knowledge of asymptotes

48. karatechopper

Also @UsukiDoll I'm surprised as to how that's popping up in Calculus. Asymptotes appeared in Conics, which absolutely blew my mind to pieces. Conics was not much of a fun subject with all that memorization.