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\[f(x)= \frac{ x^{2}+x-2 }{ x ^{2-}3x-4 }\]
Lets factor!
Do you know how to factor?

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Yes \[\frac{ (x+1)(x-2) }{ (x+1)(x-4) }\]
Cool! Okay now, cross out the common factors and what are you left with?
\[\frac{ (x-2) }{ (x-4) }\]
Right, okay. Now do you have the slightest idea of what we do now?
Not really haha
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!
Would you like me to help you make a X,Y chart?
Yes please
Alright. Let's start.
Thank you :D
|dw:1434953607858:dw| Pick 5 x-values for me. Keep them a good range. :)
-2,-1,0,1,2
Isn't that what people usually do
|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.
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.
\[\frac{ 2 }{ 3 }, -\frac{ 3 }{ 5 }, \frac{ 1 }{ 2 },-\frac{ 1 }{ 3 }, \frac{ 0 }{ -2 }\]
I'm pretty sure some of those are wrong haha
Could you show me your work for x-values -1 and 1 please?
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
(-1-2)=-3 (-1-4)=-5
Is that correct?
Negatives cancel, leaving you with positive 3/5. Same with x-value you 1, you get a positive 1/3. Also Campbell is right.
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|
Its two separate lines?
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
there are 2 parts to the curve
and there would be a point of discontinuity at x = -1... in my humble opinion
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.
pre cal
but Uki will probably clean up my solution and meake it more readable
ah...so.. ihateschool18 is getting there to asymptotes..
ha ha I need to eat dinner... no seriously...can't do math without food
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.
the great is complex... and would need a very large table to get an idea of the whole graph
@karatechopper wow that's early...I was first exposed to it in Calculus I O_O
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.
and I tend to use quick and dirty methods rather than the long and boring ones
Thank you guys so much btw
Oye, old school is cool sometimes when getting the point across. xD
Welp, I tried haha. Have a good day guys.
:)
Thanks and you too
the method is sound... just tedious... and in a question of this type.. I would expect the asker has some knowledge of asymptotes
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.

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