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anonymous
 one year ago
Graph
anonymous
 one year ago
Graph

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[f(x)= \frac{ x^{2}+x2 }{ x ^{2}3x4 }\]

karatechopper
 one year ago
Best ResponseYou've already chosen the best response.4Do you know how to factor?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes \[\frac{ (x+1)(x2) }{ (x+1)(x4) }\]

karatechopper
 one year ago
Best ResponseYou've already chosen the best response.4Cool! Okay now, cross out the common factors and what are you left with?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ (x2) }{ (x4) }\]

karatechopper
 one year ago
Best ResponseYou've already chosen the best response.4Right, okay. Now do you have the slightest idea of what we do now?

karatechopper
 one year ago
Best ResponseYou've already chosen the best response.4Alright 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!

karatechopper
 one year ago
Best ResponseYou've already chosen the best response.4Would you like me to help you make a X,Y chart?

karatechopper
 one year ago
Best ResponseYou've already chosen the best response.4Alright. Let's start.

karatechopper
 one year ago
Best ResponseYou've already chosen the best response.4dw:1434953607858:dw Pick 5 xvalues for me. Keep them a good range. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Isn't that what people usually do

karatechopper
 one year ago
Best ResponseYou've already chosen the best response.4dw:1434953723896:dw Now, go through each X value and plug each of them into the above equation we simplified to: (x2)/(x4), to find each yvalue. I will check them after you finish.

karatechopper
 one year ago
Best ResponseYou've already chosen the best response.4Also 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 2 }{ 3 }, \frac{ 3 }{ 5 }, \frac{ 1 }{ 2 },\frac{ 1 }{ 3 }, \frac{ 0 }{ 2 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm pretty sure some of those are wrong haha

karatechopper
 one year ago
Best ResponseYou've already chosen the best response.4Could you show me your work for xvalues 1 and 1 please?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.0can 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

karatechopper
 one year ago
Best ResponseYou've already chosen the best response.4Negatives cancel, leaving you with positive 3/5. Same with xvalue you 1, you get a positive 1/3. Also Campbell is right.

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.0then 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 xaxis x  2 = 0 so the xintercept is x = 2 lastly the y intercept let x = 0 and you get y = 1/2 so the graphdw:1434954017902:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Its two separate lines?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.0a 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

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.0there are 2 parts to the curve

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.0and there would be a point of discontinuity at x = 1... in my humble opinion

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0asymptotes is a calculus topic though... what if the original poster is in an elementary algebra course? Then a chart / table of values is needed.

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.0but Uki will probably clean up my solution and meake it more readable

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0ah...so.. ihateschool18 is getting there to asymptotes..

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0ha ha I need to eat dinner... no seriously...can't do math without food

karatechopper
 one year ago
Best ResponseYou've already chosen the best response.4Asymptotes is an Algebra 2 Subject for me. But that's interesting to see how its in Precalc too. I remember it being recovered there too.

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.0the great is complex... and would need a very large table to get an idea of the whole graph

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0@karatechopper wow that's early...I was first exposed to it in Calculus I O_O

karatechopper
 one year ago
Best ResponseYou've already chosen the best response.4I 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.

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.0and I tend to use quick and dirty methods rather than the long and boring ones

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you guys so much btw

karatechopper
 one year ago
Best ResponseYou've already chosen the best response.4Oye, old school is cool sometimes when getting the point across. xD

karatechopper
 one year ago
Best ResponseYou've already chosen the best response.4Welp, I tried haha. Have a good day guys.

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.0the method is sound... just tedious... and in a question of this type.. I would expect the asker has some knowledge of asymptotes

karatechopper
 one year ago
Best ResponseYou've already chosen the best response.4Also @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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