Graph

- anonymous

Graph

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- anonymous

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

- karatechopper

Lets factor!

- karatechopper

Do you know how to factor?

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## More answers

- anonymous

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

- karatechopper

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

- anonymous

\[\frac{ (x-2) }{ (x-4) }\]

- karatechopper

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

- anonymous

Not really haha

- 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!

- karatechopper

Would you like me to help you make a X,Y chart?

- anonymous

Yes please

- anonymous

@karatechopper

- karatechopper

Alright. Let's start.

- anonymous

Thank you :D

- karatechopper

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

- anonymous

-2,-1,0,1,2

- anonymous

Isn't that what people usually do

- 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.

- 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.

- anonymous

\[\frac{ 2 }{ 3 }, -\frac{ 3 }{ 5 }, \frac{ 1 }{ 2 },-\frac{ 1 }{ 3 }, \frac{ 0 }{ -2 }\]

- anonymous

I'm pretty sure some of those are wrong haha

- karatechopper

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

- 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

- anonymous

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

- anonymous

Is that correct?

- 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.

- 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|

- anonymous

Its two separate lines?

- 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

- campbell_st

there are 2 parts to the curve

- campbell_st

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

- 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.

- anonymous

pre cal

- campbell_st

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

- UsukiDoll

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

- UsukiDoll

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

- 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.

- campbell_st

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

- UsukiDoll

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

- 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.

- campbell_st

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

- anonymous

Thank you guys so much btw

- karatechopper

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

- karatechopper

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

- karatechopper

:)

- anonymous

Thanks and you too

- 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

- 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.

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