Question

Graph

Answer 1

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

Answer 2

Lets factor!

Answer 3

Do you know how to factor?

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

Not really haha

Answer 9

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!

Answer 10

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

Answer 11

Yes please

Answer 12

@karatechopper

Answer 13

Alright. Let's start.

Answer 14

Thank you :D

Answer 15

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

Answer 16

-2,-1,0,1,2

Answer 17

Isn't that what people usually do

Answer 18

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.

Answer 19

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.

Answer 20

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

Answer 21

I'm pretty sure some of those are wrong haha

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

Is that correct?

Answer 26

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

Answer 27

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