Question

find any points of discontinuity for the two rational functions. describe any vertical or horizontal asymptotes and any holes.

1. y=x-1 (over) (x+2)(x-1)

2. y= 2x^2+3 (over) x^2+2

Answer 1

please help me @jdoe0001

Answer 2

please help @whpalmer4

Answer 3

i don know um 1??? 0.0

Answer 4

-1 or x-1

Answer 5

x=1

Answer 6

when you remove the common factor the equation becomes

y = 1/(x + 2)

now for the vertical asymptote...

you can't divide by 0... so the asymptote occurs at the value of x that makes the denominator zero.

so you need to solve..

x + 2 = 0

Answer 7

x=-2

Answer 8

how did you get the point of dixcontinuity

Answer 9

please help this person didnt help me at all @whpalmer4

Answer 10

i dont understand

Answer 11

geee thanks.... I love the fact you can't identify a common factor... and then say I didn't explain it well...

so good luck with your math.

Answer 12

well im sorry your here to help me and when i dont know something i dont know so thanks for that

Answer 13

i dont know the exact definition but i know what it looks like.....i think

Answer 14

good luck...

Answer 15

oh, just a quick question.... what happens when you divide by zero..?

Answer 16

you get a error sign you cant

Answer 17

well...the 2nd part is correct... you can't... so start by finding the values of x that make the denominator zero.... in you're 1st question.

Answer 18

ok isn't that x=-2 & x=1

Answer 19

thats it... so 1 is a vertical asymptote ... the other is a point of discontinuity...

which is which... and why...?

Answer 20

ok i dont understand what you are asking

Answer 21

ok... so the denominator is

(x -1)(x + 2) if you substitute x = 1 or x = -2

you will get zero in the denominator...

so

x = 1, and x = -2 are restrictions in the domain...

one of the values is called a vertical asymptote

the other is called a point of discontinuity...

one of the tasks being asked in the question... is

which is the vertical asymptote... and which is the point of discontinuity...

and you would also need to support your answer with some mathematical reasoning...

Answer 22

once you have mastered that part... you are also asked to identify any horizontal asymptotes... and there is 1 of those you need to find...in question 1.

Answer 23

ok i think the vertical asymptote is x=-2 the horizontal is y=0

i dont know how to find the point of discontinuity

Answer 24

ok... so

the values that make the denominator zero are

x - 2 and x = 1

you're correct with x = -2 as the vertical asymptote

and the horizontal asymptote is correct...

now look at x = 1

it came from the factor (x -1)

the factor is (x -1) is in both the numerator and denominator...

so it can be removed when simplifying the equation...

does that make sense...?

Answer 25

yes i get that continue

Answer 26

ok... there are 2 versions of the equation

the original

y = (x -1)/(x +2)(x-1)

and the simplified version after the common factor is removed

y = 1/(x -2)

so the value of x = 1 is called a point of discontinuity... as it will make the denominator zero.. in the original... and has no effect on the simplified...

the reason it has no effect is that you have removed it, the common factor...

I hope that helps...

Answer 27

i get it now so the discontinuity is 1 right

Answer 28

that correct....

so discontinuity at x = 1

vertical asyptote at x = -2

horizontal asymptote at y = 0

Answer 29

the reason for the discontinuity ...

substitute x = 1 into the original equation and you get... a zero denominator... so undefined.

substitute x = 1 into the simplified equation you get y = 1/3

so the curve is inconsistent... at x = 1... depending on which version you use.

Answer 30

ok thank you i get it now. that was alot better

Answer 31

can you help me @whpalmer4

Answer 32

is number 2 look like this
(x+2)+(2x+1) (over) (x+1)+(x+1)

Answer 33

I don't understand what problem #2 is supposed to look like. Can you show it with the equation editor, please? Or draw it?