Question

What are the points of discontinuity?

Answer 1

\[y=[(x-1)(x+3)]/[x^2+3x+2]\]

Answer 2

@agent0smith

Answer 3

A. x=8
B. x=-1, x=6
C. x=1, x=6
D x=1, x=-6

Answer 4

@Opcode @StudyGurl14

Answer 5

\[\large y=\frac{(x-1)(x+3)}{x^2+3x+2}\]

Factor: \[x^2+3x+2\]

Answer 6

What next because I'm lost

Answer 7

Do you know how to factor?

Answer 8

Yes...

Answer 9

Okay, factor the denominator:
\[x^2+3x+2\]

Answer 10

(x+3)(x+?)

Answer 11

Not exactly?
\[x^2+3x+2 \implies (x+1) (x+2)\]
\[\large y=\frac{(x-1)(x+3)}{x^2+3x+2} \implies y=\frac{(x-1)(x+3)}{(x+1) (x+2)}\]

Answer 12

wow I forgot the step to add x together

Answer 13

\[\frac{ x^2+2x-3 }{ x^2+3x+3 }\]

Answer 14

Okay so we have a whole in the graph when:
\[x = -1~and~x=-2\]
I am a bit confused myself now, your answer choice don't fit :O.
A. x=8
B. x=-1, x=6
C. x=1, x=6
D x=1, x=-6

I must have done something wrong O.o

Answer 15

*hole

Answer 16

Shouldn't the points of discontinuity be at x = -1 and -2?
I learned that discontinuity happens when the denominator is equal to zero.

Answer 17

Not sure... @agent0smith

Answer 18

There are only asymptotes here.

Points of discontinuity are factors in numerator/denominator that cancel out.

Answer 19

So...

Answer 20

Are there any factors that will cancel?

Answer 21

No, right?
I don't think I can cancel anything from:
\[y=\frac{(x-1)(x+3)}{(x+1) (x+2)}\]

Answer 22

Do you have any idea how to solve this?

Answer 23

Not any more :-/.
I am just as confused as your are now, sorry about that >.<

Answer 24

Haha thanks for trying

Answer 25

Reread the above.

There are no points of discont.

Answer 26

There are two vertical asymptotes, but no p.o.d

Answer 27

Oh!
So there is asymptotic discontinuity, just not point discontinuity.
(What confused me was there was no answer choice for that though). Thanks :-).

Answer 28

A. x=8
B. x=-1, x=6
C. x=1, x=6
D x=1, x=-6

these are pretty clearly the wrong answer choices. Or the problem was written wrong.