Question

Does this function have a vertical asymptote or a removable discontinuity at x=-1?

f(x)= x^2-6x-7 / x+1

Answer 1

@jim_thompson5910 I think it has a removable disc at x=-1.. right?

Answer 2

after factoring, I got (x-7)(x+1)/x+1, the x+1 cancel, then I just have x-7... so I'm not sure

Answer 3

you are correct

Answer 4

very nice work

Answer 5

(x^2-6x-7)/(x+1) is the same as x-7 but x can't equal -1 for the domains to match up

Answer 6

Ok so I got x-7.. but how do tell if that is an asymptote at x=-1 or a removable disc

Answer 7

does x-7 have any asymptotes?

Answer 8

no

Answer 9

remember:

(x^2-6x-7)/(x+1) is the same as x-7

Answer 10

no, right?

Answer 11

so (x^2-6x-7)/(x+1) doesn't have any asymptotes

Answer 12

yes, the answer is "no"

Answer 13

Ok so its a removable discontinuity at x=-1?

Answer 14

but....it has a removable discontinuity at x = -1

Answer 15

because the two domains must match for the two expressions to be equivalent