Question

Pre-Calculus help!

Construct a rational function that has a vertical asymptote at x=3 and a removable discontinuity at x=-1. Explain how you determined your answer.

Answer 1

would you like to be freinds
and what school do you go to

Answer 2

So to create a removable discontinuity rearrange the equation x =-1 in factored form and then keep it as denominator (Hint: (x+1))
Now keep anything as numerator and add 3 to the fraction

Answer 3

@FaiqRaees I'm not sure I understand..... You weren't very clear. like the function 4/x+1 +3 doesn't have a hole, as far as I can tell

Answer 4

If there is a vertical asymptote as x = 3, then (x - 3) MUST be part of the equation.

Answer 5

Yeah, I know that.

Answer 6

a discontinuity would exist where 0 was in the denominator. A removable one means that it can be taken out. So...
\(\Large\frac{(x -3)(x+1)}{x +1}\)

Answer 7

gotcha. So part of the equation would definitely be (x-3)/(x-3), right?

Answer 8

no....the equation I gave is the answer.

Answer 9

oh. Gotcha. Does that have the asymptote in there too?

Answer 10

nvm I see it. Thanks so much!

Answer 11

yes. the (x-3) in the numerator is the vertical asymtote.

Answer 12

you're welcome. PS. I love Doctor Who

Answer 13

Ayyyy lit

Answer 14

wait @StudyGurl14 Wouldn't it be x-2 instead of x+1? Since the removable discontinuity has to be at x=-2?

Answer 15

Wait no it'd be +2. And I realizee that I wrote it wrong in the question. It should be a discontinuity at x=-2, not x=-1.

Answer 16

then it would be x + 2