Question

Ok here's a Calc. Question. I understand how to do this but keep getting the wrong answer for my homework. Please help....

f(x)= (x-8) all over ((x-1)(x+1))
Use interval notation to indicate where f(x) is continuous.

Answer 1

First, what are the x-values that make the denominator 0?

Answer 2

1 and -1.

Answer 3

Right, so if we were to graph this, those would be the 2 asymptotes.

Answer 4

yeah that's the graph that I got was asymptotes

Answer 5

Now, is there any y-asymptotes?

Answer 6

no.....

Answer 7

are you sure?

Answer 8

\[\lim_{x \rightarrow \pm \infty}\frac{ x-8 }{ x^2-1 }=\lim_{x \rightarrow \pm \infty}\frac{ \frac{ 1 }{ x }-\frac{ 8 }{ x^2 } }{ 1-\frac{ 1 }{ x^2 } }=0\]

Answer 9

As x approaches positive/negative infinity, the function approaches 0.

Answer 10

Now, is there any x-intercepts?

Answer 11

@jclark1989

Answer 12

BTW we're doing this to graph the function because once you graph it, it is very easy to see where the function is continuous.

Answer 13

yeah.. the x intercepts would be 1 and -1 right?

Answer 14

Remeber that x intercepts are the points where the y-coordinate is 0.

Answer 15

In other words, f(x)=0

Answer 16

ahh ok.

Answer 17

What did you get?

Answer 18

\[\frac{ x-8 }{ x^2-1 }=0\rightarrow x=8\]

Answer 19

ahh. so the x cord. is 1 and -1.

Answer 20

this graph is throwing me off. the y cord is 8?