# Thread: Discontinuity and Vertical Asymptotes

1. ## Discontinuity and Vertical Asymptotes

So I have a problem that I have all parts figured out except for the final part of the question.

The function is f(x)=(x^2 - 1) / (2x^2 - 2x).

The function is discontinuous at 0 and 1.

x=0 is a vertical asymptote.

The horizontal asymptote is y=1/2

Now the final part of the question is: Since there is only one vertical asymptote, indicate on the graph what occurs at the other point of discontinuity. Be specific by giving numerical values.

I am really unsure what happens here. Does the point not exist? I'm confused Any help would be appreciated

2. $\displaystyle f(x) = \frac{(x^2 - 1)}{(2x^2 - 2x)}$

Expand it & Simplify:

$\displaystyle f(x) = \frac{(x - 1)(x + 1)}{2x(x - 1)}$

So we know the degree of the numerator is the same as the degree of the denominator

There is a vertical asymptote at x = 0 because 2x = 0 and therefore x = 0

Now we have an (x - 1) in the numerator and the denominator
This means we can cancel these out giving us a point of discontinuity at x = 1
Now sub it back into the equation with the (x - 1) removed

$\displaystyle f(x) = \frac{(x+1)}{2x}$
$\displaystyle f(x) = 1$

Therefore there is a point of discontinuity at the point (1,1)

Hope this helps : )

3. thanks man appreciate it