# Find Vertical Asymptote

• Oct 20th 2008, 10:08 AM
magentarita
Find Vertical Asymptote
Here is another review question placed on the board by my teacher.

Find the vertical asymptote(s) of the function

G(x) = (x^3 - 1)/(x - x^2)

He said the only vertical asymptote is x = 0.

However, I disagree.

I say that x = 1 is another asymptote.

HERE IS MY WORK:

I factored the denominator.

x - x^2 = x(1 - x).

I then set each factor to 0 and solve for x.

One factor is x = 0

I also set 1 - x = 0 and got x = 1 as another asymptote.

My teacher said that I am wrong but gave me credit for trying.

Who is right and why?

Thanks
• Oct 20th 2008, 10:24 AM
masters
Quote:

Originally Posted by magentarita
Here is another review question placed on the board by my teacher.

Find the vertical asymptote(s) of the function

G(x) = (x^3 - 1)/(x - x^2)

He said the only vertical asymptote is x = 0.

However, I disagree.

I say that x = 1 is another asymptote.

HERE IS MY WORK:

I factored the denominator.

x - x^2 = x(1 - x).

I then set each factor to 0 and solve for x.

One factor is x = 0

I also set 1 - x = 0 and got x = 1 as another asymptote.

My teacher said that I am wrong but gave me credit for trying.

Who is right and why?

Thanks

Your teacher is correct. What you found was where the graph has point discontinuity. There's a "hole" in the graph where x=1.

When you simplify the rational expression, one of the factors divides out. When that happens, it is not an asymptote.

Point of Discontinuity
Definition of Point of Discontinuity

A function is said to have a point of discontinuity at x = a or the graph of the function has a hole at x = a, if the original function is undefined for x = a, whereas the related rational expression of the function in simplest form is defined for x = a.

$f(x)=\frac{x^3-1}{x-x^2}$

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

The above function in its original state is not defined for x=0 nor x=1.

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

In the simplified form above, the function is still not defined for x = 0.

Therefore, x = 0 is an asymptote. However, it is defined for x = 1.
Plugging x=1 into the simplified function, we find f(1)=-3.
Thus we can say that the function f(x) has a point of discontinuity at (1, -3).
• Oct 20th 2008, 12:01 PM
Shyam
Your Teacher is Right, because, the curve has Vertical Asymptote x = 0. It also has one Linear Oblique Asymptote, y = - x - 1

$f(x)=\frac{x^3-1}{x-x^2}=\frac{(x-1)(x^2+x+1)}{-x(x-1)}=\frac{x^2+x+1}{-x}, \;\;x\ne1$

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

So, Linear Oblique Asymptote,

$y = -x-1$

The value x = 1 is excluded from the graph, because it will make function undefined.

See the attached graph. Green line x = 0 is vertical Asymptote VA(y-axis)
Green dotted line (y = -x-1) is Linear Oblique Asymptote (LOA).