1. ## 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

2. 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.

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

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

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

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

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).

Your Teacher is Right, because, the curve has Vertical Asymptote x = 0. It also has one Linear Oblique Asymptote, y = - x - 1

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

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

So, Linear Oblique Asymptote,

$\displaystyle y = -x-1$

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).