# Thread: find the asymptote and holes in the graph

1. ## find the asymptote and holes in the graph

This is my last item in my assignment and all I know is the vertical asymptote. Can you guys help me with this problem.

Given f(x) = x^3 +8 / x +2

Find the asymptotes and holes of the graph. Find the x and Y axis.

2. Originally Posted by cjru
This is my last item in my assignment and all I know is the vertical asymptote. Can you guys help me with this problem.

Given f(x) = x^3 +8 / x +2

Find the asymptotes and holes of the graph. Find the x and Y axis.

I assume that you mean:

$f(x) = \dfrac{x^3+8}{x+2}$

If so:
1. Use brackets
2. Re-write the equation of the function:

$f(x) = \dfrac{(x+2)(x^2-2x+4)}{x+2}$

Since you can cancel out the factor (x + 2) the graph of the function is a parabola with a hole at H(-2, 12).

3. How did you get (-2, 12)? So what is the indication a hole if the graph? Is it the cancelation of the numerator and denominator?

Thanks for the help.

4. Originally Posted by cjru
How did you get (-2, 12)? So what is the indication a hole if the graph? Is it the cancelation of the numerator and denominator?

Thanks for the help.
The rational function in its original state is not defined at x=-2 because that would cause the denominator to be 0.

After Earboth simplified the function to $f(x)=x^2-2x+4$, x=-2 is now ok. If it were still undefined at x=-2, you would have a vertical asymptote at x = -2. But that's not the case. Therefore, you have a hole in the curve where x=-2. The corresponding y coordinate is found by using your simplified function and seeing that f(-2) = 12. This function has 'point discontinuity' at (-2, 12)

5. Thank you very much. Now I inderstand why it became like that.