Thread: 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.

Thanks in advance.

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.

Thanks in advance.

I assume that you mean:



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



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

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.

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 , 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)

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