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.
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:
$\displaystyle f(x) = \dfrac{x^3+8}{x+2}$
If so:
1. Use brackets
2. Re-write the equation of the function:
$\displaystyle f(x) = \dfrac{(x+2)(x^2-2x+4)}{x+2}$
Since you can cancel out the factor (x + 2) the graph of the function is a parabola with a hole at H(-2, 12).
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 $\displaystyle f(x)=x^2-2x+4$, 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)