Since (x − 2) cancelled out, x = 2 is the location of a hole.
Now simplify. And notice what was cancelled out.
The y-value of the hole can be found by evaluating the simplified function at x = 2.
x(x + 5) did not cancel out, so x = 0 and x = −5 are vertical asymptotes.
Now find the end behavior. You can long divide or use an approach statement: x → − ∞, y → 0 and as x → ∞, y → 0. There is a horizontal asymptote at y = 0.
Last but not least, investigate what is happening in the middle, by exploring the holes and vertical asymptotes. Use the simplified equation:
x → −5−, y → −∞
x → 0+, y → +∞
x → −5+, y → +∞
x → 0−, y → −∞
Use the eTool below to view the steps for graphing the first equation.
Click the link at right for the full version of the eTool: Calc 1-90 HW eTool