Sunday

February 1, 2015

February 1, 2015

Posted by **zama** on Sunday, November 8, 2009 at 7:48am.

- calculus -
**Reiny**, Sunday, November 8, 2009 at 8:14amf'(x) = 2 - 8/x^2

setting this equal to zero yields x = ±2

where y is then ±8

I will describe the function:

Vertical asymptote at x = 0

Minimum point at (2,8)

Maximum point at (-2,-8)

Oblique asymptote y = 2x

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