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July 30, 2014

Posted by **Dan** on Sunday, December 19, 2010 at 2:34pm.

vertical asymptotes.

- Math -
**MathMate**, Sunday, December 19, 2010 at 8:13pmStationary points are points at which f'(x)=0 AND f"(x)=0.

Vertical asymptote is where the denominator becomes zero.

Horizontal asymptote is the limit when x->∞ or x->-∞, if the limit is defined.

- Math -
**Dan**, Sunday, December 19, 2010 at 9:00pmcould you please solve the problem completely .. i am still confused.

Thanks.

- Math -
**MathMate**, Tuesday, December 21, 2010 at 10:47amDenominator becomes zero when x=3, therefore vertical asymptote is a x=3.

The limit of f(x)=[(x-5)/(x-3)]^2 as x->±∞ is 1, so the horizontal asymptote is at y=1.

Since f'(5)=0 and f"(6)=0, there are no stationary points, just a minimum at x=5.

There is a point of inflection at x=6 (where f"(6)=0).

You will need to take the time to calculate f'(x) and f"(x) from f(x), and hence sketch the graph of f(x).

A sketch of the graph can be seen here:

http://img718.imageshack.us/img718/4296/1292787276.png

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