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October 1, 2014

October 1, 2014

Posted by **Sydney** on Saturday, July 25, 2009 at 8:59pm.

- College Algebra -
**MathMate**, Saturday, July 25, 2009 at 11:10pm(6x²-6)

=6(x²-1)

=6(x+1)(x-1)

If the above expression is the denominator of a given function, there will be two vertical asymptotes, namely at the zeroes of the given expression, at x=1 and x=-1.

The expression equals zero at these two places since 6*0=0.

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