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July 28, 2014

July 28, 2014

Posted by **Ian** on Saturday, October 15, 2011 at 8:37pm.

f(x)= [(x-8)^2(x+3)(2x^2+3x-5)]/[(x+4)^2(2x+5)(x-3)(x^2-1)]

- calculus -
**Reiny**, Saturday, October 15, 2011 at 9:59pmyour expression factors to

[(-8)^2(x+3)(2x+5)(x-1)]/[(x+4)^2(2x+5)(x+1)(x-1)]

So the graph is discontinuous at

x = -4, -5/2, -1, 1

There will be asymptotes at

x=-4 and x=-1

and holes at x= -5/2 and x=1

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