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March 6, 2015

March 6, 2015

Posted by **Anonymous** on Sunday, March 24, 2013 at 8:22pm.

- college algebra -
**Reiny**, Sunday, March 24, 2013 at 8:32pmb(x) = x^3/(5x^3 - x^2 - 22x)

= x^2/(5x^2 - x - 22) , after dividing by x , x≠0

f'(x) = (2x(5x^2 - x - 22) - x^2(10x - 1) )/(5x^2-x-22)^2

to have a vertical tangent, the slope must be undefined, that is

the denominator of the above f'(x) must be zero

5x^2 - x - 22 = 0

(5x-11)(x+2) = 0

two vertical asymptotes:

x = 11/5 and x = -2

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