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April 23, 2014

April 23, 2014

Posted by **Dave** on Sunday, June 6, 2010 at 8:58pm.

f(x)= 3x^2-5x+2/6x^2-5x+1

f(x)= 2x+3/(x^2-2x+3)

- Math:use parentheses -
**MathMate**, Sunday, June 6, 2010 at 10:04pmVertical asymptotes in rational functions usually come from the denominator approaching zero.

In

f(x)= (3x^2-5x+2)/(6x^2-5x+1)

the denominator can be factorized as:

(3x-1)(2x-1)

at x=1/3 or x=1/2, the denominator becomes zero, and hence a vertical asymptote. You also need to check if the numerator vanishes at the same points, in which case the point will be undefined.

To find the horizontal asymptotes, do a long division, and the resulting leading constant (1/2) is the horizontal asymptote. Take care to determine from which side of y=1/2 the curve approaches the asymptote as x approaches ±∞.

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