Asked by Dave
How do you find the vertical and horizontal assymptotes of the graph of the rational function?
f(x)= 3x^2-5x+2/6x^2-5x+1
f(x)= 2x+3/(ãx^2-2x+3)
f(x)= 3x^2-5x+2/6x^2-5x+1
f(x)= 2x+3/(ãx^2-2x+3)
Answers
Answered by
MathMate
Vertical 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 ±∞.
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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