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How do you find vertical, horizontal and oblique asymptotes for # f(x)= (x^3 - 5x^2 + 6x) / (x^2 + x -6)#?

Answer 1

Christian Antunez

First, notice that

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

Using long division, we find that

#f(x) = x-6 + 18/(x+3)#,

from which we deduce that the vertical asymptote is #x=-3# and the oblique asymptote is #y=x-6#.

The oblique asymptote is found by letting #x# get arbitrarily larger i,e, as #x -> infty#, #18/(x+3)->0#, thus #y -> x-6#.

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Answer 2

Chloe Alexander

To find the vertical asymptotes of (f(x) = \frac{x^3 - 5x^2 + 6x}{x^2 + x - 6}), identify the values of (x) that make the denominator equal to zero.

To find horizontal asymptotes, analyze the behavior of the function as (x) approaches positive and negative infinity.

To find oblique asymptotes, perform polynomial long division between the numerator and denominator.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.