How do you find the slant asymptote of $y = \frac{x^{2}}{2 x^{2} - 8}$ $y=(x^2)/(2x^2-8)$ ?

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How do you find the slant asymptote of $y = \frac{x^{2}}{2 x^{2} - 8}$ $y=(x^2)/(2x^2-8)$ ?

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Answer 1 · 2018-04-12T03:01:43

No slant asymptote exists, but there is the horizontal asymptote $y = \frac{1}{2} .$ $y=1/2.$

Explanation:

The numerator and denominator are of the same degree (degree $2$ $2$ ) -- no slant asymptote exists in such cases.

Rather, there is a horizontal asymptote given by dividing the coefficients of the leading terms (terms of highest degree), in this case, $y = \frac{1}{2} .$ $y=1/2.$

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How do you find the slant asymptote of $y = \frac{x^{2}}{2 x^{2} - 8}$ $y=(x^2)/(2x^2-8)$ ?

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Explanation:

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How do you find the slant asymptote of $y = \frac{x^{2}}{2 x^{2} - 8}$ $y=(x^2)/(2x^2-8)$ ?