Question

find the asymptotes of the curve x^3+2x^y-xy^2-2y^3+3xy+3y^2+x+1=0​

Answer 1

Answer:

Here the line y = 0 is the asymptote parallel to X-axis whereas there is no asymptote parallel to Y-axis.  For Oblique Asymptotes:  In the given equation of curve, expression containing the third degree terms is  y3 + x2y + 2xy2 Thus, φ3(m) = m3 + 2m2 + m (by taking y = m, x = 1)  so that φ'3(m) = 3m2 + 4m + 1 and φ"3(m) = 6m + 4  Likewise, φ2(m) = 0, φ1(m) = –m   φ3 = 0 ⇒ m3 + 2m2 + m = 0 or m = –1, –1, 0 Now for equal values of m in φn(m), corresponding values of ‘c’ are obtainRead more on Sarthaks.com - https://www.sarthaks.com/495998/find-the-asymptotes-of-the-curve-y-3-x-2y-2xy-2-y-1-0

Answer 2

$x^3+2x^y-xy^2-2y^3+3xy+3y^2+x+1 = 0$ does not have any vertical or horizontal asymptotes.

• It may appear contrary to reality that a curve can approach a line arbitrarily closely without truly matching it. A line's and a curve's representations as marks on paper or as pixels on a computer screen have a positive width. So, if they were to be stretched out far enough, at least as far as the eye could see, they would appear to merge.
• However, the line and the curve are idealised conceptions whose breadth is; these are the physical representations of the corresponding mathematical entities. As a result, reasoning rather than experience is needed to comprehend the concept of an asymptote.

Here, the curve is given as,

$x^3+2x^y-xy^2-2y^3+3xy+3y^2+x+1 = 0$

Here, there are no vertical or horizontal asymptotes.

Hence, $x^3+2x^y-xy^2-2y^3+3xy+3y^2+x+1 = 0$ does not have any vertical or horizontal asymptotes.

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