Consider the graph when x and y are very very large. (x+y)(x²-xy+y²)=1 ⇒ x+y = 1/(x²-xy+y²) ≈ 0 ⇒ x≈-y
For the graph x^3+y^3=1, why is there an asymptote at y=-x?
Thanks.
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Consider the graph when x and y are very very large. (x+y)(x²-xy+y²)=1 ⇒ x+y = 1/(x²-xy+y²) ≈ 0 ⇒ x≈-y
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
When ,
Which has no solution.
HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]
_{1}∫^{(3√3)} t^{2} dt cos(3π/9) = log(^{3}√e) | Integral t^{2} dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.
making y the subject:
y = (-x^3+1)^1/3
factor out x^3
=(x^3)^1/3 * (-1+1/x^3)^1/3
=x (-1+1/x^3)^1/3
as x becomes larger; the part with 1/x^3 becomes smaller and the graph of y=x (-1 + 1/x^3 )^1/3 gets closer and closer to the graph y=x(-1+0)^1/3 or y=-x. of course there will be no solutions but that is just because it's an asymtote and those never have solutions (take for example y=1/x which has an asymtote at y=0; yet one cannot arrive at a solution; same thing with y=e^x)
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