I know it's simple, also I can create the graph. But how can I prove it logically or properly?
$x-y+2=0$ is a tangent line to the curve $(x+y)^3 = (x-y+2)^2$ at $(-1,1)$
$(x+y)^3 = (x-y+2)^2$ find tangent line on $(-1,1)$
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calculus
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0Do you know where to start? – 2017-02-03
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0Equation of a tangent line y-y1 = M(x-x1) where M is slope also need to find f'(x) ... is it right way to think? – 2017-02-03
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1you are pretty much there - you want to find $M$ how do we find that with $f'(x)$? – 2017-02-03
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0Did I made it right way below? – 2017-02-03
1 Answers
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As this is an affine algebraic curve, the tangents at the origin are given by the lowest total degree part of the equation.
At another point, you change origin and apply the previous rule. So set $X=x+1$, $Y=y-1$. You obtain $$(x+y)^3 -(x-y+2)^2=0\iff (X+Y)^3-(X-Y)^2=0.$$ Thus the tangents at $X=0, Y=0$ have equation $(X-Y)^2$, i.e. the curve has a double tangent at $x=-1, y=1$: $$(x-y+2)^2=0.$$
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0You impress me with - `affine algebraic curve` (+1), but I am not sure the OP would be as comfortable. – 2017-02-03
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0Can you explain the (X+Y)3−(X−Y)2=0 bit – 2017-02-03
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0There's nothing else than replacing $x$ with $X-1$ and $y$ with $Y+1$ in the equation. – 2017-02-03
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0Was thinking too much... Also found this way: $f'(x) = f'(-1) = 1$ Which mean slope=1. And $y-y1 = M(x-x1) <=> x-y+2 = 0$ – 2017-02-03
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0Yes. That's the differential geometry way. – 2017-02-03