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I have been trying this for a while. But I couldn't get it. How can I determine the point of intersection of the tangent line at (0, 0) on the curve $y^2 + y = x^3 + x^2$ ?

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The tangent line at point $P(x_1,y_1)$ is:

$\frac{\partial f}{\partial x}(x-x_1)+\frac{\partial f}{\partial y}(y-y_1)$

Solving differentials, we obtain:

$\frac{\partial f}{\partial x}=3x^2+2x$

$\frac{\partial f}{\partial y}=-2y-1$

For $x_1=0$ and $y_1=0$ we have:

$\frac{\partial f}{\partial x}=3x^2+2x = 0$

$\frac{\partial f}{\partial y}=-2y-1 = -1$

So, by the first equation, the tangent line at $P(0,0)$ is $-y$.

Now, to calculate the intersection point with the curve we need to solve the next equations system:

$-y=0$

$y^2 + y = x^3 + x^2$

That have two solutions: $x=-1, y=0$ and $x=0, y=0$

So, the tangent line intersect the curve at $(-1,0)$ and $(0,0)$.

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    $\begingroup$ Welcome to crypto! Note: you can use $ signs inline as well to format your formulas etc. $\endgroup$ – Maarten Bodewes Jul 27 '14 at 22:29

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