What is the tangent of 5x^3-x^2y+y^3+1=0,(1,-2) ?

The tangent of 5x^3-x^2y+y^3+1=0,(1,-2) is y=-19/11 x-3/11

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tangent of 5x^3-x^2y+y^3+1=0,(1,-2)

tangent of

5 x^{3} - x^{2} y + y^{3} + 1 = 0, (1, - 2)

y = - \frac{19}{11} x - \frac{3}{11}

Solution steps

Find the slope of

5 x^{3} - x^{2} y + y^{3} + 1 = 0 : \frac{d y}{d x} = \frac{- 15 x ^{2} + 2 x y}{3 y ^{2} - x ^{2}}

Compute the slope of

5 x^{3} - x^{2} y + y^{3} + 1 = 0

(1, - 2) : m = - \frac{19}{11}

Find the line with slope m=

- \frac{19}{11}

and passing through

(1, - 2) : y = - \frac{19}{11} x - \frac{3}{11}

y = - \frac{19}{11} x - \frac{3}{11}

y = t^{3} - 4 t - 3, (2, - 3)

x \to 2 + lim (\frac{3}{( x - 2 ) ^{3}})

\int (y e^{y}) d y

\int_{0}^{5} π (2 5 y)^{2} d y

\int \frac{e ^{x}}{36 - e ^{2 x}} d x

What is the tangent of 5x^3-x^2y+y^3+1=0,(1,-2) ?
The tangent of 5x^3-x^2y+y^3+1=0,(1,-2) is y=-19/11 x-3/11