What is the tangent of x^2+4x-11=-2y^2-3y,(2,-1) ?

The tangent of x^2+4x-11=-2y^2-3y,(2,-1) is y=8x-17

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tangent of x^2+4x-11=-2y^2-3y,(2,-1)

tangent of

x^{2} + 4 x - 11 = - 2 y^{2} - 3 y, (2, - 1)

y = 8 x - 17

Solution steps

Find the slope of

x^{2} + 4 x - 11 = - 2 y^{2} - 3 y : \frac{d y}{d x} = - \frac{2 x + 4}{3 + 4 y}

Compute the slope of

x^{2} + 4 x - 11 = - 2 y^{2} - 3 y

(2, - 1) : m = 8

Find the line with slope m=

8

and passing through

(2, - 1) : y = 8 x - 17

y = 8 x - 17

\int (3 - x)^{- \frac{1}{2}} d x

\frac{d}{d x} (x^{2} + x + 4)

\frac{d h}{d t} = 4 - 16 t^{2}, h (1) = 0

\frac{d}{d x} (ln (x^{2} - 16 x))

\int \frac{x ^{2} + 3 x - 4}{x ^{2} - 2 x - 8} d x

What is the tangent of x^2+4x-11=-2y^2-3y,(2,-1) ?
The tangent of x^2+4x-11=-2y^2-3y,(2,-1) is y=8x-17