What is the tangent of (2y+1)^3-24x=-3,(-1,-2) ?

The tangent of (2y+1)^3-24x=-3,(-1,-2) is y= 4/(9(-1)^{2/3)}x-14/9

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tangent of (2y+1)^3-24x=-3,(-1,-2)

tangent of

(2 y + 1)^{3} - 24 x = - 3, (- 1, - 2)

y = \frac{4}{9 ( - 1 ) ^{\frac{2}{3}}} x - \frac{14}{9}

Solution steps

Find the slope of

(2 y + 1)^{3} - 24 x = - 3 : \frac{d y}{d x} = \frac{4}{( 24 x - 3 ) ^{\frac{2}{3}}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{4}{9 ( - 1 ) ^{\frac{2}{3}}}

Find the line with slope m=

\frac{4}{9 ( - 1 ) ^{\frac{2}{3}}}

and passing through

(- 1, - 2) : y = \frac{4}{9 ( - 1 ) ^{\frac{2}{3}}} x - \frac{14}{9}

y = \frac{4}{9 ( - 1 ) ^{\frac{2}{3}}} x - \frac{14}{9}

\frac{\partial}{\partial x} (5 (x - 3)^{2})

\frac{d y}{d x} + 2 x y^{4} = 0

\frac{d}{d x} (\frac{1}{6 + x})

(1 + x^{2}) d y + (1 + y^{2}) d x = 0

\int (ln (x) - ln (x y)) d x

What is the tangent of (2y+1)^3-24x=-3,(-1,-2) ?
The tangent of (2y+1)^3-24x=-3,(-1,-2) is y= 4/(9(-1)^{2/3)}x-14/9