What is the tangent of y^3+xy^2-5=x+3y^2,\at x=3 ?

The tangent of y^3+xy^2-5=x+3y^2,\at x=3 is y=-1/4 x+11/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of y^3+xy^2-5=x+3y^2,\at x=3

tangent of

y^{3} + x y^{2} - 5 = x + 3 y^{2}, at x = 3

y = - \frac{1}{4} x + \frac{11}{4}

Solution steps

Find the tangent point:

(3, 2)

Find the slope of

y^{3} + x y^{2} - 5 = x + 3 y^{2} : \frac{d y}{d x} = \frac{1 - y ^{2}}{y ( - 6 + 2 x + 3 y )}

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{1}{4}

Find the line with slope m=

- \frac{1}{4}

and passing through

(3, 2) : y = - \frac{1}{4} x + \frac{11}{4}

y = - \frac{1}{4} x + \frac{11}{4}

What is the tangent of y^3+xy^2-5=x+3y^2,\at x=3 ?
The tangent of y^3+xy^2-5=x+3y^2,\at x=3 is y=-1/4 x+11/4