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

The normal of y=x^3+1,(1,2) is y=-1/3 x+7/3

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normal of y=x^3+1,(1,2)

normal of

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

y = - \frac{1}{3} x + \frac{7}{3}

Solution steps

Find the slope of

y = x^{3} + 1 : \frac{d y}{d x} = 3 x^{2}

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 = 3

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{3}

Find the line with slope m=

- \frac{1}{3}

and passing through

(1, 2) : y = - \frac{1}{3} x + \frac{7}{3}

y = - \frac{1}{3} x + \frac{7}{3}

f (x) = \frac{1}{x ^{4}}

\frac{d y}{d x} = y (2 x^{2} + 6)

e^{3 x^{3}}

\int \frac{1}{( 4 - x ^{2} ) ^{2}} d x

\frac{\partial}{\partial x} (5 ln (1 + x^{2} - y))

What is the normal of y=x^3+1,(1,2) ?
The normal of y=x^3+1,(1,2) is y=-1/3 x+7/3