The slope of the normal to the curve $y=x^3+3 x-2$ at its point of intersection with y-axis is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $-\frac{1}{3}$
$\frac{dy}{dx}=3x^2+3$
when it intersects y axis ⇒ x = 0
⇒ tangent = $\frac{dy}{dx}=3$ so normal
$=-\frac{dx}{dy}=-\frac{1}{3}$