Question

Find the equation of tangent to the curve y=x^3+3x^2-5 which is perpendicular to the line 2x-6y+1=0​

Answer 1

Answer:

y+3x+6=0

Step-by-step explanation:

y=x^3+3x^2-5

dy/dx=3x^2+6x

Given tangent perpendicular to 2x+1=6y

Slope of the above line = 1/3

required slope =m

mx(1/3)=-1

m=-3

dy/dx=-3

3x^2+6x+3=0

x^2+2x+1=0

(x+1)^2=0

x=-1

y=x^3+3x^2-5

when x=-1

y=-3

point of tangent=(-1,-3)

Required equation

(y+3)=-3(x+1)

y+3x+6=0