Find the equations of tangents to the curve y=x^3+2x+6 which is perpendicular to the line x+14y+4=0

Dear Student,

y = x³ + 2x + 6

Slope of tangent = m₁ = $\frac{d\mathrm{y}}{d\mathrm{x}}$ = 3x^​2 + 2 
Slope of perpendicular line = m₂ = $\frac{-1}{14}$
m_1 . m₂  = -1
m₁ = 14

3x² + 2 = 14
x = $\pm$2
Therefore the curve has tangents at x = 2 and x = -2
and these points also lie on the given curve

Equation of tangent - y = 14x + c 
Coordinates of points of tangency
At x = 2 , y = 2³ + 2(2) + 6 = 18 
At x = -2 , y = (-2)³ + 2(-2) + 6 = -6

18 = 14(2) + c
c = -10 

-6 = 14(-2) + c
c = 22

Equation of tangent - y = 14x - 10 and y = 14x + 22

Regards