# 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 = m$\frac{-1}{14}$
m1 . m2  = -1
m= 14

3x+ 2 = 14
x = $±$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 = 2+ 2(2) + 6 = 18
At x = -2 , y = (-2)+ 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

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