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

y = x

^{3}

^{ }+ 2x + 6

Slope of tangent = m

_{1}

_{ }= $\frac{d\mathrm{y}}{d\mathrm{x}}$ = 3x

^{2 }+ 2

Slope of perpendicular line = m

_{2 }= $\frac{-1}{14}$

m

_{1}

_{ . }m

_{2}

_{ }= -1

m

_{1 }= 14

3x

^{2 }+ 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 = 2

^{3 }+ 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

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