Equation of the tangent to the curve   ${\text{y = x}}^{\text{3}}{\text{ + 3x}}^{\text{2}}\text{ - 5 }$ and which is perpendicular to the line 2x – 6y + 1= 0 is

${\text{y\hspace{0.17em}=\hspace{0.17em}x}}^{\text{3}}{\text{\hspace{0.17em}+\hspace{0.17em}3x}}^{\text{2}}\text{\hspace{0.17em}-\hspace{0.17em}5}$

$\frac{\text{dy}}{\text{dx}}{\text{\hspace{0.17em}=\hspace{0.17em}3x}}^{\text{2}}{\text{\hspace{0.17em}+\hspace{0.17em}6x\hspace{0.17em}=\hspace{0.17em}m}}_{\text{1}}$

Slope of given line ${\text{m}}_2=\frac{1}{3}$

m₁ m₂ = -1 

$\left({\text{3x}}^{\text{2}}\text{\hspace{0.17em}+\hspace{0.17em}6x}\right)\frac{\text{1}}{\text{3}}\text{=\hspace{0.17em}-\hspace{0.17em}1}$

$\Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}}^{\text{2}}\text{\hspace{0.17em}+\hspace{0.17em}2x\hspace{0.17em}+\hspace{0.17em}1\hspace{0.17em}=\hspace{0.17em}0}$

$\Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x\hspace{0.17em}=\hspace{0.17em}-1,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y\hspace{0.17em}=\hspace{0.17em}-1\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}3-5\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}-3}$

P = (-1, -3)

Equation of tangent at P(-1, -3) is

y + 3 = -3 (x + 1)

3x + y + 6 = 0