# How do you find the equation of the tangent and normal line to the curve y=x^3 at x=2?

Mar 22, 2018

Equation of tangent is $12 x - y - 16 = 0$ and that of normal is $x + 12 y - 98 = 0$

#### Explanation:

The slope of the tangent is given by value of first derivative at that point i.e. here at $x = 2$

As $y = {x}^{3}$, we have $\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2}$

and at $x = 2$, its value is $3 \cdot {2}^{2} = 12$

When we seek a tangent at $x = 2$, it means tangent at $\left(2 , {2}^{3}\right)$ i.e. at $\left(2 , 8\right)$

As tangent passes through $\left(2 , 8\right)$ and has a slope $12$, its equation is $y - 8 = 12 \left(x - 2\right)$ or $12 x - y - 16 = 0$

As normal is perpendicular to tangent, its slope is $\frac{- 1}{12}$

and hence equation of normal is $y - 8 = - \frac{1}{12} \left(x - 2\right)$

or $x + 12 y - 98 = 0$