Example 1. Write the equations of the tangent and normal to the curve $f(x)=x^{3}$ at the point $M_{0}(2,8)$.
Solution. First of all, the point $M_{0}$ lies on the curve, since its coordinates satisfy the given equation.

We find the derivative of the given function and its value at $x_{0}=2$:

$$
f^{\prime}(x)=\left(x^{3}\right)^{\prime}=3 x^{2}, f^{\prime}\left(x_{0}\right)=f^{\prime}(2)=3 \cdot 2^{2}=12
$$

Substituting the values $x_{0}=2, y_{0}=8, f^{\prime}\left(x_{0}\right)=12$ into equations (3.1) and (3.2), we get respectively:

$$
\begin{aligned}
& y-8=12(x-2) \text { or } 12 x-y-16=0-\text { the equation of the tangent; } \\
& y-8=-\frac{1}{12}(x-2) \text { or } x+12 y-98=0 \text { - the equation of the normal. }
\end{aligned}
$$