\(f(2) = 8 \text{,}\) dus \(A (2 , 8) \text{.}\)

\(f(x) = -x^{3} + 2 x^{2} + 6 x - 4\) geeft \(f'(x) = -3 x^{2} + 4 x + 6 \text{.}\)

Stel \(l{:}\,y = a x + b\) met \(a = f'(2) = 2 \text{.}\)

\(\begin{rcases}y = 2 x + b \\ \text{door } A (2 , 8)\end{rcases} \begin{matrix}2 ⋅ 2 + b = 8 \\ 4 + b = 8 \\ b = 4\end{matrix}\)
Dus \(l{:}\,y = 2 x + 4 \text{.}\)

\(f(-2) = \frac{4}{-2^{2}}+2 \cdot -2^{3} = -15 \text{,}\) dus \(A (-2 , -15)\)

\(f(x) = \frac{4}{x^{2}}+2 x^{3}\) geeft
\(f'(x) = \frac{-8}{x^{3}}+6 x^{2}\)

\(l{:}\,y = a x + b\) met \(a = f'(-2) = \frac{-8}{-2^{3}}+6 \cdot -2^{2} = 25\)

\(\begin{rcases}y = 25 x + b \\ \text{door } A (-2 , -15)\end{rcases} \begin{matrix}25 -2 + b = -15 \\ -50 + b = -15 \\ b = 35\end{matrix}\)
Dus \(l{:}\,y = 25 x + 35 \text{.}\)