\(f(-1) = -6 \text{,}\) dus \(A (-1 , -6) \text{.}\)

\(f(x) = x^{3} - 5 x^{2} - 6 x - 6\) geeft \(f'(x) = 3 x^{2} - 10 x - 6 \text{.}\)

Stel \(l{:}\,y = a x + b\) met \(a = f'(-1) = 7 \text{.}\)

\(\begin{rcases}y = 7 x + b \\ \text{door } A (-1 , -6)\end{rcases} \begin{matrix}7 ⋅ -1 + b = -6 \\ -7 + b = -6 \\ b = 1\end{matrix}\)
Dus \(l{:}\,y = 7 x + 1 \text{.}\)

\(f(3) = \frac{1}{-3+2}-5 \cdot 3^{3} = -136 \text{,}\) dus \(A (3 , -136)\)

\(f(x) = \frac{1}{-x+2}-5 x^{3}\) geeft
\(f'(x) = \frac{1}{(-x+2)^{2}}-15 x^{2}\)

\(l{:}\,y = a x + b\) met \(a = f'(3) = \frac{1}{(-3+2)^{2}}-15 \cdot 3^{2} = -134\)

\(\begin{rcases}y = -134 x + b \\ \text{door } A (3 , -136)\end{rcases} \begin{matrix}-134 3 + b = -136 \\ -402 + b = -136 \\ b = 266\end{matrix}\)
Dus \(l{:}\,y = -134 x + 266 \text{.}\)