\(f(2) = 7 \text{,}\) dus \(A (2 , 7) \text{.}\)

\(f(x) = -x^{3} + 2 x^{2} + 5 x - 3\) geeft \(f'(x) = -3 x^{2} + 4 x + 5 \text{.}\)

Stel \(l{:}\,y = a x + b\) met \(a = f'(2) = 1 \text{.}\)

\(\begin{rcases}y = x + b \\ \text{door } A (2 , 7)\end{rcases} \begin{matrix}1 ⋅ 2 + b = 7 \\ 2 + b = 7 \\ b = 5\end{matrix}\)
Dus \(l{:}\,y = x + 5 \text{.}\)

\(f(-1) = -2 \cdot -1^{3}+{4 \over 1} = 6 \text{,}\) dus \(A (-1 , 6)\)

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

\(l{:}\,y = a x + b\) met \(a = f'(-1) = -6 \cdot -1^{2}+\frac{4}{-1^{2}} = -2\)

\(\begin{rcases}y = -2 x + b \\ \text{door } A (-1 , 6)\end{rcases} \begin{matrix}-2 -1 + b = 6 \\ 2 + b = 6 \\ b = 4\end{matrix}\)
Dus \(l{:}\,y = -2 x + 4 \text{.}\)