\(f(2) = 5 \text{,}\) dus \(A (2 , 5) \text{.}\)

\(f(x) = -2 x^{3} + 6 x^{2} + x - 5\) geeft \(f'(x) = -6 x^{2} + 12 x + 1 \text{.}\)

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

\(\begin{rcases}y = x + b \\ \text{door } A (2 , 5)\end{rcases} \begin{matrix}1 ⋅ 2 + b = 5 \\ 2 + b = 5 \\ b = 3\end{matrix}\)
Dus \(l{:}\,y = x + 3 \text{.}\)

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

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

\(l{:}\,y = a x + b\) met \(a = f'(2) = -3+3 \cdot 2^{2}+6 \cdot 2 = 21\)

\(\begin{rcases}y = 21 x + b \\ \text{door } A (2 , -4)\end{rcases} \begin{matrix}21 2 + b = -4 \\ 42 + b = -4 \\ b = -46\end{matrix}\)
Dus \(l{:}\,y = 21 x + -46 \text{.}\)