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

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

Stel \(l{:}\,y = a x + b\) met \(a = f'(2) = 3 \text{.}\)

\(\begin{rcases}y = 3 x + b \\ \text{door } A (2 , 8)\end{rcases} \begin{matrix}3 ⋅ 2 + b = 8 \\ 6 + b = 8 \\ b = 2\end{matrix}\)
Dus \(l{:}\,y = 3 x + 2 \text{.}\)

\(f(5) = 2+\sqrt{-4 \cdot 5+29} = 5 \text{,}\) dus \(A (5 , 5)\)

\(f(x) = 2+\sqrt{-4 x+29}\) geeft
\(f'(x) = \frac{-2}{\sqrt{-4 x+29}}\)

\(l{:}\,y = a x + b\) met \(a = f'(5) = \frac{-2}{\sqrt{-4 \cdot 5+29}} = {-2 \over 3}\)

\(\begin{rcases}y = {-2 \over 3} x + b \\ \text{door } A (5 , 5)\end{rcases} \begin{matrix}{-2 \over 3} 5 + b = 5 \\ {-10 \over 3} + b = 5 \\ b = {25 \over 3}\end{matrix}\)
Dus \(l{:}\,y = {-2 \over 3} x + {25 \over 3} \text{.}\)