\(f(1) = 2 \text{,}\) dus \(A (1 , 2) \text{.}\)

\(f(x) = -4 x^{3} + 5 x^{2} + 3 x - 2\) geeft \(f'(x) = -12 x^{2} + 10 x + 3 \text{.}\)

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

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

\(f(5) = -4+\sqrt{5 \cdot 5-9} = 0 \text{,}\) dus \(A (5 , 0)\)

\(f(x) = -4+\sqrt{5 x-9}\) geeft
\(f'(x) = \frac{5}{2 \sqrt{5 x-9}}\)

\(l{:}\,y = a x + b\) met \(a = f'(5) = \frac{5}{2 \sqrt{5 \cdot 5-9}} = {5 \over 8}\)

\(\begin{rcases}y = {5 \over 8} x + b \\ \text{door } A (5 , 0)\end{rcases} \begin{matrix}{5 \over 8} 5 + b = 0 \\ {25 \over 8} + b = 0 \\ b = {-25 \over 8}\end{matrix}\)
Dus \(l{:}\,y = {5 \over 8} x + {-25 \over 8} \text{.}\)