\(f(2) = 4 \text{,}\) dus \(A (2 , 4) \text{.}\)

\(f(x) = -x^{3} + 2 x^{2} + 5 x - 6\) geeft \(f'(x) = -3 x^{2} + 4 x + 5 \text{.}\)

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

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

(\(f (x) = 0\) geeft snijpunten \(x \text{-}\)as)
\({x^{2} + 11 x + 30 \over 3 x + 1} = 0\)
\(x^{2} + 11 x + 30 = 0\)

(som-productmethode)
\((x + 6) (x + 5) = 0\)
\(x = -6 ∨ x = -5\)
Dus \(A (-6 , 0)\)

\(f'(x) = \frac{(3 x+1) (2 x+11)-(x^{2}+11 x+30) \cdot 3}{(3 x+1)^{2}}\)

\(l{:}\,y = a x + b\) met \(a = f'(-6) = \frac{1}{17}\)

\(\begin{rcases}y = \frac{1}{17} x + b \\ \text{door } A (-6 , 0)\end{rcases} \begin{matrix}\frac{1}{17} ⋅ -6 + b = 0 \\ b = \frac{6}{17}\end{matrix}\)
Dus \(l{:}\,y = \frac{1}{17} x + \frac{6}{17} \text{.}\)

\(f(-3) = (5 \cdot -3+14)^{4}-5 \cdot (-3)^{3} = 136 \text{,}\) dus \(A (-3 , 136)\)

\(f(x) = (5 x+14)^{4}-5 x^{3}\) geeft
\(f'(x) = 20 (5 x+14)^{3}-15 x^{2}\)

\(l{:}\,y = a x + b\) met \(a = f'(-3) = 20 (5 \cdot -3+14)^{3}-15 \cdot (-3)^{2} = -155\)

\(\begin{rcases}y = -155 x + b \\ \text{door } A (-3 , 136)\end{rcases} \begin{matrix}-155 ⋅ -3 + b = 136 \\ 465 + b = 136 \\ b = -329\end{matrix}\)
Dus \(l{:}\,y = -155 x - 329 \text{.}\)