\(f(-1) = -6 \text{,}\) dus \(A (-1 , -6) \text{.}\)

\(f(x) = 4 x^{3} + x^{2} - 3 x - 6\) geeft \(f'(x) = 12 x^{2} + 2 x - 3 \text{.}\)

Stel \(l{:}\,y = a x + b\) met \(a = f'(-1) = 7 \text{.}\)

\(\begin{rcases}y = 7 x + b \\ \text{door } A (-1 , -6)\end{rcases} \begin{matrix}7 ⋅ -1 + b = -6 \\ -7 + b = -6 \\ b = 1\end{matrix}\)
Dus \(l{:}\,y = 7 x + 1 \text{.}\)

\(f(5) = (-2 \cdot 5+11)^{3}+5^{2} = 26 \text{,}\) dus \(A (5 , 26)\)

\(f(x) = (-2 x+11)^{3}+x^{2}\) geeft
\(f'(x) = -6 (-2 x+11)^{2}+2 x\)

\(l{:}\,y = a x + b\) met \(a = f'(5) = -6 (-2 \cdot 5+11)^{2}+2 \cdot 5 = 4\)

\(\begin{rcases}y = 4 x + b \\ \text{door } A (5 , 26)\end{rcases} \begin{matrix}4 5 + b = 26 \\ 20 + b = 26 \\ b = 6\end{matrix}\)
Dus \(l{:}\,y = 4 x + 6 \text{.}\)