\(f(-3) = 2 \text{,}\) dus \(A (-3 , 2) \text{.}\)

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

Stel \(l{:}\,y = a x + b\) met \(a = f'(-3) = 1 \text{.}\)

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

\(f(-2) = 4 \cdot (-2)^{3}+\sqrt{4 \cdot -2+12} = -30 \text{,}\) dus \(A (-2 , -30)\)

\(f(x) = 4 x^{3}+\sqrt{4 x+12}\) geeft
\(f'(x) = 12 x^{2}+\frac{2}{\sqrt{4 x+12}}\)

\(l{:}\,y = a x + b\) met \(a = f'(-2) = 12 \cdot (-2)^{2}+\frac{2}{\sqrt{4 \cdot -2+12}} = 49\)

\(\begin{rcases}y = 49 x + b \\ \text{door } A (-2 , -30)\end{rcases} \begin{matrix}49 ⋅ -2 + b = -30 \\ -98 + b = -30 \\ b = 68\end{matrix}\)
Dus \(l{:}\,y = 49 x + 68 \text{.}\)