\(f(2) = 5 \text{,}\) dus \(A (2 , 5) \text{.}\)

\(f(x) = -x^{3} + 4 x^{2} - 3 x + 3\) geeft \(f'(x) = -3 x^{2} + 8 x - 3 \text{.}\)

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

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

(\(f (x) = 0\) geeft snijpunten \(x \text{-}\)as)
\({x^{2} + x - 12 \over 6 x - 1} = 0\)
\(x^{2} + x - 12 = 0\)

(som-productmethode)
\((x + 4) (x - 3) = 0\)
\(x = -4 ∨ x = 3\)
Dus \(B (3 , 0)\)

\(f'(x) = \frac{(6 x-1) (2 x+1)-(x^{2}+x-12) \cdot 6}{(6 x-1)^{2}}\)

\(l{:}\,y = a x + b\) met \(a = f'(3) = \frac{7}{17}\)

\(\begin{rcases}y = \frac{7}{17} x + b \\ \text{door } B (3 , 0)\end{rcases} \begin{matrix}\frac{7}{17} ⋅ 3 + b = 0 \\ b = -\frac{21}{17}\end{matrix}\)
Dus \(l{:}\,y = \frac{7}{17} x + -1\frac{4}{17} \text{.}\)

\(f(3) = 5 \cdot 3^{2}+\sqrt{3+1} = 47 \text{,}\) dus \(A (3 , 47)\)

\(f(x) = 5 x^{2}+\sqrt{x+1}\) geeft
\(f'(x) = 10 x+\frac{1}{2 \sqrt{x+1}}\)

\(l{:}\,y = a x + b\) met \(a = f'(3) = 10 \cdot 3+\frac{1}{2 \sqrt{3+1}} = 30\frac{2}{8}\)

\(\begin{rcases}y = 30\frac{2}{8} x + b \\ \text{door } A (3 , 47)\end{rcases} \begin{matrix}30\frac{2}{8} 3 + b = 47 \\ 90\frac{3}{4} + b = 47 \\ b = -43\frac{3}{4}\end{matrix}\)
Dus \(l{:}\,y = 30\frac{2}{8} x + -43\frac{3}{4} \text{.}\)