Wiskunde

\(\textbf{Waardentabel} \\ \begin{array}{l|c|c|c|c|}x & -2 & \dfrac{-3}{2} & -1 & 0 \\ \hline f(x) & 0 & \dfrac{1}{2} & 0 & -4\end{array} \\\\ \underline{\text{Nulwaarden}} \\\Delta = b^2-4ac = 36-32 = 4 > 0 \\x_1 = \dfrac{-b-\sqrt{\Delta}}{2a} = -1\quad x_2 = \dfrac{-b+\sqrt{\Delta}}{2a} = -2 \\\\ \underline{\text{Snijpunt Y-as}} \\ (0;-4) \\ \\\underline{\text{Symmetrie-as en top}} \\x = -\dfrac{b}{2a} = -\dfrac{-6}{2 \cdot -2} = \dfrac{-3}{2} \\f\!\left(\dfrac{-3}{2}\right) = \dfrac{1}{2} \\\text{Top: } T\!\left(\dfrac{-3}{2} ;\ \dfrac{1}{2}\right) \\ \\\textbf{Tekenschema} \\\begin{array}{r|ccccccc} x & -\infty & & -1 & & -2 & & +\infty \\ \hline f(x) & - & - & 0 & + & 0 & - & -\\ \end{array}\\\textbf{Domein en bereik} \\ \text{dom} f = \mathbb{R} \text{ en ber} f = ]-\infty,\ \dfrac{1}{2}]\\ \\\textbf{Verloopschema} \\ \begin{array}{r|ccccc} x & -\infty & & \dfrac{-3}{2} & & +\infty \\ \hline f(x) & -\infty & \nearrow & \dfrac{1}{2} & \searrow & -\infty\\ \end{array}\\ \\\textbf{Voor leerlingen derde graad} \\ \displaystyle{ \lim_{x \to -\infty} } f(x) = -\infty \\ \displaystyle{ \lim_{x \to +\infty} } f(x) = -\infty\)

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