$x (x + 1) = 12$
$x^{2} + x - 12 = 0$
$(x - 3) (x + 4) = 0$

$x = 3$	of	$x = ‐ 4$
$y = 3 + 1 = 4$		$y = ‐ 4 + 1 = ‐ 3$
$(3,4)$		$(‐ 4, ‐ 3)$

Zie vraag a.

$2 x - 3 y + 6 = 0$
$2 x + 6 = 3 y$
$\frac{2}{3} x + 2 = y$

$x (\frac{2}{3} x + 2) = 12$
$\frac{2}{3} x^{2} + 2 x - 12 = 0$
$x^{2} + 3 x - 18 = 0$
$(x - 3) (x + 6) = 0$

$x = 3$	of	$x = ‐ 6$
$y = \frac{12}{3} = 4$		$y = \frac{12}{‐ 6} = ‐ 2$
$(3,4)$		$(‐ 6, ‐ 2)$

$x (‐ x + k) = 12$
$‐ x^{2} + k x - 12 = 0$
$x^{2} - k x + 12 = 0$
$\begin{array}{l} a = 1 \\ b = ‐ k \\ c = 12 \end{array}} D = {(‐ k)}^{2} - 4 \cdot 1 \cdot 12 = k^{2} - 48$
$k^{2} - 48 = 0$
$k^{2} = 48$
$k = \sqrt{48} = 4 \sqrt{3}$ of $k = ‐ \sqrt{48} = ‐ 4 \sqrt{3}$

Vergelijkingen van de raaklijnen:
$y = ‐ x + 4 \sqrt{3}$ en $y = ‐ x - 4 \sqrt{3}$

$x^{2} + y^{2} + 2 y = 8$
$x^{2} + {(y + 1)}^{2} - 1 = 8$
$x^{2} + {(y + 1)}^{2} = 9$

Zie vraag b.

$x - y + 2 = 0 \Rightarrow x + 2 = y$
$x + 2 = y$ en $x^{2} + {(y + 1)}^{2} = 9$

$x^{2} + {(x + 2 + 1)}^{2} = 9$
$x^{2} + {(x + 3)}^{2} = 9$
$2 x^{2} + 6 x = 0$
$2 x (x + 3) = 0$

$x = 0$	of	$x = ‐ 3$
$y = 0 + 2 = 2$		$y = ‐ 3 + 2 = ‐ 1$
$(0,2)$		$(‐ 3, ‐ 1)$

$x + 2 = y$ en $y = x^{2}$

$x + 2 = x^{2}$
$x^{2} - x - 2 = 0$
$(x - 2) (x + 1) = 0$

$x = 2$	of	$x = ‐ 1$
$y = 2 + 2 = 4$		$y = ‐ 1 + 2 = 1$
$(2,4)$		$(‐ 1,1)$

$y = x^{2}$ en $x^{2} + {(y + 1)}^{2} = 9$

$y + {(y + 1)}^{2} = 9$
$y^{2} + 3 y - 8 = 0$
$\begin{array}{l} a = 1 \\ b = 3 \\ c = ‐ 8 \end{array}} \begin{matrix} D = 9 - 4 \cdot 1 \cdot ‐ 8 = 41 \\ \sqrt{D} = \sqrt{41} \end{matrix}$
$y = \frac{‐ 3 + \sqrt{41}}{2} = ‐ 1 \frac{1}{2} + \frac{1}{2} \sqrt{41}$ of $y = \frac{‐ 3 - \sqrt{41}}{2} = ‐ 1 \frac{1}{2} - \frac{1}{2} \sqrt{41}$

Alleen $y = ‐ 1 \frac{1}{2} + \frac{1}{2} \sqrt{41}$ voldoet, want $y \geq 0$ .

$y = \frac{1}{2} {(x + 2)}^{2} - 2$
$y + 2 = \frac{1}{2} {(x + 2)}^{2}$
$2 y + 4 = {(x + 2)}^{2}$ en ${(x + 2)}^{2} + {(y - 3)}^{2} = 25$

$2 y + 4 + {(y - 3)}^{2} = 25$
$y^{2} - 4 y - 12 = 0$
$(y - 6) (y + 2) = 0$
$y = 6$ of $y = ‐ 2$

Als $y = 6$ :
${(x + 2)}^{2} + 9 = 25$
${(x + 2)}^{2} = 16$
$x + 2 = 4$ of $x + 2 = ‐ 4$
$x = 2$ of $x = ‐ 6$
Snijpunten: $(2,6)$ en $(‐ 6,6)$

Als $y = ‐ 2$ :
${(x + 2)}^{2} + 25 = 25$
${(x + 2)}^{2} = 0$
$x = ‐ 2$
Snijpunt: $(‐ 2, ‐ 2)$
Dus de parabool en de cirkel hebben in totaal drie snijpunten.

${rc}_{lijn} = \frac{3 - ‐ 1}{‐ 7 - 1} = ‐ \frac{1}{2}$
$b = ‐ 1 + \frac{1}{2} = ‐ \frac{1}{2}$
Vergelijking lijn: $y = ‐ \frac{1}{2} x - \frac{1}{2}$

${(x + 2)}^{2} + {(‐ \frac{1}{2} x - \frac{1}{2} - 3)}^{2} = 25$
${(x + 2)}^{2} + {(‐ \frac{1}{2} x - 3 \frac{1}{2})}^{2} = 25$
$1 \frac{1}{4} x^{2} + 7 \frac{1}{2} x - 8 \frac{3}{4} = 0$
$x^{2} + 6 x - 7 = 0$
$(x - 1) (x + 7) = 0$

$x = 1$	of	$x = ‐ 7$
$y = ‐ \frac{1}{2} - \frac{1}{2} = ‐ 1$		$y = 3 \frac{1}{2} - \frac{1}{2} = 3$
$(1, ‐ 1)$		$(‐ 7,3)$

$y = ‐ \frac{1}{2} x - \frac{1}{2}$ en $y = \frac{1}{2} {(x + 2)}^{2} - 2$

$‐ \frac{1}{2} x - \frac{1}{2} = \frac{1}{2} {(x + 2)}^{2} - 2$
$‐ \frac{1}{2} x - \frac{1}{2} = \frac{1}{2} x^{2} + 2 x + 2 - 2$
$\frac{1}{2} x^{2} + 2 \frac{1}{2} x + \frac{1}{2} = 0$
$x^{2} + 5 x + 1 = 0$
$\begin{array}{l} a = 1 \\ b = 5 \\ c = 1 \end{array}} \begin{matrix} D = 25 - 4 \cdot 1 \cdot 1 = 21 \\ \sqrt{D} = \sqrt{21} \end{matrix}$

$x = \frac{‐ 5 + \sqrt{21}}{2} = ‐ 2 \frac{1}{2} + \frac{1}{2} \sqrt{21}$	of	$x = ‐ 2 \frac{1}{2} - \frac{1}{2} \sqrt{21}$
$y = \frac{3}{4} - \frac{1}{4} \sqrt{21}$		$y = \frac{3}{4} + \frac{1}{4} \sqrt{21}$

Snijpunten:

(‐ 2 \frac{1}{2} + \frac{1}{2} \sqrt{21}, \frac{3}{4} - \frac{1}{4} \sqrt{21})

(‐ 2 \frac{1}{2} - \frac{1}{2} \sqrt{21}, \frac{3}{4} + \frac{1}{4} \sqrt{21})

$y = ‐ x^{2} + 2 x + 3$
$y = ‐ (x^{2} - 2 x - 3)$
$y = ‐ ({(x - 1)}^{2} - 4)$
$y = ‐ {(x - 1)}^{2} + 4$
Top $(1,4)$

1 ; 4

$y \leq 4$

$‐ x^{2} + 4 x + 5 = 0$
$x^{2} - 4 x - 5 = 0$
$(x - 5) (x + 1) = 0$
$x = 5$ of $x = ‐ 1$
Snijpunten: $(5,0)$ en $(‐ 1,0)$

Vergelijking symmetrieas: $x = \frac{5 - 1}{2} = 2$

$x = 2 \Rightarrow y = ‐ 2^{2} + 4 \cdot 2 + 5 = ‐ 4 + 8 + 5 = 9$
Top $(2,9)$

$p \cdot V = c$
$4 \cdot 5 = c$
$20 = c$

$p \cdot 3 = 20$
$p = \frac{20}{3} = 6 \frac{2}{3}$ bar

$V > \frac{20}{3} = 6 \frac{2}{3}$ bar

$C = ‐ 20 x^{2} + 60 x$ .
$C = ‐ 20 (x^{2} - 3 x)$
$C = ‐ 20 ({(x - 1 \frac{1}{2})}^{2} - 2 \frac{1}{4})$
$C = ‐ 20 {(x - 1 \frac{1}{2})}^{2} + 45$
Top $(1 \frac{1}{2},45)$

Een bergparabool.

Bij hoogte $1 \frac{1}{2}$ , de capaciteit is dan 45.

Inhoud is $π \cdot {0,3}^{2} \cdot 8 \approx 2,2619$ dm³

$c = 0,980 \cdot 2,2619 = 2,216662$

$V = π \cdot r^{2} \cdot h = π \cdot {0,3}^{2} \cdot h$
$p \cdot V = c$
$p \cdot π \cdot {0,3}^{2} \cdot h = 2,216662$
$p \cdot h = \frac{2,216662}{π \cdot {0,3}^{2}} \approx 7,84$

$h = 50$ cm $= 5$ dm
$p = \frac{7,84}{5} = 1,568$ bar

3000 millibar = 3 bar
$h = \frac{7,84}{3} \approx 2,61$ dm

parabool
cirkel
lijn
(verticale) lijn
hyperbool