1

a

Parabool A:	Parabool B:
Top $(0,2)$ en punt $(1,1)$	Top $(‐ 2,3)$ en punt $(0,1)$
$1 = c {(1 - 0)}^{2} + 2$	$1 = c {(0 + 2)}^{2} + 3$
$1 = c + 2$	$1 = 4 c + 3$
$‐ 1 = c$	$‐ 2 = 4 c$
$y = ‐ x^{2} + 2$	$‐ \frac{1}{2} = c$
	$y = ‐ \frac{1}{2} {(x + 2)}^{2} + 3$

Parabool C:	Parabool D:
Top $(‐ 1, ‐ 4)$ en punt $(0, ‐ 2)$	Top $(‐ 1,0)$ en punt $(0,1)$
$‐ 2 = c {(0 + 1)}^{2} - 4$	$1 = c {(0 + 1)}^{2} + 0$
$‐ 2 = c - 4$	$1 = c$
$2 = c$	$y = {(x + 1)}^{2}$
$y = 2 {(x + 1)}^{2} - 4$

b

A en C:
$‐ x^{2} + 2 = 2 {(x + 1)}^{2} - 4$
$‐ x^{2} + 2 = 2 x^{2} + 4 x + 2 - 4$
$3 x^{2} + 4 x - 4 = 0$
$\begin{array}{l} a = 3 \\ b = 4 \\ c = ‐ 4 \end{array}} \begin{matrix} D = 16 - 4 \cdot 3 \cdot ‐ 4 = 64 \\ \sqrt{D} = \sqrt{64} = 8 \end{matrix}$

$x = \frac{‐ 4 + 8}{6} = \frac{2}{3}$	of	$x = \frac{‐ 4 - 8}{6} = ‐ 2$
$y = ‐ {(\frac{2}{3})}^{2} + 2 = 1 \frac{5}{9}$		$y = ‐ {(‐ 2)}^{2} + 2 = ‐ 2$
$(\frac{2}{3},1 \frac{5}{9})$		$(‐ 2, ‐ 2)$

B en D:
$‐ \frac{1}{2} {(x + 2)}^{2} + 3 = {(x + 1)}^{2}$
$‐ \frac{1}{2} x^{2} - 2 x - 2 + 3 = x^{2} + 2 x + 1$
$1 \frac{1}{2} x (x + \frac{8}{3}) = 0$

$x = 0$	of	$x = ‐ \frac{8}{3}$
$y = {(0 + 1)}^{2} = 1$		$y = {(‐ \frac{8}{3} + 1)}^{2} = \frac{25}{9} = 2 \frac{7}{9}$
$(0,1)$		$(‐ \frac{8}{3},2 \frac{7}{9})$

2

3

$y = \frac{1}{4} x^{2} + 3 x + 2$	$y = ‐ 2 x^{2} + 4 x + 6$
$y = \frac{1}{4} (x^{2} + 12 x + 8)$	$y = ‐ 2 (x^{2} - 2 x - 3)$
$y = \frac{1}{4} ({(x + 6)}^{2} - 36 + 8)$	$y = ‐ 2 ({(x - 1)}^{2} - 1 - 3)$
$y = \frac{1}{4} {(x + 6)}^{2} - 7$	$y = ‐ 2 {(x - 1)}^{2} + 8$
Top $(‐ 6, ‐ 7)$	Top $(1,8)$

4

a

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

b

c

Zie vraag b.

d

$3 x = 2 {(x - 1)}^{2} - 2$
$2 x^{2} - 7 x = 0$
$2 x (x - 3 \frac{1}{2}) = 0$

$x = 0$	of	$x = 3 \frac{1}{2}$
$y = 3 \cdot 0 = 0$		$y = 3 \cdot 3 \frac{1}{2} = 10 \frac{1}{2}$
$(0,0)$		$(3 \frac{1}{2},10 \frac{1}{2})$

e

$3 x + p = 2 {(x - 1)}^{2} - 2$
$2 x {}^{2}- 7 x - p = 0$
$\begin{array}{l} a = 2 \\ b = ‐ 7 \\ c = ‐ p \end{array}} D = 49 - 4 \cdot 2 \cdot ‐ p = 49 + 8 p$
Eén oplossing als $D = 0$ .
$49 + 8 p = 0$
$8 p = ‐ 49$
$p = ‐ \frac{49}{8} = ‐ 6 \frac{1}{8}$

5

a

$t = 300 : 120 = 2 \frac{1}{2}$ uur

b

$v t = 300$

c

d

Als $v$ klein is. De grafiek is dan het steilst.

6

a

$y = a x - \frac{1}{100} x^{2}$
$0 = a \cdot 100 - \frac{1}{100} \cdot 100^{2}$
$100 a = 100$
$a = 1$

b

Vanwege symmetrie wordt de grootste hoogte bereikt als $x = 50$ .
Dan is $y = 50 - \frac{1}{100} \cdot 50^{2} = 25$ , dus 25 meter.

7

$x^{2} - 8 x + 22 = 0$
$\begin{array}{l} a = 1 \\ b = ‐ 8 \\ c = 22 \end{array}} D = 64 - 4 \cdot 1 \cdot 22 = ‐ 24$
$D < 0$ , dus geen oplossingen

$3 {(x + 2)}^{2} - 2 x = 9$
$3 x^{2} + 10 x + 3 = 0$
$\begin{array}{l} a = 3 \\ b = 10 \\ c = 3 \end{array}} \begin{matrix} D = 100 - 4 \cdot 3 \cdot 3 = 64 \\ \sqrt{D} = \sqrt{64} = 8 \end{matrix}$
$x = \frac{‐ 10 + 8}{6} = ‐ \frac{1}{3}$ of $x = \frac{‐ 10 - 8}{6} = ‐ 3$

$2 x^{2} = 5 x - 3$
$2 x^{2} - 5 x + 3 = 0$
$\begin{array}{l} a = 2 \\ b = ‐ 5 \\ c = 3 \end{array}} \begin{matrix} D = 25 - 4 \cdot 2 \cdot 3 = 1 \\ \sqrt{D} = \sqrt{1} = 1 \end{matrix}$
$x = \frac{5 + 1}{4} = 1 \frac{1}{2}$ of $x = \frac{5 - 1}{4} = 1$

$‐ 5 x^{2} + 4 x - \frac{4}{5} = 0$
$\begin{array}{l} a = ‐ 5 \\ b = 4 \\ c = ‐ \frac{4}{5} \end{array}} D = 16 - 4 \cdot ‐ 5 \cdot ‐ \frac{4}{5} = 0$
$x = ‐ \frac{4}{‐ 10} = \frac{2}{5}$

8

a

$y = ‐ 2 x^{2} + 12 x$
$y = ‐ 2 (x^{2} - 6 x)$
$y = ‐ 2 ({(x - 3)}^{2} - 9)$
$y = ‐ 2 {(x - 3)}^{2} + 18$
Top $(3,18)$

b

Voor $x = 3$ , dan $y = 18$ .

9

$x^{2} - x \sqrt{2} - 4 = 0$
$\begin{array}{l} a = 1 \\ b = ‐ \sqrt{2} \\ c = ‐ 4 \end{array}} \begin{matrix} D = 2 - 4 \cdot 1 \cdot ‐ 4 = 18 \\ \sqrt{D} = \sqrt{18} = 3 \sqrt{2} \end{matrix}$
$x = \frac{\sqrt{2} + 3 \sqrt{2}}{2} = 2 \sqrt{2}$ of $x = \frac{\sqrt{2} - 3 \sqrt{2}}{2} = ‐ \sqrt{2}$

$\sqrt{x^{2} + 3 x} = 3 \sqrt{2}$
$x^{2} + 3 x = 18$
$x^{2} + 3 x - 18 = 0$
$(x - 3) (x + 6) = 0$
$x = 3$ of $x = ‐ 6$

10

$x^{2} + 3 x + p = 0$
$\begin{array}{l} a = 1 \\ b = 3 \\ c = p \end{array}} D = 9 - 4 \cdot 1 \cdot p = 9 - 4 p$
Twee oplossingen als $D > 0$ :
$9 - 4 p > 0$
$9 > 4 p$
$2 \frac{1}{4} > p$
Dus twee oplossingen als $p < 2 \frac{1}{4}$ .

Eén oplossing als $D = 0$ :
$p = 2 \frac{1}{4}$

Geen oplossingen als $D < 0$ :
$p > 2 \frac{1}{4}$

11

a

horizontale asymptoot: $y = ‐ 3$
verticale asymptoot: $x = ‐ 2$

horizontale asymptoot: $y = ‐ 1$
verticale asymptoot: $x = 4$

horizontale asymptoot: $y = 0$
verticale asymptoot: $x = 0$

b

c

Zie vraag b.

d

$(x + 2) (y + 3) = 4$ en $y = 2 x - 1$

$(x + 2) (2 x - 1 + 3) = 4$
$(x + 2) (2 x + 2) = 4$
$2 x^{2} + 6 x = 0$
$2 x (x + 3) = 0$

$x = 0$	of	$x = ‐ 3$
$y = 0 - 1 = ‐ 1$		$y = ‐ 6 - 1 = ‐ 7$
$(0, ‐ 1)$		$(‐ 3, ‐ 7)$

$(x - 4) (y + 1) = ‐ 8$ en $y = 2 x - 1$

$(x - 4) (2 x - 1 + 1) = ‐ 8$
$2 x (x - 4) = ‐ 8$
$x (x - 4) = ‐ 4$
$x^{2} - 4 x + 4 = 0$
${(x - 2)}^{2} = 0$
$x = 2$
$y = 4 - 1 = 3$
$(2,3)$

$x y = 9$ en $y = 2 x - 1$

$x (2 x - 1) = 9$
$2 x^{2} - x - 9 = 0$
$\begin{array}{l} a = 2 \\ b = ‐ 1 \\ c = ‐ 9 \end{array}} \begin{matrix} D = 1 - 4 \cdot 2 \cdot ‐ 9 = 73 \\ \sqrt{D} = \sqrt{73} \end{matrix}$

$x = \frac{1 + \sqrt{73}}{4} = \frac{1}{4} + \frac{1}{4} \sqrt{73}$	of	$x = \frac{1 - \sqrt{73}}{4} = \frac{1}{4} - \frac{1}{4} \sqrt{73}$
$y = ‐ \frac{1}{2} + \frac{1}{2} \sqrt{73}$		$y = ‐ \frac{1}{2} - \frac{1}{2} \sqrt{73}$
$(\frac{1}{4} + \frac{1}{4} \sqrt{73}, ‐ \frac{1}{2} + \frac{1}{2} \sqrt{73})$		$(\frac{1}{4} - \frac{1}{4} \sqrt{73}, ‐ \frac{1}{2} - \frac{1}{2} \sqrt{73})$

12

horizontale asymptoot: $y = 0$
verticale asymptoot: $x = 0$
Punt op hyperbool: $(3, ‐ 3)$
$x y = 3 \cdot ‐ 3 = ‐ 9$

horizontale asymptoot: $y = 1$
verticale asymptoot: $x = ‐ 1$
Punt op hyperbool: $(2,3)$
$(x + 1) (y - 1) = c$
$(2 + 1) (3 - 1) = c$
$6 = c$
$(x + 1) (y - 1) = 6$

horizontale asymptoot: $y = ‐ 3$
verticale asymptoot: $x = 0$
Punt op hyperbool: $(2,1)$
$x (y + 3) = c$
$2 (1 + 3) = c$
$8 = c$
$x (y + 3) = 8$

13

a

b

$x y = 8$ en $y = ‐ x + k$

$x (‐ x + k) = 8$
$‐ x^{2} + k x - 8 = 0$
$\begin{array}{l} a = ‐ 1 \\ b = k \\ c = ‐ 8 \end{array}} D = k^{2} - 4 \cdot ‐ 1 \cdot ‐ 8 = k^{2} - 32$

Raken, dus $D = 0$ .
$k^{2} - 32 = 0$
$k^{2} = 32$
$k = \sqrt{32} = 4 \sqrt{2}$ of $k = ‐ \sqrt{32} = ‐ 4 \sqrt{2}$
Vergelijking raaklijnen:
$y = ‐ x + 4 \sqrt{2}$ en $y = ‐ x - 4 \sqrt{2}$

14

$2 (x + 3) {}^{2}- 4 = ‐ 3 x + 1$
$2 x^{2} + 15 x + 13 = 0$
$\begin{array}{l} a = 2 \\ b = 15 \\ c = 13 \end{array}} \begin{matrix} D = 225 - 4 \cdot 2 \cdot 13 = 121 \\ \sqrt{D} = \sqrt{121} = 11 \end{matrix}$

$x = \frac{‐ 15 + 11}{4} = ‐ 1$	of	$x = \frac{‐ 15 - 11}{4} = ‐ 6 \frac{1}{2}$
$y = ‐ 3 \cdot ‐ 1 + 1 = 4$		$y = ‐ 3 \cdot ‐ 6 \frac{1}{2} + 1 = 20 \frac{1}{2}$
$(‐ 1,4)$		$(‐ 6 \frac{1}{2},20 \frac{1}{2})$

15

Oppervlakte balk is
$2 (x (x + 4) + x (x + 3) + (x + 4) (x + 3)) =$
$2 (3 x^{2} + 14 x + 12) = 6 x^{2} + 28 x + 24$

Vergelijking:

$6 x^{2} + 28 x + 24 = 162$
$6 x^{2} + 28 x - 138 = 0$
$\begin{array}{l} a = 6 \\ b = 28 \\ c = ‐ 138 \end{array}} \begin{matrix} D = 784 - 4 \cdot 6 \cdot ‐ 138 = 4096 \\ \sqrt{D} = \sqrt{4096} = 64 \end{matrix}$
$x = \frac{‐ 28 + 64}{12} = 3$ of $x = \frac{‐ 28 - 64}{12} = ‐ 7 \frac{2}{3}$
Alleen $x = 3$ voldoet, omdat $x > 0$ moet zijn.

16

oppervlakte driehoeken $= 2 \cdot x (8 - x)$
oppervlakte driehoeken $= \frac{1}{4} \cdot 8 \cdot 8 = 16$

$2 \cdot x (8 - x) = 16$
$2 x^{2} - 16 x + 16 = 0$
$x^{2} - 8 x + 8 = 0$
$\begin{array}{l} a = 1 \\ b = ‐ 8 \\ c = 8 \end{array}} \begin{matrix} D = 64 - 4 \cdot 1 \cdot 8 = 32 \\ \sqrt{D} = \sqrt{32} = 4 \sqrt{2} \end{matrix}$
$x = \frac{8 + 4 \sqrt{2}}{2} = 4 + 2 \sqrt{2}$ cmof $x = \frac{8 - 4 \sqrt{2}}{2} = 4 - 2 \sqrt{2}$ cm
Allebei de oplossingen voldoen, want $0 < x < 8$ .

17

a

$50 t - 5 t^{2} = 0$
$5 t (10 - t) = 0$
$t = 0$ of $t = 10$
Dus de vlucht duurt $10 - 0 = 10$ sec.

b

Maximale hoogte wordt bereikt na $5$ sec.
$h = 50 \cdot 5 - 5 \cdot 5^{2} = 250 - 125 = 125$ m

c

d

$50 t - 5 t^{2} > 113,75$
$0 > 5 t^{2} - 50 t + 113,75$
$t^{2} - 10 t + 22,75 < 0$
$(t - 3,5) (t - 6,5) < 0$
$3,5 < t < 6,5$
Dus tussen de $3,5$ en $6,5$ sec. is de hoogte van de vuurpijl meer dan $113,75$ m.

18

a

${(10 + 2)}^{2} - 10 - 10 = 124$ stippen

b

${(n + 2)}^{2} - 2 n = n^{2} + 2 n + 4$

c

$n^{2} + 2 n + 4 = 10.204$
$n^{2} + 2 n - 10.200 = 0$
$(n - 100) (n + 102) = 0$
$n = 100$ of $n = ‐ 102$
Alleen $n = 100$ voldoet, omdat $n > 0$ .

19

a

$\frac{18}{45} = \frac{x}{y}$
$18 y = 45 x$
$y = 2 \frac{1}{2} x$
Lengte van de rechthoek is $45 - y = 45 - 2 \frac{1}{2} x$
$O = x (45 - 2 \frac{1}{2} x) = 45 x - 2 \frac{1}{2} x^{2}$

b

$O = 45 x - 2 \frac{1}{2} x^{2}$
$O = ‐ 2 \frac{1}{2} x^{2} + 45 x$
$O = ‐ 2 \frac{1}{2} (x^{2} - 18 x)$
$O = ‐ 2 \frac{1}{2} ({(x - 9)}^{2} - 81)$
$O = ‐ 2 \frac{1}{2} {(x - 9)}^{2} + 202 \frac{1}{2}$
De oppervlakte is maximaal als $x = 9$ .

c

De oppervlakte is dan $202 \frac{1}{2}$ .