Kwadraatafsplitsen

$y = {(x + 1)}^{2} + 3 = x^{2} + 2 x + 4$

$y = x^{2} + 4 x - 3 = {(x + 2)}^{2} - 7$

$y = 2 x^{2} + 8 x - 6$
$y = 2 (x^{2} + 4 x - 3)$
$y = 2 ({(x + 2)}^{2} - 7)$
$y = 2 {(x + 2)}^{2} - 14$

$(‐ 2, ‐ 14)$

$y = \frac{1}{2} x^{2} + 3 x + 2$
$y = \frac{1}{2} (x^{2} + 6 x + 4)$
$y = \frac{1}{2} ({(x + 3)}^{2} - 5)$
$y = \frac{1}{2} {(x + 3)}^{2} - 2 \frac{1}{2}$

$(‐ 3, ‐ 2 \frac{1}{2})$

Linkerkolom:

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

$y = ‐ 2 x^{2} + 10 x + 1$
$y = ‐ 2 (x^{2} - 5 x - \frac{1}{2})$
$y = ‐ 2 ({(x - 2 \frac{1}{2})}^{2} - 6 \frac{3}{4})$
$y = ‐ 2 {(x - 2 \frac{1}{2})}^{2} + 13 \frac{1}{2}$
Top $(2 \frac{1}{2},13 \frac{1}{2})$

Rechterkolom:

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

$y = (x + 3) (x - 7)$
snijpunten met de $x$ -as zijn: $(‐ 3,0)$ en $(7,0)$ . De symmetrieas ligt daar midden tussen.
Symmetrieas van de parabool: $x = \frac{‐ 3 + 7}{2} = 2$ ,
$y = (2 + 3) (2 - 7) = ‐ 25$
Top $(2, ‐ 25)$

abc-formule

$x = ‐ \frac{11}{6} + \sqrt{\frac{25}{36}} = ‐ 1$ of $x = ‐ \frac{11}{6} - \sqrt{\frac{25}{36}} = ‐ 2 \frac{2}{3}$

$x = ‐ \frac{9}{2 \cdot 5} + \sqrt{{(\frac{9}{2 \cdot 5})}^{2} - \frac{2}{5}}$ of $x = ‐ \frac{9}{2 \cdot 5} - \sqrt{{(\frac{9}{2 \cdot 5})}^{2} - \frac{2}{5}}$

$x = ‐ \frac{9}{2 \cdot 5} + \sqrt{{(\frac{9}{2 \cdot 5})}^{2} - \frac{2}{5}} = ‐ \frac{9}{10} + \sqrt{\frac{41}{100}} = ‐ \frac{9}{10} + \frac{1}{10} \sqrt{41}$ of
$x = ‐ \frac{9}{2 \cdot 5} - \sqrt{{(\frac{9}{2 \cdot 5})}^{2} - \frac{2}{5}} = ‐ \frac{9}{10} - \sqrt{\frac{41}{100}} = ‐ \frac{9}{10} - \frac{1}{10} \sqrt{41}$

$x = ‐ \frac{b}{2 a} + \sqrt{(\frac{b}{2 a}) {}^{2}- \frac{c}{a}}$ of $x = ‐ \frac{b}{2 a} - \sqrt{(\frac{b}{2 a}) {}^{2}- \frac{c}{a}}$

$x = ‐ \frac{‐ 5}{2 \cdot 2} + \sqrt{{(\frac{‐ 5}{2 \cdot 2})}^{2} - \frac{‐ 25}{2}} = \frac{5}{4} + \sqrt{\frac{225}{16}} = \frac{5}{4} + \frac{15}{4} = \frac{20}{4} = 5$ of
$x = ‐ \frac{‐ 5}{2 \cdot 2} - \sqrt{{(\frac{‐ 5}{2 \cdot 2})}^{2} - \frac{‐ 25}{2}} = \frac{5}{4} - \sqrt{\frac{225}{16}} = \frac{5}{4} - \frac{15}{4} = ‐ \frac{10}{4} = ‐ 2 \frac{1}{2}$

$2 \cdot 5^{2} - 5 \cdot 5 - 25 = 50 - 25 - 25 = 0$
$2 \cdot {(‐ 2 \frac{1}{2})}^{2} - 5 \cdot ‐ 2 \frac{1}{2} - 25 = 12 \frac{1}{2} + 12 \frac{1}{2} - 25 = 0$

$x^{2} + 4 x + 5 = 0$
${(x + 2)}^{2} - 4 + 5 = 0$
${(x + 2)}^{2} = ‐ 1$ en dat kan voor geen enkele $x$

$\sqrt{{(\frac{4}{2 \cdot 1})}^{2} - \frac{5}{1}} = \sqrt{4 - 5} = \sqrt{‐ 1}$ , maar $\sqrt{‐ 1}$ bestaat niet

stap 1:
haakjes uitwerken, ${(\frac{b}{2 a})}^{2} = \frac{b}{2 a} \cdot \frac{b}{2 a} = \frac{b^{2}}{4 a^{2}}$

stap 2:
breuken gelijknamig maken, $\frac{c}{a} \cdot \frac{4 a}{4 a} = \frac{4 a c}{4 a^{2}}$

stap 3:
twee breuken met dezelfde noemer optellen:
de noemer zo laten en de tellers optellen.

als $a > 0$ , dan $2 a \cdot 2 a = 4 a^{2}$ en
als $a < 0$ , dan $‐ 2 a \cdot ‐ 2 a = 4 a^{2}$

als $a > 0$	$‐ \frac{b}{2 a} + \sqrt{\frac{D}{4 a^{2}}} = ‐ \frac{b}{2 a} + \frac{\sqrt{D}}{2 a}$
of	$‐ \frac{b}{2 a} - \sqrt{\frac{D}{4 a^{2}}} = ‐ \frac{b}{2 a} - \frac{\sqrt{D}}{2 a}$

als $a < 0$	$‐ \frac{b}{2 a} + \sqrt{\frac{D}{4 a^{2}}} = ‐ \frac{b}{2 a} + \frac{\sqrt{D}}{‐ 2 a} = ‐ \frac{b}{2 a} - \frac{\sqrt{D}}{2 a}$
of	$‐ \frac{b}{2 a} - \sqrt{\frac{D}{4 a^{2}}} = ‐ \frac{b}{2 a} - \frac{\sqrt{D}}{‐ 2 a} = ‐ \frac{b}{2 a} + \frac{\sqrt{D}}{2 a}$

$D = 2^{2} - 4 \cdot 1 \cdot 3 = ‐ 8$
$D = 3^{2} - 4 \cdot ‐ 1 \cdot 2 = 17$
$D = 20^{2} - 4 \cdot 4 \cdot 25 = 0$

geen oplossingen
$x = \frac{‐ 3 + \sqrt{17}}{2 \cdot ‐ 1} = 1 \frac{1}{2} - \frac{1}{2} \sqrt{17}$ of $x = \frac{‐ 3 - \sqrt{17}}{2 \cdot ‐ 1} = 1 \frac{1}{2} + \frac{1}{2} \sqrt{17}$
$x = ‐ \frac{20}{8} = ‐ 2 \frac{1}{2}$

Nee, niet met de $a b c$ -formule, want dan moet je delen door 0 en dat kan niet.
Als $a = 0$ , dan is $b x + c = 0$ , dus $x = ‐ \frac{c}{b}$ .

Linkerkolom:

$2 x {}^{2}- 3 x - 35 = 0$
$\begin{array}{l} a = 2 \\ b = ‐ 3 \\ c = ‐ 35 \end{array}} \begin{matrix} D = 9 - 4 \cdot 2 \cdot ‐ 35 = 289 \\ \sqrt{D} = \sqrt{289} = 17 \end{matrix}$
$x = \frac{3 + 17}{4} = 5$ of $x = \frac{3 - 17}{4} = ‐ 3 \frac{1}{2}$

$2 x^{2} + 4 x - 1 = 0$
$\begin{array}{l} a = 2 \\ b = 4 \\ c = ‐ 1 \end{array}} \begin{matrix} D = 16 - 4 \cdot 2 \cdot ‐ 1 = 24 \\ \sqrt{D} = \sqrt{24} = 2 \sqrt{6} \end{matrix}$
$x = \frac{‐ 4 + 2 \sqrt{6}}{4} = ‐ 1 + \frac{1}{2} \sqrt{6}$ of $x = \frac{‐ 4 - 2 \sqrt{6}}{4} = ‐ 1 - \frac{1}{2} \sqrt{6}$

$7 x^{2} - 6 x + 2 = 0$
$\begin{array}{l} a = 7 \\ b = ‐ 6 \\ c = 2 \end{array}} D = 36 - 4 \cdot 7 \cdot 2 = ‐ 20$
$D < 0$ , dus géén oplossingen

$5 x - 3 x^{2} = 0$
$\begin{array}{l} a = ‐ 3 \\ b = 5 \\ c = 0 \end{array}} \begin{matrix} D = 25 - 4 \cdot ‐ 3 \cdot 0 = 25 \\ \sqrt{D} = \sqrt{25} = 5 \end{matrix}$
$x = \frac{‐ 5 + 5}{‐ 6} = 0$ of $x = \frac{‐ 5 - 5}{‐ 6} = \frac{10}{6} = 1 \frac{2}{3}$

Rechterkolom:

$4 x = 1 + 4 x^{2}$
$4 x^{2} - 4 x + 1 = 0$
$\begin{array}{l} a = 4 \\ b = ‐ 4 \\ c = 1 \end{array}} D = 16 - 4 \cdot 4 \cdot 1 = 0$
$x = ‐ \frac{‐ 4}{8} = \frac{1}{2}$

${(x - 3)}^{2} = 5 - 3 x$
$x^{2} - 6 x + 9 = 5 - 3 x$
$x^{2} - 3 x + 4 = 0$
$\begin{array}{l} a = 1 \\ b = ‐ 3 \\ c = 4 \end{array}} D = 9 - 4 \cdot 1 \cdot 4 = ‐ 7$
$D < 0$ , dus géén oplossingen

$\frac{1}{2} x^{2} - 3 x - 4 \frac{1}{2} = 0$
$\begin{array}{l} a = \frac{1}{2} \\ b = ‐ 3 \\ c = ‐ 4 \frac{1}{2} \end{array}} \begin{matrix} D = 9 - 4 \cdot \frac{1}{2} \cdot ‐ 4 \frac{1}{2} = 18 \\ \sqrt{D} = \sqrt{18} = 3 \sqrt{2} \end{matrix}$
$x = \frac{3 + 3 \sqrt{2}}{1} = 3 + 3 \sqrt{2}$ of $x = \frac{3 - 3 \sqrt{2}}{1} = 3 - 3 \sqrt{2}$

${(x + 3)}^{2} = 16$
$x + 3 = 4$ of $x - 3 = 4$
$x = 1$ of $x = ‐ 7$

${(x + 1)}^{2} = {(2 x + 3)}^{2}$
$x + 1 = 2 x + 3$ of $x + 1 = ‐ 2 x - 3$
$‐ 2 = x$ of $3 x = ‐ 4$
$‐ 2 = x$ of $x = ‐ \frac{4}{3} = ‐ 1 \frac{1}{3}$

$x^{2} + 6 x + 5 = 0$
$(x + 1) (x + 5) = 0$
$x = ‐ 1$ of $x = ‐ 5$

Raaklijn en raakpunt

Zie vraag a.

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

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

$x^{2} = x - 2$
$x^{2} - x + 2 = 0$
$\begin{array}{l} a = 1 \\ b = ‐ 1 \\ c = 2 \end{array}} D = 1 - 4 \cdot 1 \cdot 2 = ‐ 7$
$D < 0$ , dus geen snijpunten

Alle lijnen hebben richtingscoëfficiënt 1.

Zie stippellijn in vraag a.

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

$x^{2} = x$
$x^{2} - x = 0$
$\begin{array}{l} a = 1 \\ b = ‐ 1 \\ c = 0 \end{array}} D = 1 - 4 \cdot 1 \cdot 0 = 1$

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

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

$1 + 4 k = 0$
$4 k = ‐ 1$
$k = ‐ \frac{1}{4}$

$x^{2} = x - \frac{1}{4}$
$x^{2} - x + \frac{1}{4} = 0$
$x = ‐ \frac{b}{2 a} = ‐ \frac{‐ 1}{2} = \frac{1}{2} \Rightarrow y = {(\frac{1}{2})}^{2} = \frac{1}{4}$
Raakpunt is $(\frac{1}{2}, \frac{1}{4})$ .

10s

11s

Omdat $0 = k \cdot 0$ klopt, wat je ook voor $k$ neemt.

Als $k = 0$ , dan $y = 0$ ; $5 = k \cdot ‐ 2 \Rightarrow k = ‐ 2 \frac{1}{2}$

De verticale as, dus de $y$ -as.

Zie vraag a.

Zie de twee stippellijnen in vraag a.

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

$D = {(2 + k)}^{2} - 4 \cdot 1 \cdot 1 = {(2 + k)}^{2} - 4$

Als $k = 1$ , dan $D = 9 - 4 = 5$ , dus twee snijpunten.

${(2 + k)}^{2} - 4 = 0$
${(2 + k)}^{2} = 4$
$2 + k = 2$ of $2 + k = ‐ 2$
$k = 0$ of $k = ‐ 4$

$y = 0$ en $y = ‐ 4 x$

$‐ x^{2} + 1 = a x + 3$
$x^{2} + a x + 2 = 0$

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

Raken, dus $D = 0$ .
$a^{2} - 8 = 0$
$a^{2} = 8$
$a = \sqrt{8} = 2 \sqrt{2}$ of $a = ‐ \sqrt{8} = ‐ 2 \sqrt{2}$

$x^{2} + 2 \sqrt{2} x + 2 = 0$
$x = ‐ \frac{2 \sqrt{2}}{2} = ‐ \sqrt{2} \Rightarrow y = ‐ {(‐ \sqrt{2})}^{2} + 1 = ‐ 1$
Raakpunt is $(‐ \sqrt{2}, ‐ 1)$ .

$x^{2} - 2 \sqrt{2} x + 2 = 0$
$x = ‐ \frac{‐ 2 \sqrt{2}}{2} = \sqrt{2} \Rightarrow y = ‐ {(\sqrt{2})}^{2} + 1 = ‐ 1$
Raakpunt is $(\sqrt{2}, ‐ 1)$ .

$x^{2} + 5 x + 4 = 0$
$(x + 1) (x + 4) = 0$
$x = ‐ 1$ of $x = ‐ 4$

$x^{2} + k x + 4 = 0$
$\begin{array}{l} a = 1 \\ b = k \\ c = 4 \end{array}} D = k^{2} - 4 \cdot 1 \cdot 4 = k^{2} - 16$
Eén oplossing, dus $D = 0$ .
$k^{2} - 16 = 0$
$k^{2} = 16$
$k = 4$ of $k = ‐ 4$

$x^{2} + 4 x + 4 = 0$
${(x + 2)}^{2} = 0$
$x = ‐ 2$

13s

14s

$x^{2} + {(‐ x + k)}^{2} = 3$
$2 x^{2} - 2 k x + k^{2} - 3 = 0$
$\begin{array}{l} a = 2 \\ b = ‐ 2 k \\ c = k^{2} - 3 \end{array}} D = {(‐ 2 k)}^{2} - 4 \cdot 2 \cdot (k^{2} - 3) = 24 - 4 k^{2}$

$24 - 4 k^{2} = 0$
$24 = 4 k^{2}$
$6 = k^{2}$
$k = \sqrt{6}$ of $k = ‐ \sqrt{6}$

Als $k = \sqrt{6}$ :
$x = ‐ \frac{b}{2 a} = ‐ \frac{‐ 2 \sqrt{6}}{4} = \frac{1}{2} \sqrt{6}$
$y = ‐ \frac{1}{2} \sqrt{6} + \sqrt{6} = \frac{1}{2} \sqrt{6}$
Raakpunt is $(\frac{1}{2} \sqrt{6}, \frac{1}{2} \sqrt{6})$ .

Als $k = ‐ \sqrt{6}$ :
$x = ‐ \frac{2 \sqrt{6}}{4} = ‐ \frac{1}{2} \sqrt{6}$
$y = \frac{1}{2} \sqrt{6} - \sqrt{6} = ‐ \frac{1}{2} \sqrt{6}$
Raakpunt is $(‐ \frac{1}{2} \sqrt{6}, ‐ \frac{1}{2} \sqrt{6})$ .

$x - as \Rightarrow y = 0$ , dan:
$p x^{2} - 6 x - 1 = 0$
$\begin{array}{l} a = p \\ b = ‐ 6 \\ c = ‐ 1 \end{array}} D = 36 - 4 \cdot p \cdot ‐ 1 = 36 + 4 p$
Eén raakpunt met $x$ -as $\Rightarrow D = 0$
$36 + 4 p = 0$
$4 p = ‐ 36$
$p = ‐ 9$

$x - as \Rightarrow y = 0$ , dan:
$\frac{1}{2} x^{2} - p x + 2 = 0$
$\begin{array}{l} a = \frac{1}{2} \\ b = ‐ p \\ c = 2 \end{array}} D = {(‐ p)}^{2} - 4 \cdot \frac{1}{2} \cdot 2 = p^{2} - 4$
Twee snijpunten met $x$ -as $\Rightarrow D > 0$
$p^{2} - 4 > 0$
$p^{2} > 4$
$p < ‐ 2$ of $p > 2$

$x - as \Rightarrow y = 0$ , dan:
$2 x^{2} + 4 x + p = 0$
$\begin{array}{l} a = 2 \\ b = 4 \\ c = p \end{array}} D = 16 - 4 \cdot 2 \cdot p = 16 - 8 p$
Geen snij- of raakpunten punten met $x$ -as $\Rightarrow D < 0$
$16 - 8 p < 0$
$16 < 8 p$
$2 < p$