geg.: Funktionsgleichung $f:y=-1,5\cdot {\log }_2(1-x)-6\mathbb{G}=\mathbb{R}\times \mathbb{R}f:\; y=-1\{,\}5\backslash cdot\backslash log\_\{2\}\{(1-x)-6\}\backslash qquad\backslash mathbb\{G=R\backslash times\; R\}$

Gerade $g:y=-9g:\; y=-9$

ges.: $x-x-$Koordinate des Schnittpunktes $PP$

Ansatz und Rechnung:

Gleichsetzen der Funktion $gg$ mit Funktion $ff$, Auflösen nach $xx$

$-9=-1,5\cdot {\log }_2(1-x)-6-9=-1\{,\}5\backslash cdot\backslash log\_\{2\}\{(1-x)-6\}\backslash qquad\backslash :$|$+6\backslash textcolor\; \{blue\}\; \{+6\}$

$-3=-1,5\cdot {\log }_2(1-x)-3=-1\{,\}5\backslash cdot\; \backslash log\_\{2\}\{(1-x)\}\backslash qquad\backslash qquad$|$:(-1,5)\backslash textcolor\; \{blue\}\; \{:(-1\{,\}5)\}$

$2={\log }_2(1-x)2=\backslash log\_\{2\}\{(1-x)\}\backslash qquad$| Logarithmus in Potenz umwandeln $\mathbf{x}={\log }_{\mathbf{a}}(\mathbf{y})\leftrightarrow \mathbf{y}={\mathbf{a}}^{\mathbf{x}}\backslash mathbf\{\backslash textcolor\{red\}\{x\}=\; \backslash log\_\{\backslash textcolor\{green\}\{a\}\}\{(\backslash textcolor\{blue\}\{y\})\}\; \backslash leftrightarrow\; \backslash textcolor\{blue\}\{y\}=\backslash textcolor\{green\}\{a\}^\{\backslash textcolor\{red\}x\}\}$

$\Rightarrow \backslash Rightarrow$$\mathbf{1}-\mathbf{x}=\mathbf{2}{}^{\mathbf{2}}\backslash mathbf\{\backslash textcolor\; \{blue\}\{1-x\}=\backslash textcolor\{green\}\{2\}\; \backslash textcolor\{red\}\{^2\}\}$ $1-x=41-4=x\mathbf{x}=-\mathbf{3}\backslash qquad1-x=4\backslash qquad\; 1-4=x\backslash qquad\; \backslash mathbf\; \{x=-3\}$

$\mathbf{P}(-\mathbf{3}\mathbf{\mid }-\mathbf{9})\backslash mathbf\; \{P(-3|-9)\}$, die $xx$-Koordinate ist also $-3-3$.

geg.: Die mit $\vec{v}=\left(\begin{array}{c}2\\ -3\end{array}\right)\backslash vec\{v\}=\backslash begin\{pmatrix\}\; 2\backslash \backslash -3\; \backslash end\{pmatrix\}$ verschobene und an der x-Achse gespiegelte Funktion

$f:f:$ $y=-1,5\cdot {\log }_2(1-x)-6y=-1\{,\}5\backslash cdot\; \backslash log\_\{2\}\{(1-x)-6\}\backslash quad$ gemäß Aufgabe A 1.a

ges.: Ausgangsfunktion $f_0:y=a\cdot {\log }_2(b-x)-cf\_0:\; y=a\backslash cdot\; \backslash log\_\{2\}\{(b-x)-c\}$

Ansatz und Rechnung:

1. Rück-Verschiebung von $ff$ durch Addition mit $\vec{v}=\left(\begin{array}{c}2\\ -3\end{array}\right)\backslash color\; \{purple\}\; \backslash vec\{v\}=\backslash begin\{pmatrix\}\; 2\backslash \backslash -3\; \backslash end\{pmatrix\}$: $y^{\mathrm{\prime }\mathrm{\prime }}\Rightarrow y^{\mathrm{\prime }}y\text{'}\text{'}\backslash Rightarrow\; y\text{'}$

$\left(\begin{array}{c}x{}^{\mathrm{\prime }\mathrm{\prime }}\\ -1,5\cdot {\log }_2(1-x{}^{\mathrm{\prime }\mathrm{\prime }})-6\end{array}\right)=\left(\begin{array}{c}x{}^{\mathrm{\prime }}\\ y{}^{\mathrm{\prime }}\end{array}\right)\oplus \left(\begin{array}{c}\mathbf{2}\\ -\mathbf{3}\end{array}\right)\backslash begin\{pmatrix\}\; x\{\text{'}\text{'}\}\backslash \backslash -1\{,\}5\backslash cdot\; \backslash log\_\{2\}\{(1-x\{\text{'}\text{'}\})-6\}\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}\; x\{\text{'}\}\backslash \backslash y\{\text{'}\}\; \backslash end\{pmatrix\}\; \oplus \; \backslash color\; \{purple\}\backslash bold\; \{\backslash begin\{pmatrix\}\; 2\; \backslash \backslash -3\; \backslash end\{pmatrix\}\}$

$x{}^{\mathrm{\prime }\mathrm{\prime }}=x{}^{\mathrm{\prime }}+2x\{\text{'}\text{'}\}=x\{\text{'}\}+2$ einsetzen in $y^{\mathrm{\prime }}y\text{'}$:

$y^{\mathrm{\prime }}=-1,5\cdot {\log }_2(1-(x^{\mathrm{\prime }}+2))-6+3y\text{'}=-1\{,\}5\backslash cdot\; \backslash log\_\{2\}\{(1-(x\text{'}+2))-6+3\}$ $\backslash qquad$| zusammenfassen

${\mathbf{y}}^{\mathbf{\prime }}=-\mathbf{1},\mathbf{5}\cdot {\log }_{\mathbf{2}}(-\mathbf{1}-{\mathbf{x}}^{\mathbf{\prime }})-\mathbf{3}\backslash mathbf\{y\text{'}=-1\{,\}5\backslash cdot\; \backslash log\_\{2\}\{(-1-x\text{'})-3\}\}$ $\mathbb{G}=\mathbb{R}\times \mathbb{R}\backslash qquad\; \backslash qquad\; \backslash mathbb\{G\}=\backslash mathbb\{R\}\backslash times\; \backslash mathbb\{R\}\backslash \backslash$

2. Rückspiegelung an $x-x-$Achse durch Multiplikation des Funktionsterms mit −1:

$y^{\mathrm{\prime }}\Rightarrow yy\text{'}\backslash Rightarrow\; y$ $\mathrm{\stackrel{\frown}{=}}{f}_0\stackrel{\wedge}{=}\; f\_0$

$x{}^{\mathrm{\prime }}=xx\{\text{'}\}=x$ einsetzen in $yy$:

$-y=-1,5\cdot {\log }_2(-1-x)-3-y=-1\{,\}5\backslash cdot\; \backslash log\_\{2\}\{(-1-x)-3\}\backslash qquad\backslash qquad$|$\cdot (-1)\backslash color\; \{blue\}\backslash cdot(-1)$

$\Rightarrow \mathbf{y}=\mathbf{1},\mathbf{5}\cdot {\log }_{\mathbf{2}}(-\mathbf{1}-\mathbf{x})+\mathbf{3}\mathbf{\stackrel{\frown}{=}}{\mathbf{f}}_{\mathbf{0}}\backslash Rightarrow\; \backslash mathbf\{\; y=1\{,\}5\backslash cdot\; \backslash log\_\{2\}\{(-1-x)+3\}\}\backslash quad\; \backslash mathbf\{\stackrel{\wedge}{=}f\_0\}$

Die Funktion $f_{0}f\_0$ hat also die Gleichung $\mathbf{y}=\mathbf{1},\mathbf{5}\cdot {\log }_{\mathbf{2}}(-\mathbf{1}-\mathbf{x})+\mathbf{3}\backslash mathbf\{\; y=1\{,\}5\backslash cdot\; \backslash log\_\{2\}\{(-1-x)+3\}\}$