$\begin{array}{ll}\\ 1.&\textrm{Sebuah fungsi}\quad y=3x^{\displaystyle 2}-4\quad \textrm{ditranslasikan}\\ &\textrm{oleh}\quad \left( \begin{matrix} 0 \\ 3 \end{matrix} \right),\quad \textrm{maka hasilnya adalah}\, ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle y=3x^{\displaystyle 2}-1\\ \textrm{b}.&\displaystyle y=3x^{\displaystyle 2}+2\\ \textrm{c}.&\displaystyle y=3x^{\displaystyle 2}-18x+23\\ \textrm{d}.&\displaystyle y=4x^{\displaystyle 2}-15x+23\\ \textrm{e}.&\displaystyle y=6x^{\displaystyle 2}+12x+26 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\textrm{Perhatikan bahwa untuk translasi tersebut}\\ &\textrm{di atas, berlaku}\\ &\left( \begin{matrix} x' \\ y' \end{matrix} \right)=\left( \begin{matrix} 0 \\ 3 \end{matrix} \right)+\left( \begin{matrix} x \\ y \end{matrix} \right).\quad \textrm{Selanjutnya}\\ & \left( \begin{matrix} x \\ y \end{matrix} \right)=\left( \begin{matrix} x' \\ y' \end{matrix} \right)-\left( \begin{matrix} 0 \\ 3 \end{matrix} \right)=\left( \begin{matrix} x' \\ y'-3 \end{matrix} \right)\\ &\textrm{Sehingga bayangan garis}\quad y=3x^{\displaystyle 2}-4\\ &\textrm{adalah}:\\ &y=3x^{\displaystyle 2}-4\qquad (\textrm{mula-mula})\\ &(y'-3)=3(x')^{\displaystyle 2}-4\Leftrightarrow y'=3(x')^{2}-1\\ &\textrm{Jadi, bayangan garis}\quad y=3x^{\displaystyle 2}-4\\ &\textrm{adalah}\quad y=3x^{\displaystyle 2}-1 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 2.&\textrm{Diketahui garis}\quad 3x+4y+12=0\quad \textrm{direfleksikan}\\ &\textrm{terhadap sumbu-Y. Hasil dari refleksi garis}\\ &\textrm{tersebut adalah}\, ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 4x-3y-12=0\\ \textrm{b}.&\displaystyle 4x-3y+12=0\\ \textrm{c}.&\displaystyle 3x+4y-12=0\\ \textrm{d}.&\displaystyle 3x-4y+12=0\\ \textrm{e}.&\displaystyle 3x-4y-12=0 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\textrm{Perhatikan bahwa untuk refleksi tersebut}\\ &\textrm{dengan simulasi bayangan sebuah titik}\\ &(x,y)\overset{\displaystyle M_{\textrm{sumbu}-Y}}{\Longrightarrow }(-x,y)=(x',y')\\&\textrm{Selanjutnya}\\ & \left( \begin{matrix} x \\ y \end{matrix} \right)=\left( \begin{matrix} -x' \\ y' \end{matrix} \right)\\ &\textrm{Sehingga bayangan garis}\quad 3x+4y+12=0\\ &\textrm{adalah}:\\ &3x+4y+12=0\qquad (\textrm{mula-mula})\\ &3(-x')+4y'+12=0\Leftrightarrow -3x'+4y'+12=0\\ &\textrm{Jadi, bayangan garis}\quad 3x+4y+12=0\\ &\textrm{adalah}\quad -3y+4y+12=0\quad \textrm{atau}\\ &3x-4y-12=0\quad (\textrm{dikali dengan} -1) \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 3.&\textrm{Bayangan kurva}\quad y=2^{\displaystyle x+1}-4\quad \textrm{jika}\\ &\textrm{direfleksikan terhadap sumbu-X adalah}\, ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle y=2^{\displaystyle x+1}+4\\ \textrm{b}.&\displaystyle y=-2^{\displaystyle x+1}-4\\ \textrm{c}.&\displaystyle y=-2^{\displaystyle x+1}+4\\ \textrm{d}.&\displaystyle y=-2^{\displaystyle -x+1}+4\\ \textrm{e}.&\displaystyle y=-2^{\displaystyle -x+1}-4 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&\textrm{Perhatikan bahwa untuk refleksi tersebut}\\ &\textrm{dengan simulasi bayangan sebuah titik}\\ &(x,y)\overset{\displaystyle M_{\textrm{sumbu}-X}}{\Longrightarrow }(x,-y)=(x',y')\\&\textrm{Selanjutnya}\\ & \left( \begin{matrix} x \\ y \end{matrix} \right)=\left( \begin{matrix} x' \\ -y' \end{matrix} \right)\\ &\textrm{Sehingga bayangan kurva}\quad y=2^{\displaystyle x+1}-4\\ &\textrm{adalah}:\\ &y=2^{\displaystyle x+1}-4\qquad (\textrm{mula-mula})\\ &(-y')=2^{\displaystyle x'+1}-4\Leftrightarrow y'=-2^{\displaystyle x+1}+4\\ &\textrm{Jadi, bayangan kurva}\quad y=2^{\displaystyle x+1}-4\\ &\textrm{adalah}\quad y=-2^{\displaystyle x+1}+4 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 4.&\textrm{Bayangan kurva}\quad y=x^{\displaystyle 3}-6x^{\displaystyle 2}+4x+12\quad \textrm{jika}\\ &\textrm{dirotasikan sebesar}\quad 90^{0}\quad \textrm{searah jarum jam}\\ &\textrm{dengan pusat di}\quad (0,0)\quad \textrm{adalah}\, ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle x=y^{\displaystyle 3}+6y^{\displaystyle 2}+4y+12\\ \textrm{b}.&\displaystyle x=y^{\displaystyle 3}-6y^{\displaystyle 2}-4y+12\\ \textrm{c}.&\displaystyle x=y^{\displaystyle 3}-6y^{\displaystyle 2}-4y-12\\ \textrm{d}.&\displaystyle x=-y^{\displaystyle 3}-6y^{\displaystyle 2}-4y+12\\ \textrm{e}.&\displaystyle x=-y^{\displaystyle 3}+6y^{\displaystyle 2}-4y-12 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\textrm{Diketahui bahwa untuk rotasi tersebut}\\ &\textrm{dengan simulasi bayangan sebuah titik}\\ &(x,y)\overset{\displaystyle R_{\left[ O,90^{0} \right]}}{\Longrightarrow }(x',y')\\&\textrm{dengan matriks rotasinya adalah}:\\ &\begin{pmatrix} cos\: 90^{0} & -sin\: 90^{0} \\ sin\: 90^{0} & cos\: 90^{0} \end{pmatrix}=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\\ &\textrm{Bayangan proses hasilnya adalah}:\\ &\left( \begin{matrix} x' \\ y' \end{matrix} \right)=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\left( \begin{matrix} x \\ y \end{matrix} \right)=\left( \begin{matrix} -y \\ x \end{matrix} \right)\\ &\textrm{Selanjutnya}\\ & \left( \begin{matrix} x \\ y \end{matrix} \right)=\left( \begin{matrix} \displaystyle y' \\ -x' \end{matrix} \right)\\ &\textrm{Sehingga bayangan kurva}\\ &y=x^{\displaystyle 3}-6x^{\displaystyle 2}+4x+12\\ &\textrm{adalah}:\\ &y=x^{\displaystyle 3}-6x^{\displaystyle 2}+4x+12\qquad (\textrm{mula-mula})\\ &-x'=(y')^{\displaystyle 3}-6(y')^{\displaystyle 2}+4(y')+12\\ &\Leftrightarrow x'=-(y')^{\displaystyle 3}+6(y')^{\displaystyle 2}-4(y')-12\\ &\textrm{Jadi, bayangan kurva}\quad y=x^{\displaystyle 3}-6x^{\displaystyle 2}+4x+12\\ &\textrm{adalah}\quad x=-y^{\displaystyle 3}+6y^{\displaystyle 2}-4y-12 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 5.&\textrm{Bayangan kurva}\quad y=6^{\displaystyle x}+9\quad \textrm{jika}\\ &\textrm{didilatasikan sejajar sumbu-X dengan}\\ &\textrm{skala}\quad \displaystyle \frac{1}{3}\quad \textrm{adalah}\, ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle y=6^{\displaystyle \frac{1}{3}x}+3\\ \textrm{b}.&\displaystyle y=6^{\displaystyle \frac{1}{3}x}+9\\ \textrm{c}.&\displaystyle y=6^{\displaystyle 3x}+3\\ \textrm{d}.&\displaystyle y=3^{\displaystyle \frac{1}{3}x}+3\\ \textrm{e}.&\displaystyle y=3^{\displaystyle 3x}+3 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\begin{aligned}&\textrm{Perhatikan bahwa untuk dilatasi tersebut}\\ &\textrm{dengan simulasi bayangan sebuah titik}\\ &(x,y)\overset{\displaystyle D\left[ O,\frac{1}{3} \right]}{\Longrightarrow }(x',y')=(\frac{1}{3}x,\frac{1}{3}y)\\ &\textbf{Catatan}:\textrm{agak beda dengan translasi, refleksi}\\ &\textrm{dan rotasi dalam pengambilan bayangannya}\\\\&\textrm{Karena yang diminta hanya sejajar}\\ &\textrm{dengan sumbu-X, maka hasilnya}\\ &\textrm{adalah}:\quad \left( x',y' \right)=\left( \displaystyle \frac{1}{3}x,y \right)\\ &\textrm{Selanjutnya}\\ & \left( \begin{matrix} x' \\ y' \end{matrix} \right)=\left( \begin{matrix} \displaystyle \frac{1}{3}x \\ y \end{matrix} \right)\\ &\textrm{Sehingga bayangan kurva}\quad y=6^{\displaystyle x}+9\\ &\textrm{adalah}:\\ &y=6^{\displaystyle x}+9\qquad (\textrm{mula-mula})\\ &y'=6^{\displaystyle x'}+9\Leftrightarrow y=6^{\displaystyle \frac{1}{3}x}+9\\ &\textrm{Jadi, bayangan kurva}\quad y=6^{\displaystyle x}+9\\ &\textrm{adalah}\quad y=6^{\displaystyle \frac{1}{3}x}+9 \end{aligned} \end{array}$.
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