Contoh Soal 3 Fungsi Kuadrat (Kelas X/Fase E Semester Genap) Tahun 2024

 $\begin{array}{ll}\\ 11.&\textrm{Perhatikan gambar berikut} \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{Persamaan grafik fungsi kuadrat pada}\\ &\textrm{pada gambar tersebut di atas dengan}\\ &\textrm{titik balik}\: \: \left ( -2,\displaystyle \frac{9}{2} \right )\: \: \textrm{dan melalui}\: (-1,4)\\ &\textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle y=\displaystyle \frac{1}{2}\left (2+4x-x^{2}  \right )&&&&&\\ \textrm{B}.&\color{red}\displaystyle y=\displaystyle \frac{1}{2}\left (5-4x-x^{2}  \right )&&\\ \textrm{C}.&y=\displaystyle \frac{1}{2}\left (5-2x-x^{2}  \right )\\ \textrm{D}.&y=1-4x-x^{2}\\ \textrm{E}.&y=5+3x-x^{2} \end{array}\\\\ &\textbf{Jawab}:\textbf{B}\\ &\begin{aligned}&\textrm{Diketahui FK}:f(x)=ax^{2}+bx+c,\: \: \textrm{dengan}\\ &\textrm{koordinat}\: \: \left ( x_{ss},y_{ss} \right )=\left ( -2,\displaystyle \frac{9}{2} \right )\: \: \textrm{dan melalui}\\ &\textrm{titik}\: \: (-1,4)\: ,\: (\color{red}-5\color{black},0)\: ,\: \textrm{serta}\: \: (\color{red}1\color{black},0),\: \textrm{maka}\\ &y=a(x-x_{ss})^{2}+y_{ss}=a(x-x_{1})(x-x_{2})\\ &\textrm{Pilih salah satunya, di sini saya pilih formula}\\ &\textrm{yang kedua, yaitu}:\color{red}y=a(x-x_{1})(x-x_{2})\\ &\Leftrightarrow 4=a(-1-(-5))(-1-1)\Leftrightarrow 4=a(4)(-2)\\ &\Leftrightarrow a=-\displaystyle \frac{1}{2}\\ &\textrm{Selanjutnya kembalikan ke formula semula}\\ &\textrm{yaitu}:\\ &y=a(x-x_{1})(x-x_{2})=-\displaystyle \frac{1}{2}(x-(-5))(x-1)\\ &\Leftrightarrow y=-\displaystyle \frac{1}{2}(x+5)(x-1)=\color{red}\displaystyle \frac{1}{2}(5-4x-x^{2}) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 12.&\textrm{Gambar berikut adalah fungsi parabola}\\ &\textrm{dengan persamaan}\: \: y=ax^{2}-4x+k \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{Jika nilai minimum}\: \: y\: \: \textrm{adalah}\: \: -8\: \: \textrm{maka}\\ &\textrm{nilai}\: \: k\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle -5&&&&&\textrm{D}.&\displaystyle 2\\ \textrm{B}.&\displaystyle -4&&\textrm{C}.&\color{red}\displaystyle -2&&\textrm{E}.&5 \end{array}\\\\ &\textbf{Jawab}:\textbf{C}\\ &\begin{aligned}&\textrm{Diketahui FK}:f(x)=ax^{2}-4x+k,\: \: \textrm{dengan}\\ &\textrm{koordinat}\: \: \left ( x_{ss},y_{ss} \right )=\left ( 3,-8 \right )\: \: \textrm{maka}\\ &y=a(x-x_{ss})^{2}+y_{ss}\\ &-2=a(0-3)^{2}+(-8)\Leftrightarrow -2+8=9.a\\&6=9a\Leftrightarrow a=\displaystyle \frac{6}{9}=\frac{2}{3}\\ &\textrm{Kembali ke persamaan awal, yaitu}:\\ &y=\displaystyle \frac{2}{3}(x-3)^{2}-8=\frac{2}{3}(x^{2}-6x+9)-8\\ &\Leftrightarrow y=\displaystyle \frac{2}{3}x^{2}-4x+6-8=\color{red}\displaystyle \frac{2}{3}x^{2}-4x-2\\ &\textrm{Jadi, nilai}\: \: \: k=\color{red}-2 \end{aligned} \end{array}$.

 $\begin{array}{ll}\\ 13.&\textrm{Perhatikan gambar berikut} \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{Persamaan grafik fungsi kuadrat pada}\\ &\textrm{gambar di atas adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle y=-(x-3)^{2}-1&&&&&\\ \textrm{B}.&\displaystyle y=-(x-3)^{2}+1&&\\ \textrm{C}.&y=-\displaystyle \frac{1}{3}\left (x-3 \right )^{2}+1\\ \textrm{D}.&\color{red}y=-\displaystyle \frac{1}{3}\left (x-3 \right )^{2}-1\\ \textrm{E}.&y=-\displaystyle \frac{1}{3}\left (x+3 \right )^{2}-1 \end{array}\\\\ &\textbf{Jawab}:\textbf{D}\\ &\begin{aligned}&\textrm{Diketahui FK dengan koordinat}\\ &\left ( x_{ss},y_{ss} \right )=\left ( 3,-1 \right )\: \: \textrm{dan melalui titik}\: \: (0,-4)\\ &\textrm{maka}\: \: y=a(x-x_{ss})^{2}+y_{ss}\\ &-4=a(0-3)^{2}+(-1)\Leftrightarrow -4+1=9.a\\&-3=9a\Leftrightarrow a=-\displaystyle \frac{1}{3}\\ &\textrm{Kembali ke persamaan awal, yaitu}:\\ &y=\color{red}-\displaystyle \frac{1}{3}(x-3)^{2}-1 \end{aligned} \end{array}$.

 $\begin{array}{ll}\\ 14.&\textrm{Perhatikan gambar berikut} \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{Jika grafik fungsi kuadrat di atas adalah}\\ &y=ax^{2}+bx+c,\: \textrm{maka hasil kali dari}\\ &a.b.c\: \: \textrm{ adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle -20&&&&&\textrm{D}.&\displaystyle 3\\ \textrm{B}.&\displaystyle -6&&\textrm{C}.&\color{red}\displaystyle -3&&\textrm{E}.&20 \end{array}\\\\ &\textbf{Jawab}:\textbf{C}\\ &\begin{aligned}&\textrm{Diketahui FK}:ax^{2}+bx+c\\ &\left ( x_{ss},y_{ss} \right )=\left ( -2,-4 \right )\: \: \textrm{dan melalui titik}\: \: (0,-6)\\ &\textrm{maka}\: \: y=a(x-x_{ss})^{2}+y_{ss}\\ &-6=a(0-(-4))^{2}+(-2)\Leftrightarrow -6+2=16.a\\&-4=16a\Leftrightarrow a=-\displaystyle \frac{1}{4}\\ &\textrm{Kembali ke persamaan awal, yaitu}:\\ &y=\color{red}-\displaystyle \frac{1}{4}(x+4)^{2}-2\\ &=-\displaystyle \frac{1}{4}x^{2}-2x-6\left\{\begin{matrix} a=-\displaystyle \frac{1}{4}\\  b=-2\\  c=-6 \end{matrix}\right.\\ &\textrm{Sehingga nilai}\: \: abc=\left ( -\displaystyle \frac{1}{4} \right ).(-2).(-6)=\color{red}-3 \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Grafik fungsi kuadrat}\: \: y=f(x)=x^{2}\\ &\textrm{digeser 1 satuan ke kanan dan dilanjutkan}\\ &\textrm{1 satuan ke atas. Persamaan parabola }\\ &\textrm{yang baru adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle y+1=(x+1)^{2}&&&&&\\ \textrm{B}.&\displaystyle y+1=x^{2}+1&&\\ \textrm{C}.&y-1=x^{2}-1\\ \textrm{D}.&\color{red}y-1=(x-1)^{2}\\ \textrm{E}.&y=(x+1)^{2}+1 \end{array}\\\\ &\textbf{Jawab}:\textbf{D}\\ &\textrm{Sebagai pedoman bantuan, suatu fungsi}\\ &\textrm{di geser ke kanan berarti}:x-1\: \: \textrm{dan}\\ &\textrm{digeser ke atas berarti}:y-1\\ &\textbf{Catatan}:\\ &\textrm{Andai digeser ke kiri 1 kemudian ke bawah 1,}\\ &\textrm{maka garfik akan menjadi}:y+1=(x+1)^{2}\\&\textrm{Berikut ilustrasi grafiknya}  \end{array}$.