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

 $\begin{array}{l}\\ 11.& \text{Perhatikan gambar berikut}\end{array}$.

$.\begin{array}{l}\\ & \text{Persamaan grafik fungsi kuadrat pada}\\ & \text{pada gambar tersebut di atas dengan}\\ & \text{titik balik}(-2,{\displaystyle \frac{9}{2}})\text{dan melalui}(-1,4)\\ & \text{adalah}....\\ & \begin{array}{lllllll}\text{A}.& {\displaystyle y={\displaystyle \frac{1}{2}(2+4x-x^2)}}& & & & & \\ \text{B}.& {\displaystyle y={\displaystyle \frac{1}{2}(5-4x-x^2)}}& & \\ \text{C}.& y={\displaystyle \frac{1}{2}(5-2x-x^2)}\\ \text{D}.& y=1-4x-x^2\\ \text{E}.& y=5+3x-x^2\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{B}}\\ & \begin{array}{rl}& \text{Diketahui FK}:f(x)=a{x}^2+bx+c,\text{dengan}\\ & \text{koordinat}(x_{ss},y_{ss})=(-2,{\displaystyle \frac{9}{2}})\text{dan melalui}\\ & \text{titik}(-1,4),(-5,0),\text{serta}(1,0),\text{maka}\\ & y=a(x-x_{ss}{)}^2+y_{ss}=a(x-x_1)(x-x_2)\\ & \text{Pilih salah satunya, di sini saya pilih formula}\\ & \text{yang kedua, yaitu}:y=a(x-x_1)(x-x_2)\\ & \iff 4=a(-1-(-5))(-1-1)\iff 4=a(4)(-2)\\ & \iff a=-{\displaystyle \frac{1}{2}}\\ & \text{Selanjutnya kembalikan ke formula semula}\\ & \text{yaitu}:\\ & y=a(x-x_1)(x-x_2)=-{\displaystyle \frac{1}{2}(x-(-5))(x-1)}\\ & \iff y=-{\displaystyle \frac{1}{2}(x+5)(x-1)={\displaystyle \frac{1}{2}(5-4x-x^2)}}\end{array}\end{array}$.

$\begin{array}{l}\\ 12.& \text{Gambar berikut adalah fungsi parabola}\\ & \text{dengan persamaan}y=a{x}^2-4x+k\end{array}$.

$.\begin{array}{l}\\ & \text{Jika nilai minimum}y\text{adalah}-8\text{maka}\\ & \text{nilai}k\text{adalah}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle -5}& & & & & \text{D}.& {\displaystyle 2}\\ \text{B}.& {\displaystyle -4}& & \text{C}.& {\displaystyle -2}& & \text{E}.& 5\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{C}}\\ & \begin{array}{rl}& \text{Diketahui FK}:f(x)=a{x}^2-4x+k,\text{dengan}\\ & \text{koordinat}(x_{ss},y_{ss})=(3,-8)\text{maka}\\ & y=a(x-x_{ss}{)}^2+y_{ss}\\ & -2=a(0-3{)}^2+(-8)\iff -2+8=9.a\\ & 6=9a\iff a={\displaystyle \frac{6}{9}=\frac{2}{3}}\\ & \text{Kembali ke persamaan awal, yaitu}:\\ & y={\displaystyle \frac{2}{3}(x-3{)}^2-8=\frac{2}{3}(x^2-6x+9)-8}\\ & \iff y={\displaystyle \frac{2}{3}{x}^2-4x+6-8={\displaystyle \frac{2}{3}{x}^2-4x-2}}\\ & \text{Jadi, nilai}k=-2\end{array}\end{array}$.

 $\begin{array}{l}\\ 13.& \text{Perhatikan gambar berikut}\end{array}$.

$.\begin{array}{l}\\ & \text{Persamaan grafik fungsi kuadrat pada}\\ & \text{gambar di atas adalah}....\\ & \begin{array}{lllllll}\text{A}.& {\displaystyle y=-(x-3{)}^2-1}& & & & & \\ \text{B}.& {\displaystyle y=-(x-3{)}^2+1}& & \\ \text{C}.& y=-{\displaystyle \frac{1}{3}{(x-3)}^2+1}\\ \text{D}.& y=-{\displaystyle \frac{1}{3}{(x-3)}^2-1}\\ \text{E}.& y=-{\displaystyle \frac{1}{3}{(x+3)}^2-1}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{D}}\\ & \begin{array}{rl}& \text{Diketahui FK dengan koordinat}\\ & (x_{ss},y_{ss})=(3,-1)\text{dan melalui titik}(0,-4)\\ & \text{maka}y=a(x-x_{ss}{)}^2+y_{ss}\\ & -4=a(0-3{)}^2+(-1)\iff -4+1=9.a\\ & -3=9a\iff a=-{\displaystyle \frac{1}{3}}\\ & \text{Kembali ke persamaan awal, yaitu}:\\ & y=-{\displaystyle \frac{1}{3}(x-3{)}^2-1}\end{array}\end{array}$.

 $\begin{array}{l}\\ 14.& \text{Perhatikan gambar berikut}\end{array}$.

$.\begin{array}{l}\\ & \text{Jika grafik fungsi kuadrat di atas adalah}\\ & y=a{x}^2+bx+c,\text{maka hasil kali dari}\\ & a.b.c\text{ adalah}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle -20}& & & & & \text{D}.& {\displaystyle 3}\\ \text{B}.& {\displaystyle -6}& & \text{C}.& {\displaystyle -3}& & \text{E}.& 20\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{C}}\\ & \begin{array}{rl}& \text{Diketahui FK}:a{x}^2+bx+c\\ & (x_{ss},y_{ss})=(-2,-4)\text{dan melalui titik}(0,-6)\\ & \text{maka}y=a(x-x_{ss}{)}^2+y_{ss}\\ & -6=a(0-(-4){)}^2+(-2)\iff -6+2=16.a\\ & -4=16a\iff a=-{\displaystyle \frac{1}{4}}\\ & \text{Kembali ke persamaan awal, yaitu}:\\ & y=-{\displaystyle \frac{1}{4}(x+4{)}^2-2}\\ & =-{\displaystyle \frac{1}{4}{x}^2-2x-6\{\begin{array}{c}a=-{\displaystyle \frac{1}{4}}\\ b=-2\\ c=-6\end{array}}\\ & \text{Sehingga nilai}abc=(-{\displaystyle \frac{1}{4}}).(-2).(-6)=-3\end{array}\end{array}$.

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