Contoh Soal 6 Fungsi Kuadrat (Kelas X/Fase E Semester Genap) Tahun 2024

 $\begin{array}{l}\\ 26.& \text{Misalkan}F={\displaystyle \frac{6x^2+16x+3m}{6}\text{merupakan kuadrat}}\\ & \text{dari bentuk linear terhadap}x(ax+b).\text{Nilai}\\ & m\text{yang memungkinkan kondisi tersebut terjadi}\\ & \text{terletak di antara}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle -6\text{dan}-5}& & & & & \text{D}.& {\displaystyle \text{4 dan 5}}\\ \text{B}.& {\displaystyle -4\text{dan}-3}& & & {\displaystyle }& & \text{E}.& {\displaystyle \text{5 dan 6}}\\ \text{C}.& {\displaystyle 3\text{dan}4}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{C}}\\ & \begin{array}{rl}& \text{Diketahui FK}:{\displaystyle \frac{6x^2+16x+3m}{6}}\\ & \text{merupakan bentuk kuadrat, maka}\\ & f(x)={\displaystyle \frac{6x^2+16x+3m}{6}=x^2+{\displaystyle \frac{8}{3}x+\frac{1}{2}m}}\\ & \text{dengan}a=1,b={\displaystyle \frac{8}{3},\text{dan}c={\displaystyle \frac{1}{2}m}}\\ & \text{dan istilah bentuk linier mengarah kepada}\\ & \text{jenis akarnya real, maka}D=b^2-4ac>0\\ & {\left({\displaystyle \frac{8}{3}}\right)}^2-4.1.\left({\displaystyle \frac{1}{2}m}\right)>0\\ & \iff {\displaystyle \frac{64}{9}-2m>0}\\ & n\to \iff {\displaystyle \frac{32}{9}-m>0(\text{dikali})-1}\\ & m-{\displaystyle \frac{32}{9}>0}\\ & \iff m>{\displaystyle \frac{32}{9}\iff m>3{\displaystyle \frac{5}{9}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 27.& \text{Ekspresi}21x^2+ax+21\text{dapat dituliskan}\\ & \text{sebagai perkalian dua bentuk linear dengan}\\ & \text{koefisien bilangan bulat. Hal tersebut hanya}\\ & \text{dapat dilakukan jika}a\text{adalah}....\\ & \begin{array}{lllllll}\text{A}.& {\displaystyle \text{bilangan 0}}& & & & & \\ \text{B}.& {\displaystyle \text{beberapa bilangan ganjil}}& & \\ \text{C}.& \text{beberapa bilangan genap}\\ \text{D}.& \text{sembarang bilangan ganjil}\\ \text{E}.& \text{sembarang bilangan genap}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{C}}\\ & \text{Ekspresi dapat dituliskan dalam perkalian dua}\\ & \text{bentuk linear, maka}D=b^2-4ac\ge 0\\ & \text{Dari}21x^2+ax+21\\ & a=21,b=a,c=21,\text{maka}\\ & a^2-4.21.21\ge 0\\ & \iff a^2-{42}^2\ge 0\\ & \iff (a+42)(a-42)\ge 0\\ & \iff a\le -42ataua\ge 42\end{array}$.

Hubungan Fungsi Kuadrat dan Sebuah garis linear

$\begin{array}{l}\\ 28.& \text{Garis}2x+y=p\text{akan memotong parabola}\\ & 4x^2-y=0\text{di dua titik apabila nilai}p=....\\ & \begin{array}{llllllll}\text{A}.& p<-{\displaystyle \frac{1}{4}}& & & & & \text{D}.& {\displaystyle p<-\frac{3}{4}}\\ \text{B}.& {\displaystyle p>-{\displaystyle \frac{1}{4}}}& & & {\displaystyle }& & \text{E}.& {\displaystyle p<-1}\\ \text{C}.& {\displaystyle p<{\displaystyle \frac{1}{4}}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{B}}\\ & \begin{array}{rl}& \text{Diketahui bahwa garis}2x+y=p\Rightarrow y=p-2x\\ & \text{memotong parabola}4x^2-y=0\text{di dua titik}\\ & \text{berbeda, maka}\\ & 4x^2-y=0\iff 4x^2-(p-2x)=0\\ & 4x^2+2x-p=0,\text{dengan}a=4,b=2,c=-p\\ & \text{Syarat memotong di dua titik berbeda adalah}:\\ & D=b^2-4ac>0\\ & \iff 2^2-4.4.(-p)>0\\ & \iff 4+16p>0\iff 16p>-4\iff p>-{\displaystyle \frac{4}{16}}\\ & \iff p>-{\displaystyle \frac{1}{4}}\end{array}\end{array}$.

$\begin{array}{l}\\ 29.& \text{Jika}a{x}^2+bx+c=0\text{tidak mempunyai akar real}\\ & \text{maka grafik fungsi}y=a{x}^2+bx+c\text{akan}\\ & \text{menyinggung garis}y=x\text{apabila}....\\ & \begin{array}{llllllll}\text{A}.& b<{\displaystyle \frac{1}{2}}& & & & & \text{D}.& {\displaystyle b>2}\\ \text{B}.& {\displaystyle b>{\displaystyle \frac{1}{2}}}& & & {\displaystyle }& & \text{E}.& {\displaystyle 1<b<2}\\ \text{C}.& {\displaystyle b>1}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{A}}\\ & \begin{array}{rl}& \text{Diketahui sebuah kurva parabola}\\ & y=a{x}^2+bx+c(\text{tidak punya akar real})\\ & \text{dan}\text{garis}y=x\text{saling bersinggungan, }\\ & \text{sehingga}y=y\\ & \iff a{x}^2+bx+c=x\\ & \iff a{x}^2+bx-x+c=0\\ & \iff a{x}^2+(b-1)x+c=0\\ & \text{dengan}a=a,b=b-1,c=c\\ & \text{Selanjutnya syarat menyinggung adalah}:\\ & D=b^2-4ac=0\\ & \iff (b-1{)}^2-4ac=0\iff (b-1{)}^2=4ac.\\ & \text{Karena}a{x}^2+bx+c(\text{tidak punya akar real})\\ & \text{maka}D<0\iff b^2-4ac<0\\ & \iff b^2-(b-1{)}^2<0\iff b^2-(b^2-2b+1)<0\\ & \iff 2b-1<0\iff b<{\displaystyle \frac{1}{2}}\end{array}\end{array}$.

$\begin{array}{l}\\ 30.& \text{Grafik}y=2x^2+ax+b\text{berpotongan dengan}\\ & \text{garis}y=3x-1\text{di titik}(x_1,y_1)\text{dan}(x_2,y_2)\\ & \text{Jika}{x}_1+x_2=3\text{dan}{x}_1\times x_2=4,\text{maka}\\ & a+b=....\\ & \begin{array}{llllllll}\text{A}.& 1& & & & & \text{D}.& {\displaystyle 4}\\ \text{B}.& {\displaystyle 2}& & & {\displaystyle }& & \text{E}.& {\displaystyle 5}\\ \text{C}.& {\displaystyle 3}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{D}}\\ & \end{array}$

Contoh Soal 5 Fungsi Kuadrat (Kelas X/Fase E Semester Genap) Tahun 2024

$\begin{array}{l}\\ 21.& \text{Jika}y=\sqrt{x^2-2x+1}+\sqrt{x^2+2x+1}\\ & \text{maka nilai}y=....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle {\displaystyle 2x}}& & & & & \text{D}.& {\displaystyle |x-1|+|x+1|}\\ \text{B}.& {\displaystyle 2(x+1)}& & \text{C}.& {\displaystyle 0}& & \text{E}.& {\displaystyle \text{tidak ada yang benar}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{D}}\\ & \text{Diketahui}y=\sqrt{x^2-2x+1}+\sqrt{x^2+2x+1}\\ & \iff y=\sqrt{(x-1{)}^2}+\sqrt{(x+1{)}^2}\\ & \iff y=|x-1|+|x+1|\end{array}$.

$\begin{array}{l}\\ 22.& \text{Jika}x\text{bilangan real dan}4y^2+4xy+x+6=0,\\ & \text{maka nilai semua nilai}x\text{agar nilai}y\text{real adalah}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle x\le -2\text{atau}x\ge 3}& & & & & \text{D}.& {\displaystyle -3\le x\le 2}\\ \text{B}.& {\displaystyle x\le 2\text{atau}x\ge 3}& & & {\displaystyle }& & \text{E}.& {\displaystyle -2\le x\le 3}\\ \text{C}.& {\displaystyle x\le -3\text{atau}x\ge 2}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{A}}\\ & \begin{array}{rl}& 4y^2+(4x)y+(x+6)=0\\ & \text{agar}y\text{real, maka}D=b^2-4ac\ge 0\\ & (4x{)}^2-4(4)(x+6)\ge 0\\ & \iff 16x^2-16(x+6)\ge 0\\ & \iff x^2-x-6\ge 0\\ & \iff (x-3)(x+2)\ge 0\\ & \iff {\displaystyle x\le -2\text{atau}x\ge 3}\end{array}\\ \\ & \text{Jika kita diletakkan dalam garis bilangan}\\ & \text{akan tampak seperti berikut}\\ \\ & \begin{array}{rl}& \underset{\begin{array}{c}\Downarrow \\ \Downarrow \end{array}}{\text{Daerah Positif}}\\ & \begin{array}{c}\\ & & & & & & \\ ++& ++& --& --& --& ++& ++\\ & -{\displaystyle 2}& & & & 3& & \mathbf{\text{Sumbu-X}}\end{array}\\ & \begin{array}{c}\text{dengan}\\ \text{titik}\\ \text{uji}\\ x=0\end{array}\stackrel{\begin{array}{c}\Uparrow \\ \Uparrow \end{array}}{\text{Daerah negatif}}\stackrel{\begin{array}{c}\Uparrow \\ \Uparrow \end{array}}{\text{Daerah positif}}\end{array}\end{array}$.

$\begin{array}{l}\\ 23.& \text{Diketahui}f(x)={\displaystyle \frac{x(x-1)}{2},\text{nilai}f(x+2)=....}\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle f(x)+f(2)}& & & & & \text{D}.& {\displaystyle \frac{xf(x)}{x+2}}\\ \text{B}.& {\displaystyle (x+2)f(x)}& & & {\displaystyle }& & \text{E}.& {\displaystyle {\displaystyle \frac{(x+2)f(x+1)}{x}}}\\ \text{C}.& {\displaystyle x(x+2)f(x)}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{E}}\\ & f(x)={\displaystyle \frac{x(x-1)}{2},}\\ & f(x+1)={\displaystyle \frac{(x+1)(x)}{2}\iff {\displaystyle \frac{x+1}{2}=\frac{f(x+1)}{x},\text{maka}}}\\ & \begin{array}{rl}f(x+2)& ={\displaystyle \frac{(x+2)((x+2)-1)}{2}=\frac{(x+2)(x+1)}{2}}\\ & ={\displaystyle \frac{f(x+1)}{x}.(x+2)={\displaystyle {\displaystyle \frac{(x+2)f(x+1)}{x}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 24.& \text{Jika}{r}_1\text{dan}{r}_2\text{merupakan dua penyelesaian }\\ & \text{dari persamaan}{x}^2+px+8=0,\text{maka}....\\ & \begin{array}{llllll}\text{A}.& {\displaystyle |r_1+r_2|>4\sqrt{2}}& & & \text{D}.& {\displaystyle r_1<0\text{dan}{r}_2<0}\\ \text{B}.& {\displaystyle \left|r_1\right|>3\text{dan}\left|r_2\right|>3}& {\displaystyle }& & \text{E}.& {\displaystyle -|r_1+r_2|<4\sqrt{2}}\\ \text{C}.& {\displaystyle \left|r_1\right|>2\text{dan}\left|r_2\right|>2}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{A}}\\ & \begin{array}{rl}& \text{Diketahui PK}:x^2+px+8=0\\ & \text{dengan}\{\begin{array}{c}{r}_1\\ r_2\end{array}\text{akar-akarnya, maka}\\ & \bullet r_1+r_2=-p\\ & \bullet r_1\times r_2=8\\ & \text{Asumsikan kedua akarnya real dan berbeda,}\\ & \text{sehingga kita pilih nilai}D=b^2-4ac>0\\ & \iff (-p{)}^2-4.1.8>0\\ & \iff p^2-32>0\\ & \iff (|r_1+r_2|+\sqrt{32})(|r_1+r_2|-\sqrt{32})>0\\ & \iff |r_1+r_2|<-\sqrt{32}\text{atau}|r_1+r_2|>\sqrt{32}\\ & \stackrel{\begin{array}{c}\Downarrow \\ \Downarrow \end{array}}{\mathbf{\text{tidak mungkin}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 25.& \text{Jika titik}(1,y_1)\text{dan}(-1,y_2)\text{terletak pada}\\ & \text{grafik}y=a{x}^2+bx+c\text{dan}{y}_1-y_2=-6\\ & \text{maka nilai}b=....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle {\displaystyle -3}}& & & & & \text{D}.& {\displaystyle \sqrt{ac}}\\ \text{B}.& {\displaystyle 0}& & \text{C}.& {\displaystyle 3}& & \text{E}.& {\displaystyle {\displaystyle \frac{a+c}{2}}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{D}}\\ & \text{Diketahui}y=a{x}^2+bx+c,\text{untuk}\\ & \bullet (1,y_1)\Rightarrow y_1=a+b+c\\ & \bullet (-1,y_1)\Rightarrow y_2=a-b+c\\ & \mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}-\\ & y_1-y_2=2b=-6\Rightarrow b=-3\end{array}$.

Contoh Soal 4 Fungsi Kuadrat (Kelas X/Fase E Semester Genap) Tahun 2024

$\begin{array}{l}\\ 16.& \text{Batas nilai}m\text{yang memenuhi agar fungsi}\\ & \text{kuadrat}y=m{x}^2+(m+2)x+m\\ & \text{memotong sumbu-X di dua titik yang}\\ & \text{berbeda adalah}....\\ & \begin{array}{lllllll}\text{A}.& {\displaystyle -{\displaystyle \frac{2}{3}<m<2}}& & & & & \\ \text{B}.& {\displaystyle -2<m<{\displaystyle \frac{2}{3}}}& & \\ \text{C}.& m<-2\text{atau}m>2\\ \text{D}.& m<-2\text{atau}m>{\displaystyle \frac{2}{3}}\\ \text{E}.& m<-{\displaystyle \frac{2}{3}\text{atau}m>2}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{A}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}y=m{x}^2+(m+2)x+m\\ & \text{dengan}a=m,b=m+2,\mathrm{\&}c=m\\ & \text{memotong sumbu-X di dua titik berbeda}\\ & \text{hal ini artinya nilai Diskriminan }D>0\\ & D=b^2-4ac>0\\ & (m+2{)}^2-4m.m>0\\ & \iff m^2+4m+4-4m^2>0\\ & \iff -3m^2+4m+4>0(\text{dikali}-1)\\ & \iff 3x^2-4m-4<0\\ & \iff (m-2)(3m+2)<0\\ & \iff -{\displaystyle \frac{2}{3}<m<2}\\ & \mathbf{\text{Anda bisa menggunakan titik uji dulu}}\\ & \text{untuk memastikannya wilayah yg dimaksud}\\ & \text{misalkan pilih}m=0\\ & \text{lalu kita ujikan, yaitu}:\\ & m=0\Rightarrow (0-2)(3.0+2)=-4<0(\mathbf{\text{benar}})\\ \\ & \text{Jika kita diletakkan dalam garis bilangan}\\ & \text{akan tampak seperti berikut}\\ \\ & \begin{array}{rl}& \underset{\begin{array}{c}\Downarrow \\ \Downarrow \end{array}}{\text{Daerah Positif}}\\ & \begin{array}{c}\\ & & & & & & \\ ++& ++& --& --& --& ++& ++\\ & -{\displaystyle \frac{2}{3}}& & & & 2& & \mathbf{\text{Sumbu}}-m\end{array}\\ & \begin{array}{c}\text{dengan}\\ \text{titik}\\ \text{uji}\\ m=0\end{array}\stackrel{\begin{array}{c}\Uparrow \\ \Uparrow \end{array}}{\text{Daerah negatif}}\stackrel{\begin{array}{c}\Uparrow \\ \Uparrow \end{array}}{\text{Daerah positif}}\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 17.& \text{Batas nilai}m\text{yang menyebabkan fungsi kuadrat}\\ & y=(m-1)x^2-2(m-1)x+(2m+1)\\ & \text{definit positif adalah}....\\ & \begin{array}{lllllll}\text{A}.& {\displaystyle m>-2}& & & & & \\ \text{B}.& {\displaystyle m>1}& & \\ \text{C}.& m<1\\ \text{D}.& -2<m<1\\ \text{E}.& m<-{\displaystyle 2\text{atau}m>1}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{B}}\\ & \begin{array}{rl}& \text{Diketahui FK}:\\ & y=(m-1)x^2-2(m-1)x+(2m+1)\\ & \text{Syarat fungsi kuadrat definit positif}:\\ & \bullet a>0\\ & \bullet D=b^2-4ac<0\\ & \text{maka}\\ & a=m-1>0\iff m>1,\text{dan}\\ & D=(2(m-1){)}^2-4(m-1)(2m+1)<0\\ & \iff 4(m-1{)}^2-(m-1)(8m+4)<0\\ & \iff (m-1)(4m-4)-(m-1)(8m+4)<0\\ & \iff (m-1)(4m-4-8m-4)<0\\ & \iff (m-1)(-4m-8)<0,\mathbf{\text{dibagi}}-4\\ & \iff (m-1)(m+2)>0\end{array}\\ \\ & \text{Jika kita diletakkan dalam garis bilangan}\\ & \text{akan tampak seperti berikut}\\ \\ & \begin{array}{rl}& \underset{\begin{array}{c}\Downarrow \\ \Downarrow \end{array}}{\text{Daerah Positif}}\\ & \begin{array}{c}\\ & & & & & & \\ ++& ++& --& --& --& ++& ++\\ & -{\displaystyle 2}& & & & 1& & \mathbf{\text{Sumbu}}-m\end{array}\\ & \begin{array}{c}\text{dengan}\\ \text{titik}\\ \text{uji}\\ m=0\end{array}\stackrel{\begin{array}{c}\Uparrow \\ \Uparrow \end{array}}{\text{Daerah negatif}}\stackrel{\begin{array}{c}\Uparrow \\ \Uparrow \end{array}}{\text{Daerah positif}}\end{array}\end{array}$.

$\begin{array}{l}\\ 18.& \text{Luas}L\text{suatu segitiga}ABC\text{diketahui}\\ & x(7-x){\text{cm}}^2.\text{Luas maksimum segitiga}\\ & \text{tersebut adalah}...{\text{cm}}^2\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle 3{\displaystyle \frac{1}{2}}}& & & & & \text{D}.& {\displaystyle 10{\displaystyle \frac{1}{4}}}\\ \\ \text{B}.& {\displaystyle 5{\displaystyle \frac{1}{2}}}& & \text{C}.& {\displaystyle 7{\displaystyle \frac{1}{4}}}& & \text{E}.& {\displaystyle 12{\displaystyle \frac{1}{4}}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{E}}\\ & \text{Diketahui}\left[ABC\right]=x(7-x)=7x-x^2\\ & \text{dengan}a=-1,b=7,\text{dan}c=0\\ & \text{akan}\mathbf{\text{maksimum}},\text{saat}(x_{ss},f(x_{ss}))\\ & \text{yaitu}:\\ & x_{ss}=-{\displaystyle \frac{b}{2a}=-\frac{7}{2.(-1)}=\frac{7}{2},\text{maka}}\\ & f\left({\displaystyle \frac{7}{2}}\right)=7\left({\displaystyle \frac{7}{2}}\right)-{\left({\displaystyle \frac{7}{2}}\right)}^2={\displaystyle \frac{49}{2}-\frac{49}{4}={\displaystyle \frac{49}{4}}}\end{array}$.

$\begin{array}{l}\\ 19.& \text{Perhatikan gambar persegi ABCD dengan}\\ & \text{panjang sisinya 10 cm}\end{array}$.

$.\begin{array}{l}\\ & \text{Jika}BP=DQ=x\text{cm, maka luas }\\ & \text{maksimum segitiga}APQ\text{adalah}...{\text{cm}}^2\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle 35}& & & & & \text{D}.& {\displaystyle 75}\\ \text{B}.& {\displaystyle 50}& & \text{C}.& {\displaystyle 60}& & \text{E}.& 80\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{B}}\\ & \text{Diketahui bahwa}\\ & \begin{array}{rl}\left[APQ\right]& =L_{◻ABCD}-\left[ADQ\right]-\left[QCP\right]-\left[APB\right]\\ & ={10}^2-{\displaystyle \frac{1}{2}x.10-\frac{1}{2}.(10-x{)}^2-\frac{1}{2}x.10}\\ & =100-10x-{\displaystyle \frac{1}{2}(10-x{)}^2}\\ & =10(10-x)-{\displaystyle \frac{1}{2}(10-x{)}^2}\\ & =(10-x)(10-{\displaystyle \frac{1}{2}(10-x)})\\ & =(10-x)(5+{\displaystyle \frac{1}{2}x})\\ & =50-{\displaystyle \frac{1}{2}{x}^2}\end{array}\\ & x_{ss}=-{\displaystyle \frac{b}{2a}=-{\displaystyle \frac{0}{2.{\displaystyle \frac{1}{2}}}=0,\text{maka nilai}}}\\ & f(x_{ss})=f(0)=50-{\displaystyle \frac{1}{2}{.0}^2=50}\end{array}$.

$\begin{array}{l}\\ 20.& \text{Fungsi}f(x)=a(b-c)x^2+b(c-a)x+c(a-b)\\ & \text{akan memotong sumbu-X di titik}(1,0)\text{dan}\\ & \text{memenuhi}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle {\displaystyle \frac{b(c-a)}{a(b-c)}}}& & & & & \text{D}.& {\displaystyle {\displaystyle \frac{c(a-b)}{a(b-c)}}}\\ \\ \text{B}.& {\displaystyle {\displaystyle \frac{a(b-c)}{c(a-b)}}}& & \text{C}.& {\displaystyle {\displaystyle \frac{a(b-c)}{b(c-a)}}}& & \text{E}.& {\displaystyle {\displaystyle \frac{c(a-b)}{b(c-a)}}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{D}}\\ & \text{Diketahui bahwa FK}:\\ & f(x)=a(b-c)x^2+b(c-a)x+c(a-b)\\ & \text{maka nilai}\\ & \bullet x_1+x_2=-{\displaystyle \frac{b(c-a)}{a(b-c)}={\displaystyle \frac{b(a-c)}{a(b-c)}}}\\ & \bullet x_1\times x_2={\displaystyle \frac{c(a-b)}{a(b-c)}}\end{array}$

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

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

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

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

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

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

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

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

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

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

Contoh Soal 2 Fungsi Kuadrat (Kelas X/Fase E Semester Genap) Tahun 2024

 $\begin{array}{l}\\ 6.& \text{Fungsi kuadrat}f(x)=(2x+p{)}^2+q\\ & \text{dengan titik balik minimum}(-1,3).\\ & \text{Nilai}p+q\text{adalah}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle 2}& & & & & \text{D}.& {\displaystyle 6}\\ \text{B}.& {\displaystyle 4}& & \text{C}.& {\displaystyle 5}& & \text{E}.& 7\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{C}}\\ & \begin{array}{rl}& \text{Diketahui FK}:f(x)=(2x+p{)}^2+q,\text{dengan}\\ & \text{titik}(x_{ss},y_{ss})=(-1,3),\text{maka}\\ & f(x)=4x^2+4px+p^2+q\text{dengan}\\ & x_{ss}=-1=-{\displaystyle \frac{b}{2a}\iff 1={\displaystyle \frac{4p}{2.4}\iff p=2}}\\ & \text{Selanjutnya}\\ & f(-1)=(2.(-1)+2{)}^2+q=3\iff q=3\\ & \text{maka}\\ & p+q=2+3=5\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Jika grafik fungsi kuadrat}f(x)=a{x}^2+x+c\\ & \text{dengan titik balik minimum}(-1,3)\text{dan melalui}\\ & (2,12)\text{maka}a+b+c\text{sama dengan}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle 7}& & & & & \text{D}.& {\displaystyle 13}\\ \text{B}.& {\displaystyle 9}& & \text{C}.& {\displaystyle 11}& & \text{E}.& 15\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{A}}\\ & \begin{array}{rl}& \text{Diketahui FK}:f(x)=a{x}^2+bx+c,\text{dengan}\\ & \text{titik}(x_{ss},y_{ss})=(-1,3)\text{dan melalui titik}\\ & (2,12),\text{maka}\\ & \overline{\begin{array}{cc}\begin{array}{rl}& 12=4a+2b+c\\ & 3=a-b+c\\ & \mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}-\\ & 9=3a+3b\\ & \iff 3=a+b\end{array}& \begin{array}{rl}& -{\displaystyle \frac{b}{2a}=-1,\text{maka}}\\ & 2a-b=0,\text{dan ingat}\\ & a+b=3\\ & \mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}+\\ & 3a=3\\ & \iff a=1,\text{maka}b=2\end{array}\end{array}}\\ & \text{dan}\\ & a-b+c=3\Rightarrow 1-2+c=3\Rightarrow c=4\\ & \text{Jadi, nilai}a+b+c=1+2+4=7\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Nilai minimum grafik fungsi}f(x)=a{x}^2-2x+8\\ & \text{adalah 5. Nilai }6a\text{sama dengan}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle 1}& & & & & \text{D}.& {\displaystyle 9}\\ \text{B}.& {\displaystyle 2}& & \text{C}.& {\displaystyle 4}& & \text{E}.& 12\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{B}}\\ & \begin{array}{rl}& \text{Diketahui FK}:f(x)=a{x}^2-2x+8,\text{dengan}\\ & \text{titik}(x_{ss},y_{ss})=(-{\displaystyle \frac{b}{2a},5})\text{maka}\\ & \bullet x_{ss}=-{\displaystyle \frac{b}{2a}=-\frac{-2}{2a}=\frac{1}{a}}\\ & \bullet y_{ss}=f\left(x_{ss}\right)=a{\left({\displaystyle \frac{1}{a}}\right)}^2-2\left({\displaystyle \frac{1}{a}}\right)+8=5\\ & \iff {\displaystyle \frac{1}{a}-\frac{2}{a}=5-8\iff -{\displaystyle \frac{1}{a}=-3}}\\ & \iff a={\displaystyle \frac{1}{3}}\\ & \text{maka nilai}6a=6\left({\displaystyle \frac{1}{3}}\right)=2\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Jika kurva fungsi}f(x)=x^2+bx+c\\ & \text{memotong sumbu-X di}(1,0)\text{dan}(5,0),\\ & \text{maka nilai}{b}^2-c^2\text{sama dengan}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle -11}& & & & & \text{D}.& {\displaystyle 11}\\ \text{B}.& {\displaystyle -3}& & \text{C}.& {\displaystyle 6}& & \text{E}.& 13\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{D}}\\ & \begin{array}{rl}& \text{Diketahui FK}:f(x)=x^2+bx+c,\text{memotong}\\ & \text{sumbu-X di}(1,0)\mathrm{\&}(5,0)\text{artinya}{x}_1=1\mathrm{\&}{x}_2=5\\ & \text{maka}\\ & x_{ss}={\displaystyle \frac{-b}{2.1}={\displaystyle \frac{x_1+x_2}{2}=\frac{1+5}{2}\iff b=-6}}\\ & \text{dan kita juga memiliki}\\ & f(1)=1+b+c=0\Rightarrow c=-b-1=6-1=5\\ & \text{Sehingga}\\ & b^2-c^2=(-6{)}^2-5^2=36-25=11\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Perhatikan gambar berikut}\end{array}$.

$.\begin{array}{l}\\ & \text{Persamaan grafik fungsi kuadrat pada}\\ & \text{pada gambar tersebut di atas adalah}....\\ & \begin{array}{lllllll}\text{A}.& {\displaystyle y=6x^2-12x+18}& & & & & \\ \text{B}.& {\displaystyle y=6x^2+12x+16}& & \\ \text{C}.& y=6x^2-24x+17\\ \text{D}.& y=6x^2-24x+19\\ \text{E}.& y=6x^2-24x+29\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{D}}\\ & \begin{array}{rl}& \text{Diketahui FK}:f(x)=a{x}^2+bx+c,\text{dengan}\\ & \text{koordinat}(x_{ss},y_{ss})=(2,-5)\text{dan melalui}\\ & \text{titik}(3,1),\text{maka}\\ & y=a(x-x_{ss}{)}^2+y_{ss}\iff 1=a(3-2{)}^2+(-5)\\ & \iff 6=a{.1}^2\iff a=6\\ & \text{maka persamaan fungsi kuadratnya adalah}:\\ & y=a(x-x_{ss}{)}^2+y_{ss}\\ & \iff y=6(x-2{)}^2-5\iff y=6(x^2-4x+4)-5\\ & \iff y=6x^2-24x+19\end{array}\end{array}$

Contoh Soal 1 Fungsi Kuadrat (Kelas X/Fase E Semester Genap) Tahun 2024

$\begin{array}{l}\\ 1.& \text{Diketahui fungsi}f(x)=x^2-2x-15.\text{Jika}\\ & \text{domain}\{x|-4\le x\le 2,x\in \mathbb{R}\},\text{maka}\\ & range\text{-nya adalah}....\\ & \begin{array}{lllll}\text{A}.& -15\le f(x)\le 20& & & \\ \text{B}.& -15\le f(x)\le 9& & \\ \text{C}.& {\displaystyle -16\le f(x)\le 9}& & \\ \text{D}.& {\displaystyle -16\le f(x)\le 20}& & \\ \text{E}.& -15\le f(x)\le 5& & \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{C}}\\ & \begin{array}{rl}& \text{Diketahui FK}:f(x)=x^2-2x-15,\text{dengan}\\ & {\text{D}}_f=\{x|-4\le x\le 2,x\in \mathbb{R}\},\\ & \text{maka}range\text{fungsinya adalah}{\text{R}}_f,\text{di mana}\\ & {\text{R}}_f\text{diperoleh dengan cara di antaranya}\\ & \text{mensubstitusikan langsung ke fungsinya, yaitu}:\\ & f(-4)=(-4{)}^2-2(-4)-15=9\\ & f(-3)=(-3{)}^2-2(-3)-15=0\\ & f(-2)=(-2{)}^2-2(-2)-15=-7\\ & f(-1)=(-1{)}^2-2(-1)-15=-12\\ & f(0)=(0{)}^2-2(0)-15=-15\\ & f(1)=(1{)}^2-2(1)-15=-16\\ & f(2)=(2{)}^2-2(2)-15=-15\\ & \text{Jadi, range fungsinya}:{\text{R}}_f={\displaystyle -16\le f(x)\le 9}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Daerah hasil fungsi}f(x)=-x^2+6x-5\text{untuk}\\ & \text{daerah asal}\{x|-1\le x\le 6,x\in \mathbb{R}\}\text{dan}\\ & y=f(x)\text{adalah}....\\ & \begin{array}{lllll}\text{A}.& \{y|-5\le y\le 0,y\in \mathbb{R}\}& & & \\ \text{B}.& \{y|-12\le y\le 4,y\in \mathbb{R}\}& & \\ \text{C}.& {\displaystyle \{y|-4\le y\le 1,y\in \mathbb{R}\}}& & \\ \text{D}.& {\displaystyle \{y|-5\le y\le 4,y\in \mathbb{R}\}}& & \\ \text{E}.& \{y|-1\le y\le 6,y\in \mathbb{R}\}& & \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{B}}\\ & \begin{array}{rl}& \text{Masih sama dengan cara di atas. Diketahui FK}:\\ & f(x)=-x^2+6x-5,\text{dengan}\\ & {\text{D}}_f=\{x|-1\le x\le 6,x\in \mathbb{R}\},\\ & \text{maka}range\text{fungsinya adalah}{\text{R}}_f,\text{di mana}\\ & {\text{R}}_f\text{diperoleh dengan cara di antaranya}\\ & \text{mensubstitusikan langsung ke fungsinya, yaitu}:\\ & f(-1)=-(-1{)}^2+6(-1)-5=-12\\ & f(0)=-(0{)}^2+6(0)-5=-5\\ & f(1)=-(1{)}^2+6(1)-5=0\\ & f(2)=-(2{)}^2+6(2)-5=3\\ & f(3)=-(3{)}^2+6(3)-5=4\\ & f(4)=-(4{)}^2+6(1)-5=3\\ & f(5)=-(5{)}^2+6(5)-5=0\\ & f(6)=-(6{)}^2+6(6)-5=-5\\ & \text{Jadi, range fungsinya}:{\text{R}}_f=\{y|-12\le y\le 4,y\in \mathbb{R}\}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Titik balik parabola}y=f(x)=-3x^2-18x+2\\ & \text{adalah}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle (-3,19)}& & & & & \text{D}.& {\displaystyle (3,27)}\\ \text{B}.& {\displaystyle (-3,29)}& & \text{C}.& {\displaystyle (-3,23)}& & \text{E}.& (3,29)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{B}}\\ & \begin{array}{rl}& \text{Diketahui FK}:y=f(x)=-3x^2-18x+2\\ & \text{Koordinat titik baliknya}=(x_{ss},y_{ss})\\ & =\left({\displaystyle -\frac{b}{2a},-\frac{D}{4a}}\right)=\left({\displaystyle -\frac{b}{2a},-\frac{b^2-4ac}{4a}}\right)\text{atau}\\ & =(-{\displaystyle \frac{b}{2a},f(-{\displaystyle \frac{b}{2a}})})\\ & =(-{\displaystyle \frac{-18}{2(-3)},-{\displaystyle \frac{(-18{)}^2-4.(-3).(2)}{4(-3)}}})\\ & =(-3,29)\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Fungsi kuadrat dengan titik balik minimum}\\ & (3,-4)\text{dan melalui titik}(0,5)\text{adalah}....\\ & \begin{array}{lllll}\text{A}.& y=x^2-6x+5& & & \\ \text{B}.& y=x^2+6x+5& & \\ \text{C}.& {\displaystyle y=2x^2-6x+5}& & \\ \text{D}.& {\displaystyle y=2x^2+6x+5}& & \\ \text{E}.& y=2x^2-6x-5& & \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{A}}\\ & \begin{array}{rl}& \text{Diketahui FK}:y=f(x)=a{(x-x_{ss})}^2+y_{ss}\\ & \text{Koordinat titik baliknya}=(x_{ss},y_{ss})=(3,-4)\\ & \text{dan melalui titik}(0,5),\text{maka}\\ & 5=a(0-3{)}^2+(-4)\iff 5+4=a.9\iff a={\displaystyle \frac{9}{9}=1}\\ & \text{Sehingga Fk-nya dengan}a=1\text{adalah}:\\ & f(x)=a{(x-x_{ss})}^2+y_{ss}=1.(x-3{)}^2+(-4)\\ & =(x^2-6x+9)-4\\ & =x^2-6x+5\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Fungsi kuadrat yang melalui titik}(0,2)\text{dan}\\ & (-1,0)\text{dengan sumbu simetri garis}\\ & x={\displaystyle \frac{1}{2}\text{adalah}....}\\ & \begin{array}{lllll}\text{A}.& y=(x+1)(2-x)& & & \\ \text{B}.& y=(x-1)(x+2)& & \\ \text{C}.& {\displaystyle y=2-x-x^2}& & \\ \text{D}.& {\displaystyle y=x^2-x+2}& & \\ \text{E}.& y=-(x-1)(x+2)& & \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{A}}\\ & \begin{array}{rl}& \text{Diketahui FK}:y=f(x)=a{(x-x_{ss})}^2+y_{ss}\\ & \text{atau}y=f(x)=a(x-x_1)(x-x_2)\text{dengan}\\ & x_1\text{dan}{x}_2\text{sebagai akar-akarnya}\\ & \text{Dan diketahui pula sebagaimana keterangan}\\ & \text{dalam soal, maka},x_1=-1,x_{ss}={\displaystyle \frac{1}{2}}\\ & \text{Sehingga}\\ & x_{ss}=-{\displaystyle \frac{b}{2a}={\displaystyle \frac{x_1+x_2}{2}\iff {\displaystyle \frac{1}{2}=\frac{-1+x_2}{2}\iff x_2=2}}}\\ & \text{Selanjutnya garfik melalui}(0,2),\text{maka}\\ & y=a(x-x_1)(x-x_2)\iff 2=a(0-(-1))(0-2)\\ & \iff 2=a(1)(-2)\iff a=-1\\ & \text{Sehingga fungsi akan berupa}\\ & f(x)=a(x-x_1)(x-x_2)=-1(x+1)(x-2)\\ & =(x+1)(2-x)\end{array}\end{array}$.

Fungsi Kuadrat (Kelas X/Fase E Semester 2)

 B. Fungsi Kuadrat

B. 1 Fungsi

Silahkan lihat materi sebelumnya, cari di blog ini

di sini 1, dan di sini 2

B. 2 Fungsi Kuadrat

Perhatikan tabel berikut

$\begin{array}{|ll|}\hline \text{Pengertian}& \begin{array}{rl}& \text{Suatu fungsi yang berbentuk}\\ & f(x)=a{x}^2+bx+c\\ & a,b,c,\in \mathbb{R},a\ne 0\end{array}\\ \text{Grafik Fungsi}& \text{Keterangan}\\ \text{Titik potong sumbu x}& \text{Jika ada}\\ & \begin{array}{rl}& \text{untuk titik potong}\\ & \text{terhadap sumbu x }\\ & \text{Jika y = 0 maka }\\ & a{x}^2+bx+c=0\\ & \text{Selanjutnya tinggal}\\ & \text{menentukan nilai D}\\ & D=b^2-4ac\text{adalah}\\ & \text{nilai diskriminan}.\\ & \text{Jika}D>0\\ & \text{maka grafik}\\ & \text{memotong sumbu x}\\ & \text{di dua tempat berbeda}\\ & \text{yaitu di}(x_1,0)\text{dan}(x_2,0).\\ & \text{dan jika D = 0}\\ & \text{maka grafik}\\ & \text{ hanya menyinggung}\\ & \text{sumbu x di satu titik}\\ & \text{yaitu di }(x_1,0)\\ & \text{dan jika}D<0\\ & \text{maka grafik}\\ & \text{tidak memotong}\\ & \text{atau menyinggung sumbu x}\end{array}\\ \text{Titik potong sumbu y}& \begin{array}{rl}& \text{titik potong terhadap}\\ & \text{sumbu y, jika x = 0}\\ & y=f(x)=a{x}^2+bx+c\\ & y=f(0)=a(0{)}^2+b(0)+c\\ & y=c\end{array}\\ \text{Sumbu Simetri (SS)}& x={\displaystyle \frac{-b}{2a}}\\ \text{Titik Puncak}& \left({\displaystyle \frac{-b}{2a},{\displaystyle \frac{D}{-4a}}}\right)\\ \text{Posisi grafik}& \text{Jika}a>0\text{maka}\\ & \text{grafik terbuka ke atas}\\ & \text{Dan jika nilai}a<0\text{maka}\\ & \text{grafik terbuka ke bawah}\\ \hline\end{array}$.

Selanjutnya cara membuat grafik fungsi kudratnya adalah sebagai berikut:

$\begin{array}{|cc|}\hline \text{Jika memotong sumbu}-\text{X}& \text{Jika menyinggung sumbu}-\text{X}\\ \text{di titik}(x_1,0)\text{dan}(x_2,0)& \text{di titik}(x_1,0)\text{dan melalui}\\ \text{dan melalui sebuah titik lain}& \text{sebuah titik lain}\\ & \\ y=f(x)=a(x-x_1)(x-x_2)& y=f(x)=a{(x-x_1)}^2\\ & \\ \text{Jika grafik fungsi itu melalui}& \text{Jika grafik fungsi itu melalui}\\ \text{Titik puncak}P(x_p,y_p)\text{dan}& \text{tiga buah titik yaitu}(x_1,y_1)\\ \text{sebuah titik lain}& (x_2,y_2)\text{dan}(x_3,y_3)\\ & \\ y=f(x)=a{(x-x_p)}^2+y_p& y=f(x)=a{x}^2+bx+c\\ & \\ \hline\end{array}$.

B. 3 Masalah yang Melibatkan Fungsi Kuadrat

$\begin{array}{rl}& y=f(x)=a{x}^2+bx+c\\ & \text{dengan}a,b,c\in \mathbb{R},a\ne 0\end{array}$.

B.3.1 Titik Stasioner

$\begin{array}{rl}& y_{ekstrim}=-{\displaystyle \frac{D}{4a}\Rightarrow \{\begin{array}{c}{y}_{\text{minimum}},\text{jika}a>0\\ y_{\text{maksimum}},\text{jika}a<0\end{array}}\\ & y_{\text{ekstrim}}\text{tercapai saat}x=-{\displaystyle \frac{b}{2a}}\\ & \mathbf{\text{Sehingga titik stasionernya adalah}}\\ & =(x_{ss},y_{ss})=(-{\displaystyle \frac{b}{2a},-{\displaystyle \frac{D}{4a}}})\end{array}$.

B.3.2 Definit Positif dan Definit Negatif

$\begin{array}{rl}& \text{Jika}D<0\text{dan}\{\begin{array}{c}a>0,\text{maka}y\text{akan selalu positif}\\ a<0,\text{maka}y\text{akan selalu negatif}\end{array}\\ & \text{untuk setiap nilai}x\end{array}$.

Perhatikan tambahan penjelasan berikut

$\begin{array}{rl}& \text{Tentang definit positif dan negatif}\\ & \begin{array}{c}\\ a>0.D<0& \text{Gambar}& \cup & \mathbf{\text{Sumbu-X}}\\ a<0,D<0& \text{Gambar}& \cap & \\ & \end{array}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Jika}f\text{adalah fungsi linear dengan}\\ & f(2)-f(-2)=8,\\ & \text{maka nilai dari}f(4)-f(-2)\text{adalah}....\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}:\\ & f(x)=ax+b\\ & f(2)-f(-2)\\ & =(a(2)+b)-(a(-2)+b)=8\\ & 8=2a+2a\\ & 8=4a\\ & 2=a\\ & f(x)=2x+b,\text{dengan}b\text{konstan}\\ & \text{Sehingga nilai}\\ & f(4)-f(-2)=(2(4)+b)-(2(-2)+b)\\ & =8+b+4-b\\ & =12\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Ubahlah}8-6x-x^2\text{ke dalam bentuk}\\ & a-(x+b{)}^2,\text{selanjutnya tentukan}\\ & \text{daerah hasil dari}f(x)=8-6x-x^2\\ & \text{untuk}x\text{bilangan real}\\ & (\mathit{\text{NTU Entrance Examination AO-level}})\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|cl|}\hline 1.& \text{Diketahui}\\ & \begin{array}{rl}\text{Misal}& \\ 8-6x-x^2& =f(x)\\ f(x)& =-x^2-6x+8\\ & =-(x^2+6x-8)\\ & =-(x^2+6x+9-17)\\ & =-((x+3{)}^2-17)\\ & =-(x+3{)}^2+17\\ & \end{array}\\ 2.& \text{Mencari koordinat}(x_{SS},y_{SS})\\ & \begin{array}{rl}f(x)& =-x^2-6x+8\{\begin{array}{c}a=-1\\ b=-6\\ c=8\end{array}\\ \text{Maka}& \\ x_{SS}& =\frac{-b}{2a}={\displaystyle \frac{-(-6)}{2(-1)}}\\ & =-3\\ y_{SS}& =f(-3)=-{(-3+3)}^2+17=17\\ \therefore & (x_{SS},y_{SS})=(-3,17)\end{array}\\ 3.& \text{Nilai fungsi}\\ & \begin{array}{rl}\text{Karena}& a=-1<0\\ \text{maka f}& \text{ungsi menghadap}\\ \mathbf{\text{ke ba}}& \mathbf{\text{wah}},\text{sehingga}\\ \text{daerah}& \text{hasilnya}\left(R_f\right)\\ \text{adalah}& :\\ & \{-\mathrm{\infty }<y\le 17\}\\ & \\ & \text{Berikut ilustrasinya}\end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Jika}\alpha \text{dan}\beta \text{adalah akar-akar dari }\\ & \text{persamaan kuadrat}{x}^2+mx+m=0,\\ & \text{maka nilai}m\text{yang menyebabkan }\\ & \text{jumlah kuadrat akar-akar mencapai}\\ & \text{minimum adalah}....\\ & \mathbf{\text{(UM UNDIP 2014 Mat Das)}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui}{x}^2+mx+m=0\\ & \mathbf{\text{persamaan kuadrat}}\text{dalam}x,\\ & \text{maka}\\ & x^2+mx+m=x^2-(\alpha +\beta )x+(\alpha \beta )=0\\ & \{\begin{array}{ll}\alpha +\beta & =-m\\ & \\ \alpha \beta & =m\end{array}\\ & \text{Selanjutnya}\\ & {\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta \\ & =(-m{)}^2-2m\text{dan dapat kita tuliskan sebagai}\\ & f(m)=m^2-2m\{\begin{array}{ll}a& =1\\ b& =-2\\ c& =0\end{array}\\ & \text{fungsi kuadrat dalam}m,\\ & \text{sehingga kita perlu mencari titik}(m_{SS},f\left(m_{SS}\right)),\\ & \text{tetapi yang kita perlukan}\\ & \text{cuma}m-\text{nya saja, yaitu}:m=m_{SS},\\ & \text{dengan}{m}_{SS}={\displaystyle \frac{-b}{2a}=\frac{-(-2)}{2.1}=1}\end{array}\end{array}$.

Sumber Referensi:

1. Budhi, Wono S. 2014. Bupena Matematika Kelompok Wajib untuk SMA/MA Kelas X. Jakarta: ERLANGGA.
2. Kurnianingsih, Sri, Kuntarti & Sulistiyono. 2007. Matematika SMA dan MA untuk Kelas X Semester 1. Jakarta: ESIS.
3. Noormandiri. 2022. Matematika untuk SMA/MA Kelas X.Jakarta: ERLANGGA
4. Sobirin. 2005. Kompas Matematika: Strategi Praktis Menguasai Tes Matematika SMA Kelas 1. Jakarta: KAWAN PUSTAKA.