$\begin{array}{ll}\\ 1.&\textrm{Diketahui fungsi}\: \: f(x)=x^{2}-2x-15. \: \: \textrm{Jika}\\ &\textrm{domain}\: \: \left \{ x|-4\leq x\leq 2,x\in \mathbb{R} \right \}\: ,\: \textrm{maka}\\ &range\textrm{-nya adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&-15\leq f(x)\leq 20&&&\\ \textrm{B}.&-15\leq f(x)\leq 9&&\\ \textrm{C}.&\color{red}\displaystyle -16\leq f(x)\leq 9&&\\ \textrm{D}.&\displaystyle -16\leq f(x)\leq 20&&\\ \textrm{E}.&-15\leq f(x)\leq 5&&\\ \end{array}\\\\ &\textbf{Jawab}:\textbf{C}\\ &\begin{aligned}&\textrm{Diketahui FK}:f(x)=x^{2}-2x-15,\: \: \textrm{dengan}\\ &\textrm{D}_{f}=\left \{ x|-4\leq x\leq 2,x\in \mathbb{R} \right \},\\ &\textrm{maka}\: \: range\: \: \textrm{fungsinya adalah}\: \: \textrm{R}_{f},\: \: \textrm{di mana}\\ &\textrm{R}_{f}\: \: \textrm{diperoleh dengan cara di antaranya}\\ &\textrm{mensubstitusikan langsung ke fungsinya, yaitu}:\\ &f(-4)=(-4)^{2}-2(-4)-15=\color{red}9\\ &f(-3)=(-3)^{2}-2(-3)-15=0\\ &f(-2)=(-2)^{2}-2(-2)-15=-7\\ &f(-1)=(-1)^{2}-2(-1)-15=-12\\ &f(0)=(0)^{2}-2(0)-15=-15\\&f(1)=(1)^{2}-2(1)-15=\color{red}-16\\&f(2)=(2)^{2}-2(2)-15=-15\\ &\textrm{Jadi, range fungsinya}:\textrm{R}_{f}=\color{red}\displaystyle -16\leq f(x)\leq 9 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 2.&\textrm{Daerah hasil fungsi}\: \: f(x)=-x^{2}+6x-5 \: \: \textrm{untuk}\\ &\textrm{daerah asal}\: \: \left \{ x|-1\leq x\leq 6,x\in \mathbb{R} \right \}\: \: \textrm{dan}\\ &y=f(x)\: \: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\left \{ y|-5\leq y\leq 0,y\in \mathbb{R} \right \}&&&\\ \textrm{B}.&\color{red}\left \{ y|-12\leq y\leq 4,y\in \mathbb{R} \right \}&&\\ \textrm{C}.&\displaystyle \left \{ y|-4\leq y\leq 1,y\in \mathbb{R} \right \}&&\\ \textrm{D}.&\displaystyle \left \{ y|-5\leq y\leq 4,y\in \mathbb{R} \right \}&&\\ \textrm{E}.&\left \{ y|-1\leq y\leq 6,y\in \mathbb{R} \right \}&&\\ \end{array}\\\\ &\textbf{Jawab}:\textbf{B}\\ &\begin{aligned}&\textrm{Masih sama dengan cara di atas. Diketahui FK}:\\ &f(x)=-x^{2}+6x-5,\: \: \textrm{dengan}\\ &\textrm{D}_{f}=\left \{ x|-1\leq x\leq 6,x\in \mathbb{R} \right \},\\ &\textrm{maka}\: \: range\: \: \textrm{fungsinya adalah}\: \: \textrm{R}_{f},\: \: \textrm{di mana}\\ &\textrm{R}_{f}\: \: \textrm{diperoleh dengan cara di antaranya}\\ &\textrm{mensubstitusikan langsung ke fungsinya, yaitu}:\\ &f(-1)=-(-1)^{2}+6(-1)-5=\color{red}-12\\ &f(0)=-(0)^{2}+6(0)-5=-5\\ &f(1)=-(1)^{2}+6(1)-5=0\\ &f(2)=-(2)^{2}+6(2)-5=3\\ &f(3)=-(3)^{2}+6(3)-5=\color{red}4\\&f(4)=-(4)^{2}+6(1)-5=3\\&f(5)=-(5)^{2}+6(5)-5=0\\ &f(6)=-(6)^{2}+6(6)-5=-5\\ &\textrm{Jadi, range fungsinya}:\textrm{R}_{f}=\color{red}\left \{ y|-12\leq y\leq 4,y\in \mathbb{R} \right \} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 3.&\textrm{Titik balik parabola}\: \: y=f(x)=-3x^{2}-18x+2\\ &\textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle (-3,19)&&&&&\textrm{D}.&\displaystyle (3,27)\\ \textrm{B}.&\color{red}\displaystyle (-3,29)&&\textrm{C}.&\displaystyle (-3,23)&&\textrm{E}.&(3,29) \end{array}\\\\ &\textbf{Jawab}:\textbf{B}\\ &\begin{aligned}&\textrm{Diketahui FK}:y=f(x)=-3x^{2}-18x+2\\ &\textrm{Koordinat titik baliknya}=\color{red}\left ( x_{ss},y_{ss} \right )\\ &=\left ( \displaystyle -\frac{b}{2a},-\frac{D}{4a} \right )=\left ( \displaystyle -\frac{b}{2a},-\frac{b^{2}-4ac}{4a} \right )\: \: \textrm{atau}\\ &=\left ( -\displaystyle \frac{b}{2a},f\left ( -\displaystyle \frac{b}{2a} \right ) \right )\\ &=\left ( -\displaystyle \frac{-18}{2(-3)},-\displaystyle \frac{(-18)^{2}-4.(-3).(2)}{4(-3)} \right )\\ &=\color{red}\left (-3,29 \right ) \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 4.&\textrm{Fungsi kuadrat dengan titik balik minimum}\\ &(3,-4)\: \: \textrm{dan melalui titik}\: \: (0,5)\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\color{red}y=x^{2}-6x+5&&&\\ \textrm{B}.&y=x^{2}+6x+5&&\\ \textrm{C}.&\displaystyle y=2x^{2}-6x+5&&\\ \textrm{D}.&\displaystyle y=2x^{2}+6x+5&&\\ \textrm{E}.&y=2x^{2}-6x-5&&\\ \end{array}\\\\ &\textbf{Jawab}:\textbf{A}\\ &\begin{aligned}&\textrm{Diketahui FK}:y=f(x)=a\left ( x-x_{ss} \right )^{2}+y_{ss}\\ &\textrm{Koordinat titik baliknya}=\color{red}\left ( x_{ss},y_{ss} \right )=(3,-4)\\ &\textrm{dan melalui titik}\: \: (0,5),\: \textrm{maka}\\ &5=a(0-\color{red}3\color{black})^{2}+(\color{red}-4\color{black})\Leftrightarrow 5+4=a.9\Leftrightarrow a=\displaystyle \frac{9}{9}=1\\ &\textrm{Sehingga Fk-nya dengan}\: \: a=1\: \: \textrm{adalah}:\\ &f(x)=a\left ( x-x_{ss} \right )^{2}+y_{ss}=1.(x-3)^{2} +(-4)\\ &\qquad =(x^{2}-6x+9)-4\\ &\qquad=\color{red}x^{2}-6x+5 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 5.&\textrm{Fungsi kuadrat yang melalui titik}\: \: (0,2)\: \: \textrm{dan}\\ &(-1,0)\: \: \textrm{dengan sumbu simetri garis}\\ &x=\displaystyle \frac{1}{2}\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\color{red}y=(x+1)(2-x)&&&\\ \textrm{B}.&y=(x-1)(x+2)&&\\ \textrm{C}.&\displaystyle y=2-x-x^{2}&&\\ \textrm{D}.&\displaystyle y=x^{2}-x+2&&\\ \textrm{E}.&y=-(x-1)(x+2)&&\\ \end{array}\\\\ &\textbf{Jawab}:\textbf{A}\\ &\begin{aligned}&\textrm{Diketahui FK}:y=f(x)=a\left ( x-x_{ss} \right )^{2}+y_{ss}\\ &\textrm{atau}\: \: y=f(x)=a(x-x_{1})(x-x_{2})\: \: \textrm{dengan}\\ &x_{1}\: \: \textrm{dan}\: \: x_{2}\: \: \textrm{sebagai akar-akarnya}\\ &\textrm{Dan diketahui pula sebagaimana keterangan}\\ &\textrm{dalam soal, maka},\: \: x_{1}=-1,\: x_{ss}=\displaystyle \frac{1}{2}\\ &\textrm{Sehingga}\\ &x_{ss}=-\displaystyle \frac{b}{2a}=\displaystyle \frac{x_{1}+x_{2}}{2}\Leftrightarrow \displaystyle \frac{1}{2}=\frac{-1+x_{2}}{2}\Leftrightarrow x_{2}=2\\ &\textrm{Selanjutnya garfik melalui}\: \: (0,2),\: \textrm{maka}\\ &y=a(x-x_{1})(x-x_{2})\Leftrightarrow 2=a(0-(-1))(0-2)\\ &\Leftrightarrow 2=a(1)(-2)\Leftrightarrow a=-1\\ &\textrm{Sehingga fungsi akan berupa}\\ &f(x)=a(x-x_{1})(x-x_{2})=-1(x+1)(x-2)\\ &\qquad =\color{red}(x+1)(2-x) \end{aligned} \end{array}$.
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