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Question-189436




Question Number 189436 by 073 last updated on 16/Mar/23
Commented by 073 last updated on 16/Mar/23
solution please
$$\mathrm{solution}\:\mathrm{please} \\ $$
Answered by cortano12 last updated on 16/Mar/23
 let  { ((B(x_1 ,0))),((A(x_2 ,0))) :}; x_1 >0 , x_2 <0  ⇒∣B−A∣= x_1 −x_2 = 6    and x_1 +x_2 =4 then  { ((x_1 =5)),((x_2 =−1)) :}  so ∴ f(x)=a(x−2)^2 +k ; (0,−5),(5,0)  ⇒ { ((−5=4a+k)),((    0=9a+k)) :} ⇒ { ((a=1)),((k=−9)) :}
$$\:\mathrm{let}\:\begin{cases}{\mathrm{B}\left(\mathrm{x}_{\mathrm{1}} ,\mathrm{0}\right)}\\{\mathrm{A}\left(\mathrm{x}_{\mathrm{2}} ,\mathrm{0}\right)}\end{cases};\:\mathrm{x}_{\mathrm{1}} >\mathrm{0}\:,\:\mathrm{x}_{\mathrm{2}} <\mathrm{0} \\ $$$$\Rightarrow\mid\mathrm{B}−\mathrm{A}\mid=\:\mathrm{x}_{\mathrm{1}} −\mathrm{x}_{\mathrm{2}} =\:\mathrm{6}\: \\ $$$$\:\mathrm{and}\:\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} =\mathrm{4}\:\mathrm{then}\:\begin{cases}{\mathrm{x}_{\mathrm{1}} =\mathrm{5}}\\{\mathrm{x}_{\mathrm{2}} =−\mathrm{1}}\end{cases} \\ $$$$\mathrm{so}\:\therefore\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{a}\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{k}\:;\:\left(\mathrm{0},−\mathrm{5}\right),\left(\mathrm{5},\mathrm{0}\right) \\ $$$$\Rightarrow\begin{cases}{−\mathrm{5}=\mathrm{4a}+\mathrm{k}}\\{\:\:\:\:\mathrm{0}=\mathrm{9a}+\mathrm{k}}\end{cases}\:\Rightarrow\begin{cases}{\mathrm{a}=\mathrm{1}}\\{\mathrm{k}=−\mathrm{9}}\end{cases} \\ $$
Commented by 073 last updated on 16/Mar/23
nice solution
$$\mathrm{nice}\:\mathrm{solution} \\ $$

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