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Question Number 189436 by 073 last updated on 16/Mar/23

Commented by 073 last updated on 16/Mar/23

$$\mathrm{solution}\mathrm{please}$$

Answered by cortano12 last updated on 16/Mar/23

$$\mathrm{let}\{\begin{array}{l}\mathrm{B}({\mathrm{x}}_1,0)\\ \mathrm{A}({\mathrm{x}}_2,0)\end{array};{\mathrm{x}}_1>0,{\mathrm{x}}_2<0$$$$\Rightarrow \mid \mathrm{B}-\mathrm{A}\mid ={\mathrm{x}}_1-{\mathrm{x}}_2=6$$$$\mathrm{and}{\mathrm{x}}_1+{\mathrm{x}}_2=4\mathrm{then}\{\begin{array}{l}{\mathrm{x}}_1=5\\ {\mathrm{x}}_2=-1\end{array}$$$$\mathrm{so}\therefore \mathrm{f}\left(\mathrm{x}\right)=\mathrm{a}{(\mathrm{x}-2)}^2+\mathrm{k};(0,-5),(5,0)$$$$\Rightarrow \{\begin{array}{l}-5=4\mathrm{a}+\mathrm{k}\\ 0=9\mathrm{a}+\mathrm{k}\end{array}\Rightarrow \{\begin{array}{l}\mathrm{a}=1\\ \mathrm{k}=-9\end{array}$$

Commented by 073 last updated on 16/Mar/23

$$\mathrm{nice}\mathrm{solution}$$

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