[tex]f(x)~~gdy~~ x\in <1,6>\\\\
Z~~wykresu~~odczytam~~wspolrzedne~~wierzcholka:\\
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W=(3,-2)~~\Rightarrow ~~x=3~~nalezy~~do przedzialu~~<1,6>\\
f(3)=-2\\
\\[/tex]
[tex]wyznacze~~funkcje~~f(x)\\
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f(x)=ax^{2} +bx+c~~szukana~~funkcja~~kwadratowa\\
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W=(3,-2)\\
M_{0} =\{1,5 \}~~\Rightarrow~~m_{1} =(1,0),~~m_{2} =(5,0),~~m_{1} \in f(x),~~m_{2} \in f(x)\\[/tex]
[tex]f(x)=a\cdot (x-p)^{2} +q~~postac~~kanonicza~~szukanej ~~funkcji\\
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f(x)=a\cdot (x-3)^{2} -2~~\land~~m_{1} \in f(x) ~~\land ~~m_{1} =(1,0)\\
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a\cdot (1-3)^{2} -2=0\\
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4a-2=0\\
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4a=2~~\mid \div 4\\
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a=\dfrac{1}{2} \\
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f(x)=\dfrac{1}{2} \cdot (x-3)^{2} -2~~dla~~x\in <1,6>\\
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f(1)=0\\
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f(6)=\dfrac{1}{2} \cdot 3^{2} -2=4\dfrac{1}{2}-2=2\dfrac{1}{2}\\
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f(3)=-2~~-wierzcholek~~gdzie~~p=3~~\Rightarrow ~~p\in <1,6>\\
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WNIOSEK:\\
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y_{najm} =-2~~dla~~x=3\\
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y_{najw} =2\frac{1}{2}~~dla ~~x=6 [/tex]
