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rozwiązanie w załączniku
[tex]\\a)\\f(x)=-\sqrt{2} x+3\\P(\sqrt{2} ,7)\\g_{\parallel}(x)=-\sqrt{2} x+b\\7=-\sqrt{2} \cdot\sqrt{2} +b\\b=9\\g_{\parallel}(x)=-\sqrt{2} x+9\\b)\\f(0)=-\sqrt{2} \cdot0+3\\P_{OY}(0,3)\\\displaystyle P(4\frac{1}{2} ,-6)\\a=\frac{-6-3}{4\frac{1}{2}-0 } =\frac{-9}{\frac{9}{2} } =-9\cdot\frac{2}{9} =-2\\g(x)=-2x+b\\3=-2\cdot0+b\\b=3\\g(x)=-2x+3[/tex]
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Odpowiedź:
rozwiązanie w załączniku
Odpowiedź:
[tex]\\a)\\f(x)=-\sqrt{2} x+3\\P(\sqrt{2} ,7)\\g_{\parallel}(x)=-\sqrt{2} x+b\\7=-\sqrt{2} \cdot\sqrt{2} +b\\b=9\\g_{\parallel}(x)=-\sqrt{2} x+9\\b)\\f(0)=-\sqrt{2} \cdot0+3\\P_{OY}(0,3)\\\displaystyle P(4\frac{1}{2} ,-6)\\a=\frac{-6-3}{4\frac{1}{2}-0 } =\frac{-9}{\frac{9}{2} } =-9\cdot\frac{2}{9} =-2\\g(x)=-2x+b\\3=-2\cdot0+b\\b=3\\g(x)=-2x+3[/tex]