Odpowiedź:
postać kanoniczna
f(x)= a(x-p)²+q
q=9
[tex]\displaystyle p=\frac{-3+5}{2} =1\\A(-3,-7)\qquad B(5,-7)\\f(x)=a(x-1)^2+9\qquad B(5,-7)\\-7=a(5-1)^2+9\\16a=-16/:16\\a=-1\\f(x)=-(x-1)^2+9\\(x-1)^2=9/\sqrt{} \\|x-1|=3\quad \Rightarrow x-1=3\quad \lor\quad x-1=-3\\x_1=4\qquad x_2=-2\\\text{iloczynowa}\\f(x)=-(x-4)(x+2)\\f(x)=-(x^{2} -2x-8)=-x^{2} +2x+8\\\text{ogolna}\\f(x)=-x^{2} +2x+8[/tex]
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Odpowiedź:
postać kanoniczna
f(x)= a(x-p)²+q
q=9
[tex]\displaystyle p=\frac{-3+5}{2} =1\\A(-3,-7)\qquad B(5,-7)\\f(x)=a(x-1)^2+9\qquad B(5,-7)\\-7=a(5-1)^2+9\\16a=-16/:16\\a=-1\\f(x)=-(x-1)^2+9\\(x-1)^2=9/\sqrt{} \\|x-1|=3\quad \Rightarrow x-1=3\quad \lor\quad x-1=-3\\x_1=4\qquad x_2=-2\\\text{iloczynowa}\\f(x)=-(x-4)(x+2)\\f(x)=-(x^{2} -2x-8)=-x^{2} +2x+8\\\text{ogolna}\\f(x)=-x^{2} +2x+8[/tex]