Odpowiedź:
[tex]\displaystyle f(x)=\frac{1}{2}( x+4)(x-6) \\\text{miejsca zerowe} \quad x_1=-4\quad x_2=6\\\text{os symetrii paraboli}\quad x=\frac{-4+6}{2} =1\qquad x=1\\f(x)=\frac{1}{2}( x+4)(x-6)=\frac{1}{2}(x^{2} -2x-24)=\underline{\frac{1}{2} x^{2} -x-12}[/tex]
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Odpowiedź:
[tex]\displaystyle f(x)=\frac{1}{2}( x+4)(x-6) \\\text{miejsca zerowe} \quad x_1=-4\quad x_2=6\\\text{os symetrii paraboli}\quad x=\frac{-4+6}{2} =1\qquad x=1\\f(x)=\frac{1}{2}( x+4)(x-6)=\frac{1}{2}(x^{2} -2x-24)=\underline{\frac{1}{2} x^{2} -x-12}[/tex]