Question

Даю 100 баллов............​

Answer 1

$\displaystyle f(x)=-3x^2+2x\\\\1)\\f(1)=-3\cdot 1^2+2\cdot 1=-3\cdot 1+2=-3+2=-1\\\\f(0)=-3\cdot 0^2+2\cdot 0=-3\cdot 0+0=0\\\\f(\frac{1}{3})=-3\cdot (\frac{1}{3})^2+2\cdot \frac{1}{3}=-3\cdot \frac{1}{9}+\frac{2}{3}=-\frac{1}{3}+\frac{2}{3}=\frac{1}{3}\\\\f(-2)=-3\cdot (-2)^2+2\cdot (-2)=-3\cdot 4-4=-12-4=-16$

$\displaystyle f(x)=-3x^2+2x\\\\2)\\-3x^2+2x=0\\\\-3x(x-\frac{2}{3})=0\\\\-3x=0\ \ \ |:(-3)\ \ \ \ \ \ \ \ \ \ x-\frac{2}{3}=0\\\\x=0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=\frac{2}{3}$

відповідь:  $\displaystyle 0,\ \frac{2}{3}$

$\displaystyle -3x^2+2x=-1\\\\-3 x^{2} + 2x + 1 =0\\\\ a=-3 ,\ \ b=2 ,\ \ c=1\\\\ D = b^2 - 4ac = 2^2 - 4\cdot( - 3)\cdot1 = 4 + 12 = 16\\\\\sqrt{D} =\sqrt{16} = 4\\\\ x_1=\frac{-b-\sqrt{D}}{2a}=\frac{-2-4}{2\cdot(-3)}=\frac{-6 }{-6 }=1\\\\ x_2=\frac{-b+\sqrt{D}}{2a}=\frac{-2+4}{2\cdot(-3)}=\frac{2}{-6}=- \frac{ 1 }{ 3 }$

Відповідь: $\displaystyle - \frac{ 1 }{ 3 },\ 1$

$\displaystyle -3x^2+2x=-56-3x^2+2x+56=0-3 x^{2} + 2 x + 56 =0\\\\ a=-3 ,\ \ b=2 ,\ \ c=56\\\\ D = b^2 - 4ac = 2^2 - 4\cdot( - 3)\cdot56 = 4 + 672 = 676\\\\\sqrt{D} =\sqrt{676} = 26\\\\ x_1=\frac{-b-\sqrt{D}}{2a}=\frac{-2-26}{2\cdot(-3)}=\frac{-28 }{-6 }= \frac{ 14 }{ 3 }=4\frac{2}{3} \\\\ x_2=\frac{-b+\sqrt{D}}{2a}=\frac{-2+26}{2\cdot(-3)}=\frac{24}{-6}=-4$

Відповідь: $\displaystyle -4 ,\ 4\frac{2}{3}$