Question

Решите 12 номер пожалуйста дам 50 баллов

Answer 1

1)

$f(x) = 3 {x}^{2} + 2x$

$f(0) = - 3 \times {0}^{2} + 2 \times 0 = 0$

$f( - 2) = 3 \times ( - 2) {}^{2} + 2 \times ( - 2) = \\ 3 \times 4 - 4 = 12 - 4 = 8$

$f(1) = - 3 \times 1 {}^{2} + 2 \times 1 = \\ - 3 + 2 = - 1$

$f( \frac{1}{3} ) = - 3( \frac{1}{3} ) {}^{2} + 2 \times \frac{1}{3} = - 3 \times \frac{1}{9} + \frac{2}{3} = \\ - \frac{1}{3} + \frac{2}{3} = \frac{1}{3}$

$f( - \frac{1}{2} ) = - 3( - \frac{1}{2} ) {}^{2} + 2 \times ( - \frac{1}{2} ) = \\ - 3 \times \frac{1}{4} - 1 = - \frac{3}{4} - 1 = - 1 \frac{3}{4} = - 1.75$

2)

$f( - 1) = - 5 \\ f( - 1) = - 3 \times {( - 1)}^{2} + 2 \times ( - 1) = \\ - 3 \times 1 - 2 = - 3 - 2 = - 5$

верно

$f( \frac{2}{3} ) = 0\\ f( \frac{2}{3} ) = - 3 \times ( \frac{2}{3} ) {}^{2} + 2 \times \frac{2}{3} = \\ - 3 \times \frac{4}{9} + \frac{4}{3} = - \frac{4}{3} + \frac{4}{3} = 0$

верно

$f( \frac{1}{6} ) = \frac{1}{3} \\ f( \frac{1}{6} ) = - 3 \times ( \frac{1}{6} ) {}^{2} + 2 \times \frac{1}{6} = \\ - 3 \times \frac{1}{36} + \frac{1}{3} = - \frac{1}{12} + \frac{1}{3} = \\ - \frac{1}{12} + \frac{4}{12} = \frac{3}{12} = \frac{1}{4}$

неверно

$f( - 2) = - 16 \\ f( - 2) = 8$

неверно

$f(4) = - 42 \\ f(4) = - 3 \times {4}^{2} + 2\times 4 = \\ -3 \times 16 + 8 = - 48 + 8 = - 40$

неверно