Question

помогите пожалуйста ​

Answer 1

2.

$y = 3x^{2} + 4x - 4$

а)

$x = -3$

$3*(-3)^{2} + 4*3 - 4 = 3 * 9 + 4 * 3 - 4 = 27 + 12 - 4 = 35$

б)

$y = -4[tex]</p><p>[tex]3x^{2} + 4x - 4 = -4$

$3x^{2} + 4x = 0$

$x(3x + 4) = 0$

$\left \{ {{x = 0} \atop {3x + 4 = 0}} \right.$

$\left \{ {{x = 0} \atop {3x = -4}} \right.$

$\left \{ {{x=0} \atop {x=\frac{-4}{3} }} \right.$

в)

$3x^2 + 4x - 4 =0$

$D = b^2 - 4ac$

$D = 16 + 4*3*4 = 64$

$\sqrt{D} = 8$

$x_{1} = \frac{-b-\sqrt{D}}{2a}\\x_{2} = \frac{-b+\sqrt{D}}{2a}$

$x_{1} = \frac{-4-8}{6} = \frac{-12}{6} = -2\\x_{2} = \frac{-4+8}{6} = \frac{4}{6} = \frac{2}{3}$

3. построй сам график пожалуйста, там парабола будет ветвями вверх. я тебе нули щас найду

$\sqrt{D} = \sqrt{4^2 - 4 * 3} = 2\\x_{1} = \frac{-4-2}{2} = -3\\x_{2} = \frac{-4+2}{2} = -1$

а)$(-\infty; -3) u (-1; +\infty)$

б)$(-2; +\infty)$

4. $\sqrt{D} = \sqrt{25 + 24} = \sqrt{49} = 7\\x_{1} = \frac{-b-\sqrt{D}}{2a} = \frac{-5-7}{2} = -6\\x_{2} = \frac{-b+\sqrt{D}}{2a} = \frac{-5+7}{2} = 1\\(-6; 1)$