Odpowiedź:
zad 1
x³ - 6x = x²
x³ - x² -6x = 0
x(x² - x - 6) = 0
x = 0 ∨ x² - x - 6 = 0
x² - x - 6 = 0
a = 1 , b = - 1 , c = - 6
Δ = b² - 4ac = (- 1)² - 4 * 1 * (- 6) = 1 + 24 = 25
√Δ = √25 = 5
x₁ = ( - b - √Δ)/2a = ( 1 - 5)/2 = - 4/2 = - 2
x₂ =(- b + √Δ)/2a = (1 + 5)/2 = 6/2 = 3
x = { - 2 , 0 , 3 }
zad 2
y = - 3(x - 7)² + 5 = - 3(x² - 14x + 49) + 5 = - 3x² + 42x - 147 + 5 =
= - 3x² + 42x - 142
y = - 7
- 3x² + 42x - 142 = - 7
- 3x² + 42x - 142 + 7 = 0
- 3x² + 42x - 135 = 0
a = - 3 , b = 42 , c = - 135
Δ = b² - 4ac = 42² - 4 * (- 3) * ( - 135) = 1764 - 1620 = 144
√Δ = √144 = 12
x₁ = (- b - √Δ)/2a = ( -42- 12)/(- 6) = - 54/(- 6) = 54/6 = 9
x₂ = (- b + √Δ)/2a = (- 42 + 12)/(- 6) = - 30/(- 6) = 30/6 = 5
Parabola ma dwa punkty wspólne z prostą ; współrzędne x punktów wspólnych = 5 i 9
wykres w załączniku
zad 3
y = 3x² - 6
3x² - 6 < 0
3(x² - 2) < 0
x² - 2 < 0
(x - √2)(x + √2) < 0
x - √2 > 0 ∧ x + √2 < 0 ∨ x - √2 < 0 ∧ x + √2 > 0
x > √2 ∧ x < -√2 ∨ x < √2 ∧ x > - √2
x > - √2 ∧ x < √2
x ∈ ( - √2 , √2)
1.
[tex]x^{3}-6x=x^{2}\\\\x^{3}-x^{2}-6x = 0\\\\x(x^{2}-x-6) = 0\\\\\boxed{x_{o} = 0}\\\\lub\\\\x^{2}-x-6=0\\\\a = 1, \ b = -1, \ c = -6\\\\\Delta = b^{2}-4ac = (-1)^{2}-4\cdot1\cdot(-6) = 1+24 = 25\\\\\sqrt{\Delta} = \sqrt{25} = 5\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{1-5}{2} = \frac{-4}{2} =\boxed{-2}\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{1+5}{2} = \frac{6}{2} = \boxed{3}\\\\x\in\{-2,0,3\}[/tex]
2.
[tex]y = -3(x-7)^{2}+5 = -3(x^{2}-14x+49)+5 = -3x^{2}+42x-147+5 =\\\\=-3x^{2}+42x-142\\\\oraz\\\\y = -7\\\\\\-3x^{2}+42x-142 = -7\\\\-3x^{2}+42x-142+7 = 0\\\\-3x^{2}+42x-135 = 0 \ \ /:(-3)\\\\\underline{x^{2}-14x+45 = 0}\\\\a = 1, \ b = -14, \ c = 45\\\\\Delta = b^{2}-4ac = (-14)^{2}-4\cdot1\cdot45 = 196-180 = 16\\\\\sqrt{\Delta} = \sqrt{16} = 4\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{14-4}{2} = \frac{10}{2} =\boxed{ 5}\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{14+4}{2} = \frac{18}{2} = \boxed{9}[/tex]
Parabola z daną prostą ma dwa punkty wspólne.
3.
[tex]y = 3x^{2}-6\\\\3x^{2}-6 < 0 \ \ /:3\\\\x^{2}-2 < 0\\\\M. \ zerowe\\\\(x+\sqrt{2})(x-\sqrt{2})=0\\\\x+\sqrt{2} = 0 \ \ \vee \ \ x-\sqrt{2} = 0\\\\x = -\sqrt{2} \ \ \vee \ \ x = \sqrt{2}[/tex]
a > 0, to parabola zwrócona jest ramionami do góry wówczas:
x ∈ (-√2; √2)