1) f : ℝ → ℝ, f(x) = 2x²+3x-1
[tex]\\[/tex]
a) f(0)·f(1)·f(2) = (2·0²+3·0-1)·(2·1²+3·1-1)·(2·2²+3·2-1) =
= (0+0-1)·(2+3-1)·(2·4+6-1) = (-1)·4·(13) = -52
[tex]\\[/tex]
b) y = 2x²+3x-1
Gf ∩ Ox:
y = 0 ⇒ f(x) = 0 ⇒ 2x²+3x-1 = 0
(Ecuația e de forma ax²+bx+c = 0)
Δ = b²-4ac = 3²-4·2·(-1) = 9+8 = 17
[tex]x_{1,2} = \dfrac{-b\pm \sqrt{\Delta}}{2a} = \dfrac{-3\pm \sqrt{17}}{2\cdot 2} = \dfrac{-3\pm \sqrt{17}}{4}[/tex]
Punctele sunt de forma (x₀, y₀).
Prin urmare, punctele sunt:
[tex]\left(\dfrac{-3-\sqrt{17}}{4},0\right)[/tex] și [tex]\left(\dfrac{-3+\sqrt{17}}{4},0\right)[/tex]
[tex]\\[/tex]
2) f : ℝ → ℝ, f(x) = -2x²+x+5
Valoarea extremă a funcției este dată de ordonata vărfului parabolei:
[tex]V\left(-\dfrac{b}{2a},-\dfrac{\Delta}{4a}\right)[/tex], adică [tex]f_{max} = -\dfrac{\Delta}{4a}[/tex]
[tex]\begin{aligned} \Rightarrow f_{max} &= -\dfrac{b^2-4ac}{4a} = -\dfrac{1^2-4\cdot(-2)\cdot 5}{4\cdot (-2)} = \\ &= -\dfrac{1+40}{-8} = \boxed{\dfrac{41}{8}}\end{aligned}[/tex]
[tex]\\[/tex]
3) x²+6x-9 = 0
Relațiile lui Viète:
[tex]x_1+x_2 = -\dfrac{b}{a} = -\dfrac{6}{1} = -6[/tex]
[tex]x_1\cdot x_1 = \dfrac{c}{a} =\dfrac{-9}{1} = -9[/tex]
Astfel, expresia cerută va fi:
[tex]x_1+x_2+x_1\cdot x_2 = -6+(-9) = -6-9 = \boxed{-15}[/tex]
[tex]\\[/tex]
4) AB = AC = √2, m(∢A) = 30°
Folosesc formula următoare:
[tex]\begin{aligned}A_{\triangle} &=\dfrac{bc\cdot \sin(b;c)}{2} = \dfrac{AC\cdot AB\cdot \sin(AC;AB)}{2} =\\ &=\dfrac{\sqrt{2}\cdot \sqrt{2}\cdot \sin (30^{\circ})}{2} = \dfrac{2\cdot \dfrac{1}{2}}{2} = \boxed{\dfrac{1}{2}}\end{aligned}[/tex]
[tex]\\[/tex]
5) AB = 6, AC = 8, BC = 10
Formula lui Heron:
[tex]p = \dfrac{a+b+c}{2} = \dfrac{BC+AC+AB}{2} = \dfrac{10+8+6}{2} = \dfrac{24}{2}= 12[/tex]
[tex]\begin{aligned}A_{\triangle} &= \sqrt{p(p-a)(p-b)(p-c)} = \sqrt{12(12-10)(12-8)(12-6)}=\\ &=\sqrt{12\cdot 2\cdot 4\cdot 6} = \sqrt{24\cdot 24} = \boxed{24}\end{aligned}[/tex]
