Odpowiedź:
Szczegółowe wyjaśnienie:
1.
f(x) = -3x² - 12x -17
p = -b/2a = 12/(-6) = -2
q = f(p) = f(-2) =-3*(-2)² - 12*(-2) -17 = -12 + 24 - 17= -5
Postać kanoniczna:
f(x) = -3(x+2)² - 5
2.
a)
4x²-12x+9 = 0
Δ = 144 - 144 = 0
x1=x2= 12/8 = 3/2
b)
x²-8=7-2x
x²+2x-1 = 0
Δ = 4 + 4 = 8 -> √8 = 2√2
x1= (-2+2√2)/2 = -1 + √2
x2= (-2-2√2)/2 = -1 - √2
3.
a) 1 - 9x² ≥ 0 /*(-1)
9x² - 1 ≥ 0
(3x-1)(3x+1) ≥ 0
x1 = 1/3
x2 = -1/3
x∈ (-∞; -1/3> lub <1/3; +∞)
b)
4x² - 5x > x² + 4x
3x² - 9x >0
3x(x-3) > 0
x1 = 0
x2 = 3
x ∈ (-∞; 0) lub (3; +∞)
4.
f(x) = 2x² + bx + c
A = (3; -1) , B=(4; 5)
podstawiamy punkty:
-1 = 18 + 3b + c -> c = -19 - 3b
5 = 32 + 4b + c
4b + c = -27
4b -19 - 3b = -27
b = -27 + 19 = -8
c = -19 + 24 = 5
f(x) = 2x² -8x + 5
5.
A=(-5;7)
B = (4; -5)
wektor AB = [9, -12]
|AB| = √(81+144) = √225 = 15
6.
k: -2x + 3y -3 = 0
3y = 2x + 3 /:3
y = 2/3 x + 1
nasza prosta:
y = ax + b
a = 2/3
czyli:
y = 2/3 x + b
podstawiamy punkt:
-1 = 2/3 *6 + b
-1 = 4 + b
b = -5
prosta:
y = 2/3x - 5
