1) Zapisz wzór funkcji w postaci kanonicznej
f(x)=-x²-6x-7
f(x)=2x²-4x+12
f(x)=½x²+4x+5
2) Wyznacz miejsca zerowe funkcji kwadratowej
f(x)=-2x²-8x+10
f(x)=3x²+2x-1
f(x)=9x²-81
3) Rozwiąż równanie
a) -4x²+20=0
b) x²-10x+25=0
c) 3x²+5x=2
4) Rozwiąż nierówność
a) 4x²+9x>0
b) 81x²≥25
c) ½(x+1)(x-1)≥0
5) Zapisz wzór funkcji w postaci iloczynowej o ile to możliwe
f(x)=½x²+6x+10
f(x)=⅔x²+6x+12
f(x)=-4x²+40x-36
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1.
f(x) = a(x-p)²+q - postac kanoniczna
a)
f(x) = -x²-6x-7
p = -b/2a = 6/(-2) = -3
q = f(p) = f(-3) = -(-3)²-6(-3)-7 = -9+18-7 = 2
f(x) = -(x+3)²+2
===========
b)
f(x) = 2x²-4x+12
p = 4/4 = 1
q = f(1) = 2*1²-4*1+12 = 2-4+12 = 10
f(x) = 2(x-1)²+10
============
c)
f(x) = ½ x²+4x+5
p = -4/(2*½) = -4
q = f(-4) = ½ *(-4)²+4*(-4)+5 = 8-16+5 = -3
f(x) = ½(x+4)²-3
===========
2.
a)
f(x) = -2x²-8x+10
-2(x²+4x-5) = 0 /:(-2)
x²+4x-5 = 0
Δ = b²-4ac = 16+20 = 36
√Δ = 6
x₁ = (-b-√Δ)/2a = (-4-6)/2 = -5
x₂ = (-b+√Δ)/2a = (-4+6)/2 = 1
MZ: {-5; 1}
b)
f(x) = 3x²+2x-1
3x²+2x-1 = 0
Δ = 4+12 = 16
√Δ = 4
x₁ = (-2-4)/6 = -1
x₂ = (-2+4)/6 = 1/3
MZ: {-1; 1/3}
c)
f(x) = 9x²-81
9x²-81 = 0 /:9
x²-9 = 0
(x+3)(x-3) = 0
x = -3 v x = 3
MZ: {-3;3}
3.
a)
-4x²+20 = 0 /:(-4)
x²-5 = 0
x² = 5
x = -√5 v x = √5
b)
x²-10x+25 = 0
(x-5)² = 0
x = 5
c)
3x²+5x = 2
3x²+5x-2 = 0
Δ = 25+24 = 49
√Δ = 7
x₁ = (-5-7)/6 = -12/6 = -2
x₂ = (-5+7)/6 = 2/6 = 1/3
Odp. x = -2 v x = 1/3
4.
a)
4x²+9x > 0
x(4x+9) = 0
x = 0
lub
4x+9 = 0 => x = -9/4
a = 4 > 0, parabola ramionami skierowana w górę
x ∈ (-∞; -9/4) u (0; +∞)
b)
81x² ≥ 25
81x²-25 ≥ 0
(9x+5)(9x-5) ≥ 0
9x = -5 => x = -5/9
9x = 5 => x = 5/9
a = 81 > 0, ramiona paraboli skierowane w górę
x ∈ (-∞; -5/9> u <5/9; +∞)
c)
½(x+1)(x-1) ≥ 0
x = -1 v x = 1
a = 1/2 > 0, ramiona paraboli skierowana w górę
x ∈ (-∞; -1> u <1; +∞)
5.
f(x) = a(x-x₁)(x-x₂) - postać iloczynowa
a)
f(x) = ½ x²+6x+10
½ x²+6x+10 = 0
Δ = 36-4(1/2)*10 = 16
√Δ = 4
x₁ = (-6-4)/1 = -10
x₂ = (-6+4)/1 = -2
f(x) = ½(x+1)(x+2)
b)
f(x) = ⅔ x²+6x+12
⅔ x²+6x+12 = 0
Δ = 36-2*(2/3)*12 = 36-32 = 4
√Δ = 2
x₁ = (-6-2)/(4/3) = -6
x₂ = (-6+2)/(4/3) = -3
f(x) = ⅔(x+6)(x+3)
c)
f(x) = -4x²+40x-36
-4(x²-10x+9) = 0 /:(-4)
x²-10x+9 = 0
Δ = 100-36 = 64
√Δ = 8
x₁ = (10-8)/2 = 1
x₂ = (10+8)/2 = 9
f(x) = -4(x-1)(x-9)
1) f(x) = - x² -6x -7
p = -b/2a = 6/(-2) = -3
q = f(p) = f(-3) = -(-3)² -6·(-3) -7 = -9 +18 -7 = 2
Postać kanoniczna: f(x) = a(x -p)² + q , f(x) = - (x +3)² + 2
f(x) = 2x² - 4x + 12
p = -b/2a = 4/4 = 1,
q = f(p) = f(1) = 2 -4 +12 = 10 , f(x) = 2( x -1)² + 10
f(x) = ½ x² + 4x +5
p = -b/2a = -4/1 = -4,
q = f(p) = f(-4) = ½·(-4)² + 4·(-4) + 5 = 8 -16 +5 = -3, f(x) = ½( x+4)² -3
2) f(x) = -2x² -8x + 10
Δ = b²-4ac = 64 -4·(-2)·10 = 64 + 80 = 144, √Δ = 12
x₁ = (8-12)/ (-4)= -4/(-4) = 1, x₂ = (8+12)/(-4) = 20/(-4) = -5
Miejsca zerowe: x = 1 i x = -5.
f(x) = 3x² +2x -1
Δ = 4 - 4·3·(-1) = 4 + 12 = 16, √Δ = 4
x₁ = (-2-4)/6 = -6/6 = -1, x₂ = (-2+4)/6 = 2/6 = ⅓
Miejsca zerowe: x = -1 i x = ⅓ .
f(x) = 9x² - 81
f(x) = 9(x² - 9) = 9(x -3) (x+3)
x-3 =0 , x+3 =0
x = 3, x = -3
3) a) -4x² + 20 = 0
-4( x² - 5) = 0
-4( x -√5)( x +√5) = 0
x - √5 = 0 ∨ x + √5 = 0
x = √5 x = - √5
b) x² - 10x + 25 = 0
(x - 5)² = 0
x - 5 = 0 ⇒ x = 5
c) 3x² + 5x = 2
3x² +5x -2 = 0
Δ = b²-4ac = 25 -4·3·(-2) = 25 + 24 = 49, √Δ = 7
x₁ = (-5-7)/6 = -12/6 = -2, x₂ = (-5+7)/6 = 2/6 = ⅓
Pierwiastki równania to: x = -2 i x = ⅓ .
4) a ) 4x²+ 9x > 0
x(4x + 9) > 0
x = 0, 4x +9=0
4x = -9 /:4
x = -2¼ Z wykresu paraboli skierowanej ramionami w górę
odczytujemy przedziały wartości dodatnich:
x ∈ ( -∞, -2¼ ) U (0, ∞)
b) 81x² ≥ 25
81 x² - 25 ≥ 0
(9x -5) (9x +5) ≥ 0
9x -5 = 0 , 9x +5 =0
9x =5 9x = -5
x = ⁵/₉ x = - ⁵/₉ Z wykresu paraboli skierowanej ramionami w górę
odczytujemy przedziały wartości nieujemnych:
x∈ (-∞, -⁵/₉ > U < ⁵/₉ , ∞)
c) ½ (x-1)(x+1) ≥ 0
x -1 =0 , x+1 =0
x = 1 x = -1 Z wykresu paraboli skierowanej ramionami w górę
odczytujemy przedziały wartości nieujemnych:
x∈ (-∞, -1 > U < 1, ∞ )
5) a) f(x) = ½x² + 6x + 10
Δ = b² -4ac = 36 -4·½ ·10 = 36 - 20 = 16, √Δ = 4
x₁ = (-6-4)/1 = -10, x₂ = (-6+4)/1 = -2
Postać iloczynowa: f(x) = a(x-x₁)(x-x₂) =½ (x +10) (x +2)
b) f(x) = ⅔x² + 6x + 12
Δ = 36 - 4·⅔·12 = 36 - 32 = 4, √Δ = 2
x₁ = (-6 -2)/ ⁴/₃ = -8·¾ = -6, x₂ = (-6+2)/ ⁴/₃ = -4· ¾ = -3
Postać iloczynowa: f(x) = ⅔( x + 6) (x + 3).
c) f(x) = -4x² + 40x - 36 = -4 (x² -10x +9)
x² -10x +9 =0
Δ = 100 - 4·9 = 100 -36 = 64, √Δ = 8
x₁ = (10-8)/2 = 2/2 = 1, x₂ = (10+8)/2= 18/2 = 9
Postać iloczynowa: f(x) = -4 (x - 1) ( x - 9)