Question

Prosze o rozwiazanie zadan....

Answer 1

ZADANIE 19

 

a) y = x^2 - 2x + 5

 

delta = b^2 - 4ac = (-2)^2 - 4 * 1 * 5 = 4 - 20 = -16

 

x_w = -b/2a = -(-2)/2*1 = 2/2 = 1

y_w = 1^2 - 2*1 + 5 = 1 - 2 + 5 = 4

 

postać kanoniczna: y = (x - 1)^2 + 4

 

b) y = x^2 + 6x + 3

 

delta = 6^2 - 4 * 1 * 3 = 36 - 12 = 24

 

x_w = -6/2*1 = -6/2 = -3

y_w = (-3)^2 + 6*(-3) + 3 = 9 - 18 + 3 = -6

 

postać kanoniczna: y = (x + 3)^2 - 6

 

c) y = -x^2 + x - 1

 

delta = 1^2 - 4 * (-1) * 1 = 1 + 4 = 5

 

x_w = -1/2*(-1) = -1/-2 = 1/2y_w = -(1/2)^2 + 1/2 - 1 = -1/4 + 1/2 - 1 = -1/4 + 2/4 - 1 = -3/4

 

postać kanoniczna: y = -(x - 1/2)^2 - 3/4

 

ZADANIE 20

 

f(x) = 3x^2 - 6x + 1

 

x_w = -b/2a = -(-6)/2*3 = 6/6 = 1

y_w = f(1) = 3*(1)^2 - 6*1 + 1 = 3 - 6 + 1 = -2

 

W = (1, -2)

 

ZADANIE 21

 

/wykresy w załączniku/

 

a) y = x^2 + 8x + 15

 

x_w = -b/2a =-8/2 = -4

y_w = (-4)^2 + 8*(-4) + 15 = 16 - 32 + 15 = -1

 

W = (-4, -1)

 

postać kanoniczna: f(x) = (x + 4)^2 + 7

 

b) y = -x^2 + 4x - 6

 

x_w =-4/2*(-1) = -4/-2 = 2

y_w = -(2)^2 + 4*2 - 6 = -4 + 8 - 6 = -2

 

W = (2, -2)

 

postać kanoniczna: f(x) = -(x - 2)^2 - 2

 

c) y = 1/2x^2 + 2x + 2

 

x_w = -2/2*(1/2) = -2/1 = -2

y_w = 1/2*(-2)^2 + 2*(-2) + 2 = 1/2*4 - 4 + 2 = 2 - 2 = 0

 

W = (-2, 0)

 

postać kanoniczna: f(x) = 1/2(x + 2)^2

 

ZADANIE 16

 

W_1 = (-4, 1), f_1(x) = (x + 4)^2 + 1

W_2 = (-1, -2), f_2(x) = (x + 1)^2 - 2

W_3 = (4, 2), f_3(x) = (x - 4)^2 + 2