Proyecto Modular 4 Mate

Resuelve y grafica las siguientes ecuaciones cuadráticas aplicando el método que se indica:

a) Tabulación 2x2 – 3X = 9

2x2 – 3X – 9 = 0

A = 2

B = -3

C = -9

x = -b ± √ b2 – 4 ac / 2a

x = +3b ± √ (-3)2 – 4 (2) (-9) / 2 (2)

x = +3b ± √ 9 + 72 / 4

x = +3b ± 9 / 4

x1 = 3 + 9 / 4 = 12 / 4 = 3

x2 = 3 – 9 = -6 / 4 = -1.5

2x2 – 3x – 9 = 0

2 (0)2 – 3 (0) – 9 = 0

(0, -9)

x -3 -2 -1 0 1 2 3 4

y 18 5 -4 -9 -10 -7 0 11

b) tabulación m2 – 2m – 3 = 0

a = 1

b = -2

c = -3

-b ± √ b2 – 4ac / 2ª

+ 2 ± √ (-2)2 – 4 (1) (-3)

+2 ± √ 4 + 12 / 2

X1 = 2± 4 / 2 = 6 / 2 = 3

X2 = 2 – 4 = -2 / 2 = -1

x -3 -2 -1 0 1 2 3 4 5

y 12 5 0 -3 -4 -3 0 5 12

c) Cinco puntos 14a2 – 9a – 18 = 0

a = 14

b = -9

c = -18

-b ± √ b2 – 4ac / 2a

9 ± √ 81 – 4 (14) (-18)

81 + 1008

1089

x = 9 ± 3 / 28 = 42 / 28 = 21 / 14 = 3 / 2 = 1.5

x = 9 – 33 / 28 = -24 / 28 = -6 / 7

= -0.857142857

Coordenadas del vértice con el método estándar

h = -b / 2a

9 / 2 (14) = 9 / 23 = 0.32

Para obtener “k” se sustituye el valor de “h” en la función:

14a2 – 9a – 18 = 0

14(.32)2 – 9(.32) – 18 = 0

1.43 – 2.88 – 18 = 0

k = -19.45

v (0.32 , -19 .45)

v = a (x – h)2 + k

v = 14 (x-0.32)2 – 19.45

Dos puntos

14a2 – 9a – 18 = 0

14 (-1.85)2 – 9(-1.85) – 18 = 0

47.915 + 16.56 – 18

(-1.85 , 46.56) = 47

14 (2.5)2 – 9 (2.5) – 18 = 0

87.5 – 22.5 – 18 = 47

(2.5 , 47)

x1 = 1.5

x2 = -0.85

Vértice (0.32 ,

-19.45)

12n2 + 12 = 25n

12n2 – 25n + 12

A = 12

B = -25

C = 12

x = -b ± √ b2 – 4ac/2ª

x = 25 ± √ (25)2 – 4(12)(12)/2(12)

x = 25 ± √625 – 576/24

x = 25 ± 7/24

x1 = 25 + 7/24 = 32/24 = 16/12 = 8/6 = 4/3 = 1.333

x2 = 25 – 7/24 = 18/24 = 9/12 = 3/4 = 0.75

Coordenada del vértice:

d) Cinco puntos 12n2 – 25n +12 = 0

a = 12

b = -25

c = 12

x-b/2a =25/2(12) = 25/24 = 1.04

y = 4ac – b2/4a

y = 4(12)(12) – (25)2/ 4(12)

y = 576 – 625

y = -49/48 = -1.02

y = 12x2

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