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proyecto modular IV


matematicas 1b








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



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



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







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

Dospuntos
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: 

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 – 14x + 2
a = 12
b = -14 
c = 2

x = -b ± √ b2 – 4ac / 2ª
x = 14 ± √ (14)2 – 4 (12) (2) / 2 (12)
x = 14 ± √ 196 – 96 / 24 
x = 14 ± 10 / 24
x1 = 14 +...