Miejsca zerowe:
Δ=5²-4*1*4=25-16=9
√Δ=3
x₁=(-5+3)/2=-2/2=-1
x₂=(-5-3)/2=-8/2=-4
p=-b/2a
p=-5/2=-2½
q=-Δ/4a
q=-9/4=-2¼
W(-2½ ;- 2¼)
f(x) = x^2 + 5x + 4
a = 1
b = 5
c = 4
zatem
p = -b/ (2a) = -5 /2 = -2,5
delta = b^2 - 4ac = 5^2 - 4*1*4 = 25 - 16 = 9
q = - delta/ (4a) = - 9/4 = -2,25
zatem wierzchołek
W = ( p; q) = ( -2,5 ; - 2,25 )
=============================
Miejsca zerowe:x1 , x2
p(delty) = p ( 9) = 3
x1 = [ -b - p(delty)]/(2a)
x2 = [ - b + p(delty)]/(2a)
x1 = [ -5 - 3]/2 = - 8/2 = - 4
x2 = [ - 5 + 3]/2 = -2/2 = - 1
============================
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Miejsca zerowe:
Δ=5²-4*1*4=25-16=9
√Δ=3
x₁=(-5+3)/2=-2/2=-1
x₂=(-5-3)/2=-8/2=-4
p=-b/2a
p=-5/2=-2½
q=-Δ/4a
q=-9/4=-2¼
W(-2½ ;- 2¼)
f(x) = x^2 + 5x + 4
a = 1
b = 5
c = 4
zatem
p = -b/ (2a) = -5 /2 = -2,5
delta = b^2 - 4ac = 5^2 - 4*1*4 = 25 - 16 = 9
q = - delta/ (4a) = - 9/4 = -2,25
zatem wierzchołek
W = ( p; q) = ( -2,5 ; - 2,25 )
=============================
Miejsca zerowe:x1 , x2
p(delty) = p ( 9) = 3
x1 = [ -b - p(delty)]/(2a)
x2 = [ - b + p(delty)]/(2a)
zatem
x1 = [ -5 - 3]/2 = - 8/2 = - 4
x2 = [ - 5 + 3]/2 = -2/2 = - 1
============================