Question

Obtenha o vértice das parábolas abaixo:

g) y=(x-1)²+3

h) y=(2-x)²

i) y=-2x²+60x

j) x²-4x+8

k) y=(x+1)(2+x)

Answer 1

g) y=(x-1)²+3
y=x²-2x+1+3
y=x²-2x+4

Delta

∆= b²-4ac
∆=(-2)²-4×1×4
∆= 4-4×4
∆= 4-16
∆= -12

Vértice

xV= -b/2a
xV= -(-2)/2×1
xV= 2/2
xV= 1

yV= -∆/4a
yV= -(-12)/4×1
yV= 12/4
yV= 3

Logo, as coordenadas do vértice são:
V(1, 3)

h) y= (2-x)²
y= x²-4x+4

Delta

∆= b²-4ac
∆= (-4)²-4×1×4
∆= 16-4×4
∆= 16-16
∆= 0

Vértice

xV= -b/2a
xV= -(-4)/2×1
xV= 4/2
xV= 2

yV= -∆/4a
yV= -(0)/4×1
yV= 0/4
yV= 0

Logo, as coordenadas do vértice são:
V(2, 0)

i) y= -2x²+60x

Delta

∆= b²-4ac
∆= 60²-4×(-2)×0
∆= 3600-4×0
∆= 3600

Vértice

xV= -b/2a
xV= -60/2×(-2)
xV= -60/-4
xV= 60/4
xV= 15

yV= -∆/4a
yV= -3600/4×(-2)
yV= -3600/-8
yV= 3600/8
yV= 450

Logo, as coordenadas do vértice são:
V(15, 450)

j) x²-4x+8

Delta

∆= b²-4ac
∆= (-4)²-4×1×8
∆= 16-4×8
∆= 16-32
∆= -16

Vértice


xV= -b/2a
xV= -(-4)/2×1
xV= 4/2
xV= 2

yV= -∆/4a
yV= -(-16)/4×1
yV= 16/4
yV= 4

Logo, as coordenadas do vértice são:
V(2, 4)

k) y= (x+1)(2+x)
y= 2x+x²+2+x
y= 3x+x²+2
y= x²+3x+2

Delta

∆= b²-4ac
∆= 3²-4×1×2
∆= 9-4×2
∆= 9-8
∆= 1

Vértice

xV= -b/2a
xV= -3/2×1
xV= -3/2
xV= -1,5

yV= -∆/4a
yV= -1/4×1
yV= -1/4
yV= -0,25

Logo, as coordenadas do vértice são:
V(-1,5, -0,25)

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