Question

calcule os vértices da função quadratica:
a)y=x^2-3x+2
b)y=-x^2+6x-9
c)y=-x^2+4x-3
d)y=x^2-4x+4

Answer 1

a) y=x^2-3x+2
(x-2)(x-1)=0
x1-2=0》x1=2
x2-1=0》x2=1

b)y=x^2+6x-9
a=1. b= 6 c=-9
delta=b^2-4ac
delta=6^2-4(1)(-9)
delta=36-4(-9)
delta=36+36
delta= 72
x=-b+,-\/delta/2a
x=-6+,- 8.48/2
x1=-6+8,48/2》x1=2,48/2》x1=1,24
x2=-6-8,48/2》x2=-14,48/2》x2=7,24

c)y=x^2+4x-3
a=1. b=4. c=-3
delta=b^2-4ac
delta=4^2-4(1)(-3)
delta=16-4(-3)
delta=16+12
delta=28
x=-b+,-\/felta/2a
x= - 4 +,-\/28/2(1)
x=-4+,-5,29/2
x1=-4+5,29/2》x=129/2》x1=0,65
x2=-4-5,29/2》x=-9,29/2》x2=-4,65
x3-1=0》x2=1

d)y=x^2-4x+4
(x-4)(x-1)=0
x1-4=0》x1=4
x2-1=0》x2=1