Question

determine os zeros as coordenadas do vértice das Parabolas e esboce os Gráficos das Seguintes funções quadráticas:
a) F (x)= x²+6x+8
b) f (x)= -x²+ 4x-3
c) f (x)= x²+ 2x+ 1
d) f (x)= -x² - 2x-5

Answer 1

a) f(x)=x^2+6x+8
x^2+6x+8=0
delta=b^2-4ac
delta=6^2-4.1.8
delta=36-32
delta=4
x=-b-/+ raiz de delta/2a
x=-6-2/2.1
x=-8/2
x=-4
x=-6+2/2.1
x=-4/2
x=-2
xv=-b/2a
xv=-6/2.1
xv=-6/2
xv=-3
yv=-delta/4a
yv=-4/4.1
yv=-4/4
yv=-1
b) -x^2+4x-3=0
delta=b^2-4ac
delta=4^2-4.(-1).(-3)
delta=16-12
delta=4
x=-b-/+ raiz de delta/2a
x=-4-2/2.(-1)
x=-6/-2
x=3
x=-4+2/2.(-1)
x=-2/-2
x=1
xv=-b/2a
xv=-4/2.(-1)
xv=-4/-2
xv=2
yv=-delta/4a
yv=-4/4.(-1)
yv=-4/-4
yv=1
c) x^2+2x+1=0
delta=b^2-4ac
delta=2^2-4.1.1
delta=4-4
delta=0
x=-b-/+ raiz de delta/2a
x=-2-0/2.1
x=-2/2
x=-1
xv=-b/2a
xv=-(-2)/2.1
xv=2/2
xv=1
yv=-delta/4a
yv=-0/4.1
yv=-0/4
yv=0
d) -x^2-2x-5=0
delta=b^2-4ac
delta=(-2)^2-4.(-1).(-5)
delta=4-20
delta=-16
delta negativo, não há raízes reais.
xv=-b/2a
xv=-(-2)/2.(-1)
xv=2/-2
xv=-1
yv=-delta/4a
yv=-(-16)/4.(-1)
yv=16/-4
yv=-4