Question

18) determine o vertice dq funçao quadratica f(x)=-10x^2-20-40

Answer 1

-10x²-20-40=0
∆=b²-4ac
∆=(-20)²-4.(-10).(-40)
∆=400-1600
∆=-1200
xv=-b/2a
xv=-(-20)/2.(-10)
xv=20/-20
xv=-1
yv=-∆/4a
yv=-(-1200)/4.(-10)
yv=1200/-40
yv=-30
(xv= -1 ; yv=-30)