Question

Determine o vértice da função quadrática f(x)= 10X*-20X-40

Answer 1

f(x)=10X*-20X-40 (simplifica por 5)
f(x)=2X*-4X-8
a=2;-b=-4; c=-8
Xv=-b/2a
Xv=-(-4)/2.2
Xv=4/4
Xv=1

Yv=-∆/4a
∆=b*-4ac
∆=(-4)*-4.2.(-8)
∆=16+64
∆=80
Yv=-80/4.2
Yv=-80/8
Yv=-10

V{1,-10}

Answer 2

$f(x) = 10x^2-20x-40 \\ \\ a=10, b=-20 e c=-40 \\ \\ V(- \frac{b}{2a} , -\frac{delta}{4a} ) \\ \\ delta = b^2-4ac \\ \\ delta =(-20)^2-4.10.(-40)=400+1600=2000 \\ \\ V( -\frac{(-20)}{2.10} } , - \frac{2000}{4.10} ) \\ \\ V=(-( -\frac{20}{20}) ,- \frac{2000}{40} ) \\ \\ V(1,-50)$