Question

determine as coordenadas do vértice da parábola f(x)=1/3x²-x+2 onde ∆=b²-4.a.c, Xv= -b/2a e Yv= -/4a

Answer 1

Fórmula geral da equação de segundo grau :

f(x)= ax²+bx+c

f(x)= 1/3x²-x+2

a= 1/3
b= -1
c= 2

Coordenadas do vértice: (Xv,Yv)
Xv= x do vértice
Yv= y do vértice

Xv= -b/2a

Xv= -(-1)/2(1/3)
Xv=1/2/3 (Divisão de fração)
Xv= 3/2

Yv= -∆/4a
Yv= -(b²-4.a.c)/4a
Yv= -[(-1)²-4.(1/3).2]/4.(1/3)
Yv= -(1-8/3)/4/3
Yv= -(-5/3)/4/3
Yv= -(-5/4)
Yv= +5/4

(Xv,Yv)

(3/2,5/4)

Answer 2

$f(x)=\frac{1}{3}{x}^{2}-x+2$

$\boxed{\Delta={b}^{2}-4ac}$

$\Delta={(-1)}^{2}-4.\frac{1}{3}.2$

$\Delta=1-\frac{8}{3}=\frac{3-8}{3}=-\frac{5}{3}$

$x_{v}=-\frac{b}{2a}$

$x_{v}=-\frac{-1}{2.\frac{1}{3}}$

$x_{v}=\frac{1}{\frac{2}{3}}=1.\frac{3}{2}$

$\boxed{\boxed{x_{v}=\frac{3}{2}}}$

$y_{v}=-\frac{\Delta}{4a}$

$y_{v}=-\frac{-\frac{5}{3}}{4\frac{1}{3}}$

$y_{v}=\frac{\frac{5}{\cancel{3}}}{\frac{4}{\cancel{3}}}$

$\boxed{\boxed{y_{v}=\frac{5}{4}}}$