Question

1- Determine as coordenadas do vértice da parábola definida pelas funções quadrática:

a) y= -x²-8x+16

b) y= 2x²+6x

c) y=x²-16 

Me Ajudem!!! Por gentileza.

Answer 1

Xv=$- \frac{b}{2a}$

Yv= -Δ/4a

A)
a=-1
b=-8
c=16
Δ=b² -4ac
Δ=64+64
Δ=128

Xv=$\frac{8}{-2} =-4$

Yv=$\frac{-128}{-4} =+32$

V(4,32)

b)

a=2
b=6
c=0

Δ=b²-4ac
Δ=36-4(2)(0)
Δ=36

Xv=$- \frac{b}{2a} =- \frac{6}{4} =- \frac{3}{2}$

Yv=-Δ/4a

Yv=$- \frac{36}{8} =- \frac{9}{2}$

V($\frac{3}{2} , \frac{9}{2}$)

d)
a=1
b=0
c=-16

Δ= b²-4ac
Δ=0²+64
Δ=64

Xv=$- \frac{b}{2a} = \frac{0}{2} =0$

Yv=-Δ/4a

Yv=$- \frac{36}{4} =-9$

V(0,-9)