Question

dada a função quadratica f:R-R definida por f(x)= x² -6x+8, Determine: F(1), f(0), f(-2) e f(1/2)

Answer 1

f(x)=x²-6x+8

f(1)=1²-6.1+8
f(1)=1-6+8
f(1)=3

f(0)=0²-6.0+8
f(0)=8

f(-2)=(-2)²-6(-2)+8
f(-2)=4+12+8
f(-2)=24

f(1/2)=(1/2)²-6.1/2+8
f(1/2)=1/4-3+8
f(1/2)=1/4+5
f(1/2)=1/4+20/4
f(1/2)=21/4