Question

dada a função quadrática f (x)=x2-4x determine :
f (1)
f (raiz 2)
f (h+1)

Answer 1

f(x)=x²-4x
f(1)=1²-4*1
f(1)=1-4
f(1)=-3

f(√2)=√2²-4*√2
f(√2)= 2-4√2
f(√2)=-2√2

f(h+1)=(h+1)²-4*(h+1)
f(h+1)=(h+1)*(h+1)-4h+4
f(h+1)=h²+h+h+1-4h+4
f(h+1)=h²+h+h-4h=-1-4
f(h+1)==h²+2h-4h=-5
f(h+1)=h²-2h=-5
f(h+1)= h²-2h+5=0
a=1 b=-2 c=5

x=-b±√b²-4*a*c/2*a
x=-(-2)±√-2²-4*1*5/2*1
x=2±√4-20/2
x=2±√-16/2