Question

To mandando 50 pontos nessas questoes.

Considere a funcao f:IR_IR definida por f(x)=2x(2)-3x+10.Determine:

A)f(2)

B)f(-5)

C)f(0)

D)f(1 sobre 2)

E)f(5 sobre 2)

F)(3 sobre 5)

G)f(-3)

Me ajudem é pra amanha.

Answer 1

$f(x)=2x^2-3x+10\\\\\\ f(2) = 2*2^2-3*2+10\\ f(2)=2*4-6+10\\ f(2)=8+4\\ f(2)=12\\\\ f(-5)=2*(-5)^2-3*-5+10\\ f(-5)=2*25+15+10\\ f(-5)=50+25\\ f(-5)=75\\\\ f(0)=2*0^2-3*0+10\\ f(0)=10\\\\ f(1/2)=2*(1/2)^2-3*(1/2)+10\\ f(1/2)=2*1/4-3/2+10\\ f(1/2)=1/2-3/2+10\\ f(1/2)=-1+10\\ f(1/2)=9\\\\ f(5/2)=2*(5/2)^2-3*(5/2)+10\\ f(5/2)=2*25/4-15/2+10\\ f(5/2)=25/2-15/2+10\\ f(5/2)=10/2+10\\ f(5/2)=15$

$f(3/5)=2*(3/5)^2-3*(3/5)+10\\ f(3/5)=2*9/25-9/5+10\\ f(3/5)=18/25-9/5+10\\ f(3/5)=18/25-45/25+10\\ f(3/5)=-27/25+10\\ f(3/5)=-27/25+250/25\\ f(3/5)=223/25\\\\ f(-3)=2*(-3)^2-3*(-3)+10\\ f(-3)=2*9+9+10\\ f(-3)=18+19\\ f(-3)=37$