Question

Dada a função f(x) = 2x²+3x-5 , determine: f(0); f(-1); f(2);f(3/2)

Answer 1

função: f(x)=2x²+3x-5

f(0)= 2.0²+3.0-5
f(0)= 0+0-5
f(0)= -5

f(-1)= 2.(-1)²+3.(-1)²-5
f(-1)= 2+3-5
f(-1)= 0

f(2)= 2.2²+3.2²-5
f(2)= 2.4+3.4-5
f(2)= 8+12-5
f(2)= 15

f(3/2)= 2.(3/2)²+3.(3/2)²-5
f(3/2)= 2.9/4+3.9/4-5
f(3/2)= 18/4+27/4-5
f(3/2)= 45/4-5
f(3/2)= 45-20/4
f(3/2)= 25/4