Question

(2 pontos) – Dada a função f(x) = 2x² – 3x + 5, calcule:
a) f (2)
$b)f(\frac{1}{2} )$
c) f (– 3)

$d)f( - \frac{2}{3} )$
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Answer 1

A) f(x) = 2x^2 - 3x + 5
f(2) = 2.(2)^2 -3 (2) + 5
f(2) = 2.4 -6 + 5
f(2) = 8 -1
f(2) = 7

B) f(1/2)= 2(1/2)^2 -3 (1/2) + 5
f(1/2) = 2 (1/4) -3 (1/2) + 5
f(1/2) = 1/2 - 3/2 + 5
f(1/2) = (1-3)/2 + 5
f(1/2) = -1 + 5
f(1/2) = 4

C) f(-3) = 2(-3)^2 - 3(-3) +5
f(-3) = 2(9) + 9 + 5
f(-3) = 18 + 9 + 5
f(-3) = 32

D) f(-2/3) = 2(-2/3)^2 -3(-2/3) +5
f(-2/3) = 2.(4/9) +2 +5
f(-2/3) = 8/9 + 2 + 5
f(-2/3) = 8/9 + 7
f(-2/3) = (8 + 63)/9
f(-2/3) = 71/9