Question

01 -Dada a função quadrática y = 2x² – 3x + 1, calcule f(-1), f(-2), f(2) e f(10).

Answer 1

y = 2x² – 3x + 1

f(-1), f(-2), f(2), f(10):

f(-1) = 2x² – 3x + 1

f(-1) = 2(-1)² – 3(-1) + 1

f(-1) = 2 + 3 + 1

f(-1) = 6

f(-2) = 2x² – 3x + 1

f(-2) = 2(-2)² – 3(-2) + 1

f(-2) = 2 . 4 + 6 + 1

f(-2) = 8 + 6 + 1

f(-2) = 15

f(2) = 2x² – 3x + 1

f(2) = 2(2)² – 3(2) + 1

f(2) = 2 . 4 – 6 + 1

f(2) = 8 – 6 + 1

f(2) = 9 – 6

f(2) = 3

f(10) = 2x² – 3x + 1

f(10) = 2(10)² – 3(10) + 1

f(10) = 2 . 100 – 30 + 1

f(10) = 200 – 30 + 1

f(10) = 201 – 30

f(10) = 171