Question

dada a função f(x)=4x2-3x+1 determine a)f(1)
b)f(0)
c)f(-2)
d)f(a+1)

Answer 1

Questão a)

$f(x)=4x^{2} -3x+1\\ \\ f(1)=4\cdot1^{2} -3\cdot 1+1\\ \\ f(1)=4\cdot1-3+1\\ \\ f(1)=4-3+1\\ \\ \boxed{f(1)=2}$

Questão b)

$f(x)=4x^{2} -3x+1\\ \\ f(0)=4\cdot0^{2} -3\cdot 0+1\\ \\ f(1)=4\cdot0-0+1\\ \\ f(1)=0-0+1\\ \\ \boxed{f(0)=1}$

Questão c)

$f(x)=4x^{2} -3x+1\\ \\ f(-2)=4\cdot(-2)^{2} -3\cdot (-2)+1\\ \\ f(-2)=4\cdot4+6+1\\ \\ f(-2)=16+6+1\\ \\ \boxed{f(-2)=23}$

Questão d)

$f(x)=4x^{2} -3x+1\\ \\ f(a+1)=4\cdot(a+1)^{2} -3\cdot (a+1)+1\\ \\ f(a+1)=4\cdot(a^{2}+2\cdot a\cdot 1+1^{2} )-3a-3+1\\ \\ f(a+1)=4a^{2} +8a+4-3a-3+1\\ \\ \boxed{f(a+1)=4a^{2} +5a+2}$

:)