Question

Uma função f do 2º grau é tal que f(0) = 6, f(1) = 2 e f(-2) = 20. Determine o valor de f(1/2). *​

Answer 1

Uma função do 2º grau é do tipo :

$\text{a.x}^2+\text{b.x}+\text c$

Temos :

f(0) = 6, f(1) = 2 e f(-2) = 20.

ou seja :

*

$\text a\text x^2+\text b.\text x+\text c \\\\ \text{f(0)}=6 \\\\ \text a.0^2+\text b.0 + \text c = 6 \\\\ \boxed{\text c= 6}$

*

$\text a.\text x^2+\text b.\text x+\text c \\\\ \text{f(1)} = 2 \\\\ \text a.1^2+\text b.1+6 = 2 \\\\ \text a+ \text b = -4 }$

*

$\text{a.x}^2+\text{b.x}+\text c \\\\ \text{f(-2)}=20 \\\\ \text a.(-2)^2+\text b.(-2)+6=20 \\\\ 4\text a-2\text b = 14 \\\\ 2\text a-\text b=7$

Agora temos os seguinte sistema :

$\text a+\text b = -4 \\\\ \underline {2\text a-\text b = 7 } \ + \\\\\\\ \text a+2\text a +\text b-\text b = -4+7 \\\\ 3\text a = 3 \\\\ \boxed{\text a = 1}$

Achando o b :

$\text a+\text b=-4 \\\\ 1+\text b = -4 \\\\ \boxed{\text b = -5}$

Portanto a função do 2º é :

$\text x^2-5\text x+6$

A questão pede  f(1/2), ou seja :

$\displaystyle (\frac{1}{2})^2-5.\frac{1}{2}+6 \\\\\\ \frac{1}{4}- \frac{5}{2}+6\\\\\\\ \underline{\text{tirando o MMC}} :\\\\ \frac{1-2.5+4.6}{4} \\\\\\ \frac{1-10+24}{4} \\\\\\ \boxed{\frac{15}{4}}$

Portanto :

$\huge\boxed{\text f\ (\frac{1}{2}) = \frac{15}{4}}\checkmark$