Question

dada a função f(x) =ax + b...​

Answer 1

$\text{f(x) = ax+ b}$

Informações :

f(1) = - 1 e f(4) = 5

Então temos :

$\text{f(x) = ax+b } \\\\ \text{f(1) = -1} \\\\ \text{a.1+b} =-1 \\\\\ \text e \\\\\\\ \text{f(x) = ax+b } \\\\ \text{f(4) = 5} \\\\ \text{a.4+b} = 5$

Resolvendo o sistema :

$\displaystyle \text {a+b}=-1 \\\\ \underline {\text{4a+b} = 5} \ - \\\\ \text a - 4\text a + \text b -\text b = -1-5 \\\\ -3\text a = -6 \\\\\ \boxed{\text a = 2} \\\\\\\ \text a+\text b = -1 \\\\\\\ 2+ \text b = - 1 \\\\\\ \boxed{\text b = -3}$

Portanto :

$\huge\boxed{\text a= 2 \ ; \ \text b = -3}\checkmark$

OU

Podemos fazer assim :

equação da reta :

$\text y -\text y_o = \text m (\text x- \text x_o)$

Pontos :

$\text f(1)=-1 \to (1,-1) \\\\ \text f(4)=5 \to (4,5)$

Coeficiente angular :

$\displaystyle \text m=\frac{\text y_2-\text y_1}{\text x_2-\text x_1} \\\\\\ \text m = \frac{5-(-1)}{4-1} \\\\\\ \text m =\frac{5+1}{3} \\\\\\ \boxed{\text m = 2}$

Substituindo o ponto (1,-1) na equação da reta :

$\text y -\text y_o = \text m (\text x- \text x_o)$

$\text y-(-1) = 2(\text x-1) \\\\ \text y +1=2\text x-2 \\\\ \text y = 2\text x-3 \\\\\\\ \text{Portanto} : \\\\\\ \huge\boxed{\text a = 2 \ ; \ \text b = -3 }\checkmark$