Question

AJUDEM PFVR, TENHO QUE ENTREGAR AMANHÃ :C

Answer 1

Temos que, $f(x)=x^2-5x+7$

a) $f(4)=4^2-5\cdot4+7$

$f(4)=16-20+7$

$f(4)=3$

b) $f(x)=1$

$x^2-5x+7=1$

$x^2-5x+6=0$

$x=\dfrac{(-5)\pm\sqrt{(-5)^2-4\cdot1\cdot6}}{2}$

$x=\dfrac{5\pm\sqrt{1}}{2}=\dfrac{5\pm1}{2}$

$x'=\dfrac{5+1}{2}=3$ e $x"=\dfrac{5-1}{2}=2$

c) $f(3)=3^2-5\cdot3+7=9-15+7$

$f(3)=1$

d) $f(3)=1$