f(x)=(a+2)x+a
a) P(4, 3) =>
f(4) = 3
4(a+2)+a=3 =>
4a+8+a=3 =>
5a=-5 =>
a=-1
b)
Dacă a=-1 =>
f(x)=(-1+2)x+-1 =>
f(x)=x-1;
x-1=0=>x=1 => A(1; 0)
f(0)=0-1=-1 => A(0; -1)
c) A=|1 x (-1)| / 2 = 1/2
d)
M(a, -a) (dar ar merge şi M(-a, a))
f(a) =a-1
a-1 = -a
2a=1 =>a=1/2
=>
M(1/2, -1/2)