[tex]\text{Verificarea prin care vrem sa aflam daca un punct apartine graficului } \\
\text{unei functii, se face in felul urmator: } \\
\text{Variabila x a functiei primeste valoarea coordonatei x a punctului. } \\
\text{Se calculeaza f(x)} \\
=text{Daca f(x) = cu coordonata y a functiei, atunci punctul} \\
\text{apartine graficului functiei}[/tex]
[tex]f(x)=2x^2 \\
a)~~A(2; ~-2),~~f(2\times 2^2) = 8;~~ 8 \neq A_y=-2 ~~\Rightarrow A \notin G(f) \\
b)~~B(1; ~2),~~f(2\times 1^2) = 2;~~ 2 = B_y=2 ~~\Rightarrow B \in G(f) \\
c)~~C(-1; ~0,5),~~f(2\times (-1)^2) = 2;~~ 2 \neq C_y=0,5 ~~\Rightarrow C \notin G(f) [/tex]
[tex]d)~~D(1; ~-0,5),~~f(2\times 1^2) = 2;~~ 2 \neq D_y=-0,5 ~~\Rightarrow D \notin G(f) \\
e)~~E(0,5; ~0,5),~~f(2\times (0,5)^2) = 0,5;~~ 0,5 = E_y=0,5 ~~\Rightarrow E \in G(f) \\[/tex]
[tex]f(x)=-0,5x^2 \\
a)A(2; ~-2),~~f(-0,5\times 2^2) = -2;~ -2 = A_y=-2 \\ \Rightarrow A \in G(f) \\
b)B(1; ~2),~~f(-0,5\times 1^2) = -0,5;~ -0,5 \neq B_y=2 \\ \Rightarrow A \notin G(f) \\
c)C(-1; ~0,5),~~f(-0,5\times (-1)^2) = -0,5;~ -0,5 \neq C_y=0,5 \\ \Rightarrow A \notin G(f) \\
d)D(1; ~-0,5),~~f(-0,5\times 1^2) = -0,5;~ -0,5 = D_y=-0,5 \\ \Rightarrow D \in G(f) \\
e)E(0,5; ~0,5),~~f(-0,5\times (0,5)^2) = -0,125;~ -0,125 \neq E_y=0,5\\
\Rightarrow E \notin G(f) \\[/tex]
[tex]f(x)=0,5x^2 \\
a)A(2; ~-2),~~f(0,5\times 2^2) = 2;~ 2 \neq A_y=-2 ~\Rightarrow A \notin G(f) \\
b)B(1; ~2),~~f(0,5\times 1^2) = 0,5;~ 0,5 \neq B_y=2 ~\Rightarrow A \notin G(f) \\
c)C(-1; ~0,5),~~f(0,5\times (-1)^2) = 0,5;~ 0,5 = C_y=0,5 ~\Rightarrow A \in G(f) \\
d)D(1; ~-0,5),~~f(0,5\times 1^2) = 0,5;~ 0,5 \neq D_y=-0,5 ~\Rightarrow D \notin G(f) \\
e)E(0,5; ~0,5),~~f(0,5\times (0,5)^2) = 0,125;~ 0,125 \neq E_y=0,5 \\
\Rightarrow E \notin G(f) \\[/tex]
