The value which staisfy the equation are the coordinate of the f(x)
[tex]f(x)=3x^2-4[/tex]A) Table :
[tex]\begin{gathered} at\text{ (-2,-8)} \\ f(-2)=3(-2)^2-4 \\ f(-2)=12-4 \\ f(-2)=8 \\ \text{but we have y =-8} \\ -8\ne8,\text{ } \end{gathered}[/tex]So, the given table is not the solution of f(x)
B) Table :
[tex]\begin{gathered} at\text{ (-2,9)} \\ f(-2)=3(-2)^2-4 \\ f(-2)=12-4 \\ f(-2)=8 \\ \text{but we have y =}9 \\ -8\ne9,\text{ } \end{gathered}[/tex]
the given table is not the solution of f(x)
C) table:
[tex]\begin{gathered} at\text{ (-2,9)} \\ f(-2)=3(-2)^2-4 \\ f(-2)=12-4 \\ f(-2)=8 \\ and\text{ we have y =}8 \\ 8=8,\text{ } \end{gathered}[/tex]
now, for next coordinate :
[tex]\begin{gathered} at\text{ (-1,-1)} \\ f(-1)=3(-1)^2-4 \\ f(-1)=3-4 \\ f(-1)=-1 \\ and\text{ we have y =}-1 \\ -1=-1,\text{ } \end{gathered}[/tex][tex]\begin{gathered} at\text{ (0,-4)} \\ f(0)=3(0)^2-4 \\ f(0)=0-4 \\ f(0)=-4 \\ and\text{ we have y =}-4 \\ -4=-4,\text{ } \end{gathered}[/tex]
[tex]\begin{gathered} at\text{ (1,-1)} \\ f(-1)=3(1)^2-4 \\ f(-1)=3-4 \\ f(-1)=-1 \\ and\text{ we have y =}-1 \\ -1=-1,\text{ } \end{gathered}[/tex]
[tex]\begin{gathered} at\text{ (2,}8\text{)} \\ f(2)=3(2)^2-4 \\ f(2)=12-4 \\ f(2)=8 \\ and\text{ we have y =}8 \\ 8=8\text{ } \end{gathered}[/tex]
Since all the value staisfy the equation
Thus, The table is solution of the f(x)
D) table :
[tex]\begin{gathered} at\text{ (-2,6)} \\ f(-2)=3(-2)^2-4 \\ f(-2)=12-4 \\ f(-2)=8 \\ but\text{we have y =}6 \\ 6\ne8\text{ } \end{gathered}[/tex]
Thus, the given table is not the solution of f(x)
Answer : C)
