Given the quadratic function formula
[tex]\begin{gathered} y=ax^2+bx+c \\ \text{where a, b and c are real numbers and a}\ne0 \end{gathered}[/tex]
Use two points on the table to find the values of the unkonwns in the above equation
[tex]\begin{gathered} \text{when x=2, y=11} \\ 11=a(2)^2+b+c \\ 11=4a+2b+c_{}-----\text{eqn (1)} \end{gathered}[/tex][tex]\begin{gathered} \text{when x=4, y=47} \\ 47=a(4^2)+bx+c \\ 47=16a+4b+c\text{ -------eqn(2)} \end{gathered}[/tex]
Also
[tex]\begin{gathered} 107=a(6)^2+6b+c \\ 107=36a+6b+c------\text{eqn}(3) \end{gathered}[/tex][tex]\begin{bmatrix}{4} & {2} & {1} \\ {16} & {4} & {1} \\ {36} & {6} & {1}\end{bmatrix}\begin{bmatrix}{} & {a} & {} \\ {} & {b} & {} \\ {} & {c} & \end{bmatrix}=\begin{bmatrix}{} & {11} & {} \\ {} & {47} & {} \\ {} & {107} & \end{bmatrix}[/tex]
Thus, a = 3, b= 0 and c= -1
Substitute the above values in the general quadratic equation formula
[tex]\begin{gathered} y=3x^2+0(x)+(-1) \\ y=3x^2-1 \end{gathered}[/tex]
Hence, the fucntion that best describe the relationship in the table is y=3x² - 1
