Points: 0 of 1 Save Find a quadratic function expressed in vertex form that models the data in the given table. (Hint: Let the first data point in the table be the vertex.) \begin{tabular}{|c|c|c|c|c|} \hline $\mathbf{x}$ & 1980 & 1990 & 2000 & 2010 \\ \hline $\mathbf{y}$ & 6 & 66 & 242 & 555 \\ \hline \end{tabular} The vertex form of the equation is $y=\square$. (Use integers or decimals for any numbers in the expression.)
Solution
Step 1 ：The vertex form of a quadratic function is given by \(y=a(x-h)^2+k\), where \((h,k)\) is the vertex of the parabola. In this case, the vertex is given as the first data point in the table, which is (1980, 6). We can substitute this into the equation to get \(y=a(x-1980)^2+6\).
Step 2 ：To find the value of \(a\), we can use another point from the table. Let's use the point (1990, 66). Substituting these values into the equation gives us \(66=a(1990-1980)^2+6\). We can solve this equation to find the value of \(a\).
Step 3 ：Solving the equation gives us \(a=\frac{3}{5}\).
Step 4 ：Therefore, the quadratic function that models the data in the table is \(y=\frac{3}{5}(x-1980)^2+6\).
Step 5 ：Final Answer: The vertex form of the equation is \(\boxed{y=\frac{3}{5}(x-1980)^2+6}\).
