Copy and complete the table of value and draw the graph of \( y=x^{2}+3 x-1 \) \( | \)\begin{tabular}{c|c|c|c|c|c|c|c|}\( x \) & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ \( y \) & 35 & 4 \\ \hline\end{tabular}

Let's start by calculating the values of \( y \) for each \( x \) value in the table using the formula \( y = x^{2} + 3x - 1 \). 1. For \( x = -3 \): \[ y = (-3)^{2} + 3(-3) - 1 = 9 - 9 - 1 = -1 \] 2. For \( x = -2 \): \[ y = (-2)^{2} + 3(-2) - 1 = 4 - 6 - 1 = -3 \] 3. For \( x = -1 \): \[ y = (-1)^{2} + 3(-1) - 1 = 1 - 3 - 1 = -3 \] 4. For \( x = 0 \): \[ y = (0)^{2} + 3(0) - 1 = 0 - 0 - 1 = -1 \] 5. For \( x = 1 \): \[ y = (1)^{2} + 3(1) - 1 = 1 + 3 - 1 = 3 \] 6. For \( x = 2 \): \[ y = (2)^{2} + 3(2) - 1 = 4 + 6 - 1 = 9 \] 7. For \( x = 3 \): \[ y = (3)^{2} + 3(3) - 1 = 9 + 9 - 1 = 17 \] Now, we can complete the table: \[ | \begin{tabular}{c|c|c|c|c|c|c|c|} \( x \) & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline \( y \) & -1 & -3 & -3 & -1 & 3 & 9 & 17 \\ \hline \end{tabular} | \] To draw the graph of \( y = x^{2} + 3x - 1 \), plot the points from the completed table: - (-3, -1) - (-2, -3) - (-1, -3) - (0, -1) - (1, 3) - (2, 9) - (3, 17) Connect these points with a smooth curve to depict the parabola opening upwards, as it's a quadratic function.