Let's break down the two parts of the question:

Part (a)

We are given the equation $y = x^2 - 2x - 3$ and a table of values for $x$ and $y$. The goal is to fill in the missing values for $y$ when $x$ is -2, -1, 0, 1, 2, 3, and 4.

To do this, substitute each $x$-value into the equation and solve for $y$.

Let's calculate the values for each $x$:

• When $x = -2$: $y = (-2)^2 - 2(-2) - 3 = 4 + 4 - 3 = 5$
• When $x = -1$: $y = (-1)^2 - 2(-1) - 3 = 1 + 2 - 3 = 0$
• When $x = 0$: $y = (0)^2 - 2(0) - 3 = -3$
• When $x = 1$: $y = (1)^2 - 2(1) - 3 = 1 - 2 - 3 = -4$
• When $x = 2$: $y = (2)^2 - 2(2) - 3 = 4 - 4 - 3 = -3$
• When $x = 3$: $y = (3)^2 - 2(3) - 3 = 9 - 6 - 3 = 0$
• When $x = 4$: $y = (4)^2 - 2(4) - 3 = 16 - 8 - 3 = 5$

So the completed table looks like this:

Part (b)

For the graph, plot the points based on the table we just completed:

• Plot the points (-2, 5), (-1, 0), (0, -3), (1, -4), (2, -3), (3, 0), and (4, 5).
• After plotting the points, draw a smooth curve to represent the quadratic function $y = x^2 - 2x - 3$.

Would you like more detailed help with plotting the graph or with any other aspect of this question?