The problem provides the quadratic function $y = x^2 - 2x - 3$ and an incomplete table of values for $x$ and $y$. (i) Find the value of $y$ when $x = 1$. (ii) Draw the graph of the function using a suitable scale on graph paper. (iii) Write the equation of the axis of symmetry using the graph. (iv) Describe the behavior of $y$ in the interval $1 < x < 3$. (v) Express the given function in the form $y = (a - x)^2 - b$, where $a$ and $b$ are constants.

1. Problem Description

The problem provides the quadratic function $y = x^2 - 2x - 3$ and an incomplete table of values for $x$ and $y$.

(i) Find the value of $y$ when $x = 1$.

(ii) Draw the graph of the function using a suitable scale on graph paper.

(iii) Write the equation of the axis of symmetry using the graph.

(iv) Describe the behavior of $y$ in the interval $1 < x < 3$.

(v) Express the given function in the form $y = (a - x)^2 - b$, where $a$ and $b$ are constants.

2. Solution Steps

(i) To find the value of $y$ when $x = 1$, substitute $x = 1$ into the equation $y = x^2 - 2x - 3$:

$y = (1)^2 - 2(1) - 3 = 1 - 2 - 3 = -4$

(ii) To draw the graph, plot the points given in the table, including the point (1, -4) we just found:

(-2, 5), (-1, 0), (0, -3), (1, -4), (2, -3), (3, 0), (4, 5)

(iii) The axis of symmetry is the vertical line that passes through the vertex of the parabola. The vertex is at $x=1$, so the equation of the axis of symmetry is $x = 1$.

(iv) In the interval $1 < x < 3$, the value of $y$ increases. The function decreases from $x=1$ to $x=2$, reaching its minimum $y=-4$, then increases from $x=2$ to $x=3$, where $y=0$.

(v) To express the given function in the form $y = (a - x)^2 - b$, we need to complete the square for the given quadratic:

$y = x^2 - 2x - 3 = (x^2 - 2x + 1) - 1 - 3 = (x - 1)^2 - 4 = (-x + 1)^2 - 4 = (1 - x)^2 - 4$.

Therefore, $y = (1 - x)^2 - 4$, which is in the form $y = (a - x)^2 - b$, with $a = 1$ and $b = 4$.

3. Final Answer

(i) $y = -4$ when $x=1$.

(iii) Equation of the axis of symmetry: $x = 1$

(iv) In the interval $1 < x < 3$, y increases.

(v) $y = (1 - x)^2 - 4$