Question

1. (a) complete the table for $y = x^2 - 2x - 4$

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y$	$4$		$-4$	$-5$		$-1$

(2)
(b) on the grid, draw the graph of $y = x^2 - 2x - 4$
grid with x-axis from -2 to 4 and y-axis from -5 to 7, with grid lines

Explanation:

Part (a)

Step 1: Find y when x = -1

Substitute \( x = -1 \) into \( y = x^2 - 2x - 4 \).
\( y = (-1)^2 - 2(-1) - 4 = 1 + 2 - 4 = -1 \)

Step 2: Find y when x = 2

Substitute \( x = 2 \) into \( y = x^2 - 2x - 4 \).
\( y = (2)^2 - 2(2) - 4 = 4 - 4 - 4 = -4 \)

Step 3: Find y when x = 4

Substitute \( x = 4 \) into \( y = x^2 - 2x - 4 \).
\( y = (4)^2 - 2(4) - 4 = 16 - 8 - 4 = 4 \)

Answer:

The completed table is:

\( x \)	-2	-1	0	1	2	3	4

Part (b)

To draw the graph of \( y = x^2 - 2x - 4 \):

1. Plot the points from the completed table: \((-2, 4)\), \((-1, -1)\), \((0, -4)\), \((1, -5)\), \((2, -4)\), \((3, -1)\), \((4, 4)\).
2. Draw a smooth curve through these points. The graph is a parabola opening upwards (since the coefficient of \( x^2 \) is positive). The vertex of the parabola can be found using the formula \( x = -\frac{b}{2a} \) for \( y = ax^2 + bx + c \). Here, \( a = 1 \), \( b = -2 \), so \( x = -\frac{-2}{2(1)} = 1 \). Substituting \( x = 1 \) into the equation gives \( y = -5 \), so the vertex is at \((1, -5)\).

(Note: Since this is a drawing task, the actual graph would be plotted on the provided grid with the points above connected by a smooth parabolic curve.)