Question

1. (a) complete the table of values for $y = x^2 + x$.\

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$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	\
$y$	$6$	$2$		$0$		$6$		\

(2)\
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(b) on the grid, draw the graph of $y = x^2 + x$.\
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grid with x-axis from -3 to 3 and y-axis from -12 to 12, with grid lines.\
(total 4 marks)

Explanation:

Part (a)

Step1: For \( x = -1 \)

Substitute \( x = -1 \) into \( y = x^2 + x \).
\( y = (-1)^2 + (-1) = 1 - 1 = 0 \)

Step2: For \( x = 1 \)

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

Step3: For \( x = 3 \)

Substitute \( x = 3 \) into \( y = x^2 + x \).
\( y = (3)^2 + 3 = 9 + 3 = 12 \)

Answer:

The completed table is:

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

Part (b)

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

1. Plot the points from the table: \((-3, 6)\), \((-2, 2)\), \((-1, 0)\), \((0, 0)\), \((1, 2)\), \((2, 6)\), \((3, 12)\).
2. Connect the points with a smooth curve (since it's a quadratic function, the graph will be a parabola). The vertex of the parabola \( y = ax^2 + bx + c \) is at \( x = -\frac{b}{2a} \). For \( y = x^2 + x \), \( a = 1 \), \( b = 1 \), so \( x = -\frac{1}{2(1)} = -\frac{1}{2} \). The graph opens upwards (because \( a > 0 \)).

(Note: Since this is a drawing task, the key is to plot the points accurately and draw a smooth curve through them. The grid provided has appropriate axes, so mark the points and draw the curve.)