Question

ted to reproduce for classroom use.
complete the function table. then graph the function.

1. $y = x + 3$

$x$	$x + 3$	$y$	$(x, y)$
$0$	$0 + 3$	$-3$	$(0, -3)$
$1$	$1 + 3$	$4$	$(1, 4)$
$2$	$2 + 3$	$5$	$(2, 5)$

(there is also a coordinate grid image at the bottom.)

Explanation:

Step1: For \( x = -2 \)

Substitute \( x = -2 \) into \( y = x + 3 \), we get \( y=-2 + 3=1 \), so the point is \( (-2,1) \).

Step2: For \( x = 0 \)

Substitute \( x = 0 \) into \( y = x + 3 \), we get \( y = 0+3 = 3 \), so the point is \( (0,3) \).

Step3: For \( x = 1 \)

Substitute \( x = 1 \) into \( y = x + 3 \), we get \( y=1 + 3 = 4 \), so the point is \( (1,4) \).

Step4: For \( x = 2 \)

Substitute \( x = 2 \) into \( y = x + 3 \), we get \( y=2 + 3 = 5 \), so the point is \( (2,5) \).

Answer:

The completed function table (correcting the errors in the original table) is:

\( x \)	\( -2 \)	\( 0 \)	\( 1 \)	\( 2 \)
\( y \)	\( 1 \)	\( 3 \)	\( 4 \)	\( 5 \)
\( (x,y) \)	\( (-2,1) \)	\( (0,3) \)	\( (1,4) \)	\( (2,5) \)

To graph the function, plot the points \( (-2,1) \), \( (0,3) \), \( (1,4) \), \( (2,5) \) on the coordinate plane and draw a straight line through them (since \( y=x + 3 \) is a linear function with a slope of 1 and y - intercept of 3).