Question

graph the line.
$y = x + 4$
grid with x-axis from 0 to 12 and y-axis from 0 to 12

Explanation:

Step1: Identify the y - intercept

The equation of the line is in the slope - intercept form \(y = mx + b\), where \(m\) is the slope and \(b\) is the y - intercept. For the equation \(y=x + 4\), the y - intercept \(b = 4\). This means the line crosses the y - axis at the point \((0,4)\). We can plot this point on the graph.

Step2: Determine the slope

The slope \(m\) of the line \(y=x + 4\) is \(1\) (since the coefficient of \(x\) is \(1\)). The slope \(m=\frac{\text{rise}}{\text{run}}=\frac{1}{1}\). From the y - intercept \((0,4)\), we can find another point by moving up \(1\) unit (rise) and to the right \(1\) unit (run). So, from \((0,4)\), moving up \(1\) and right \(1\) gives the point \((1,5)\). We can also move down \(1\) and left \(1\) from the y - intercept to get \((- 1,3)\) (but since our graph has non - negative \(x\) values shown, \((1,5)\) is more convenient for graphing here).

Step3: Draw the line

After plotting the points \((0,4)\) and \((1,5)\) (and we can plot more points using the slope if needed, like \((2,6)\), \((3,7)\) etc.), we draw a straight line passing through these points.

(Note: Since the question is about graphing, the final answer is the graph of the line \(y = x+4\) with a y - intercept at \((0,4)\) and a slope of \(1\), passing through points like \((0,4)\), \((1,5)\), \((2,6)\) etc.)

Answer:

The line \(y = x + 4\) is graphed by plotting the y - intercept \((0,4)\) and using the slope \(1\) (rise \(1\), run \(1\)) to plot additional points (e.g., \((1,5)\), \((2,6)\)) and drawing a straight line through them.