Question

graph this line using the slope and y-intercept:
y = 3x + 3
click to select points on the graph.

Explanation:

Step1: Identify y-intercept

The equation is in slope - intercept form \(y = mx + b\), where \(b\) is the y - intercept. For \(y=3x + 3\), \(b = 3\). So the y - intercept is at the point \((0,3)\).

Step2: Use slope to find another point

The slope \(m = 3=\frac{3}{1}\), which means for a run of \(1\) (increase in \(x\) by \(1\)), the rise is \(3\) (increase in \(y\) by \(3\)). Starting from \((0,3)\), if we move \(x = 0+1 = 1\), then \(y=3 + 3=6\). So another point on the line is \((1,6)\). We can also use a run of \(- 1\) (decrease in \(x\) by \(1\)) and a rise of \(-3\) (decrease in \(y\) by \(3\)). Starting from \((0,3)\), \(x = 0-1=-1\), \(y = 3-3 = 0\), so the point \((-1,0)\) is also on the line.

Step3: Plot the points and draw the line

Plot the points \((0,3)\), \((1,6)\) (or \((-1,0)\)) on the graph and then draw a straight line passing through these points.

To graph the line \(y = 3x+3\):

1. Plot the y - intercept: Mark the point \((0,3)\) on the coordinate plane (where the \(x\) - coordinate is \(0\) and the \(y\) - coordinate is \(3\)).
2. Use the slope: From \((0,3)\), move \(1\) unit to the right (increase \(x\) by \(1\)) and \(3\) units up (increase \(y\) by \(3\)) to get the point \((1,6)\). Or move \(1\) unit to the left (decrease \(x\) by \(1\)) and \(3\) units down (decrease \(y\) by \(3\)) to get the point \((-1,0)\).
3. Draw a straight line through the plotted points.

(Note: Since the question is about graphing, the key points to plot are \((0,3)\) and another point like \((1,6)\) or \((-1,0)\) and then drawing the line through them.)

Answer:

To graph \(y = 3x + 3\), plot the y - intercept \((0,3)\) and use the slope \(m = 3\) to find another point (e.g., \((1,6)\) or \((-1,0)\)) and draw a straight line through these points. The line passes through points like \((0,3)\), \((1,6)\), \((-1,0)\) etc.