Question

graph the line $y = 3x$.

Explanation:

Step1: Identify the slope and y-intercept

The equation \( y = 3x \) is in the slope - intercept form \( y=mx + b \), where \( m \) is the slope and \( b \) is the y - intercept. Here, \( m = 3\) (which means for every 1 unit we move to the right along the x - axis, we move up 3 units along the y - axis) and \( b = 0\), so the line passes through the origin \((0,0)\).

Step2: Find another point

Let's choose \( x = 1\). Substitute \( x = 1\) into the equation \( y=3x\). Then \( y=3\times1 = 3\). So we have another point \((1,3)\) on the line.

Step3: Plot the points and draw the line

We already know the line passes through \((0,0)\) and \((1,3)\). Plot these two points on the coordinate plane. Then, using a straight - edge, draw a line passing through these two points. The line should have a steep positive slope (since the slope \( m = 3>0\)) and pass through the origin.

(Note: If we want to find more points, we can choose other values of \( x\). For example, if \( x=- 1\), then \( y = 3\times(-1)=-3\), so the point \((-1,-3)\) is also on the line.)

Answer:

To graph \( y = 3x\):

1. Recognize it has a slope of \( 3\) and y - intercept \( 0\) (passes through \((0,0)\)).
2. Find another point: when \( x = 1\), \( y=3\) (point \((1,3)\)).
3. Plot \((0,0)\) and \((1,3)\) (and/or other points like \((-1,-3)\)) and draw a straight line through them. The line will have a steep positive slope and pass through the origin.