How do you graph the line #f(x) = 3-2x #?

1 Answer

Explanation below.


Start off by rearranging the equation to make it into y=mx+b form (m = slope, b = y-intercept).

So, #y=-2x+3#

To find the starting point to this graph, we can use the y-intercept. In this case, the y-intercept is 3 (the line crosses the y-axis at 3), so the starting point would be at #(0,3)#

We can now use the slope to find the rest of the points to graph this line.

The slope here will be #-2/1#

As we know, the slope is "rise over run"; "rise" meaning we would go up/down a certain number of units and "run" meaning going horizontally to the left/right.

In this case, we would go 2 units down because it is a negative slope, and 1 unit to the right. Keep doing this to find the rest of the points, plot them, and draw a straight line. Extending the line in both directions.

graph{-2x+3 [-8.89, 8.89, -4.444, 4.445]}

The graph shows the starting point; #(0,3)# as well as other points from the slope such as; #(1,1), (2, -1), (3, -3)#, etc

Hope this helped!