How do you graph the equation #y=-3x+2#?

Question

Kennedy Adams · Accepted Answer

See below:

This equation is in slope-intercept form, which is my favourite for graphing lines. The slope-intercept form is in the general form of: #y=mx+b# where #m# is the slope and #b# is the #y#-intercept. Let's graph the #y#-intercept first. The #y#-intercept is 2. This means that that point is #(0,2)#. So let's graph that: graph{(x-0)^2+(y-2)^2-.3^2=0} So that's one point. Now let's plot another point (and then we can use a straightedge to join them). The slope is #-3#. Slope is calculated by #"rise"/"run"# - in other words the change in #y# divided by the change in #x#. In this case, we will drop 3 spots for every 1 spot we move right. So that point is #(0+1, 2-3)=(1,-1)#. Let's graph that: graph{((x-0)^2+(y-2)^2-.3^2)((x-1)^2+(y+1)^2-.3^2)=0} And now connect them up! graph{((x-0)^2+(y-2)^2-.3^2)((x-1)^2+(y+1)^2-.3^2)(y+3x-2)=0}

Isaiah Anthony · Answer

undefined To graph the equation $ y = -3x + 2 $:

1. Plot the y-intercept, which is $ (0, 2) $.
2. Determine another point by using the slope. Since the slope is -3, from the y-intercept, move down 3 units and to the right 1 unit to get another point.
3. Draw a straight line passing through both points to represent the graph of $ y = -3x + 2 $.