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

1 Answer

Answer:

See below:

Explanation:

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}