How do you graph the linear function #f(x)=2/3x+1#?

1 Answer
May 9, 2017

#y=2/3x+1#

We need to know three things to graph a linear equation: #x#-intercept, #y#-intercept, and slope.

We can tell, thanks to #y=color(red)(m)x+color(orange)(b)#, that the #color(red)(slope)# is #color(red)(2/3)# and the #color(orange)(y-i n t ercept)# is #color(orange)(1)#. We have two out of the three required pieces of information, but we still need the #x-#intercept.

To find the #x#-intercept, we need to set #y# equal to zero and solve for #x#:

#0=2/3x+1#

subtract #1# on both sides

#-1=2/3x#

multiply by #3/2# on both sides

#3/2*-1/1=cancel(3)/cancel(2)*cancel(2)/cancel(3)x#

#-3/2=x#

So, the #x#-intercept is #(-3/2,0)#, the #y#-intercept is #(0,1)#, and the slope is #2/3#. That means we begin at #(0,1)#, and go up #2#, over #3#, and also down #2#, over #3#. We keep doing that forever (that's the nature of a line), and if everything is correct, the line should pass through the point #(-3/2,0)#.

Let's check our work
graph{y=2/3x+1}