How do you graph #y=-x+3/2# using the slope and intercept?

2 Answers
May 16, 2017

graph{y=-x+3/2 [-10, 10, -5.21, 5.21]}
This is the line that you should get from graphing #y=-x+3/2#

Explanation:

The format of this equation is #y=mx+b#, where #m# is the slope and #b# is the #y#-intercept. Given this, since we know that #b=3/2#, the graph begins at point #(0, 3/2)#, #3/2# above the origin #(0,0)#. From here, since the slope is #-1#, then you can just
begin counting down one, and right one point. From there, you should create a graph similar to the one shown in the answer.

Helpful Sources:

https://weteachscience.org/mentoring/resources/lesson-plans/algebra-1-%E2%80%93-how-to-graph-a-linear-equation-using-slope-and-y

https://www.khanacademy.org/math/algebra/two-var-linear-equations/graphing-slope-intercept-equations/v/graphing-a-line-in-slope-intercept-form

May 16, 2017

See Explanation

Explanation:

The function #y=mx+b# is defined as the equation of a line where #m# is the slope and #b# is the #y#-intercept.

In the function you gave

#y=-x+3/2#

The slope is #-1# and the #y#-intercept is #3/2#

To graph this, you begin at the point #(0,3/2)# because this is your #y#-intercept. From there, your slope which is #"rise"/"run"# which is #-1# means you go down #1# unit and then right #1# unit.

Essentially, the graph looks like this: (You can interact with the graph to get the exact points to plot)

graph{-x+3/2 [-10, 10, -5, 5]}