How do you graph #x+y=5# using intercepts?

1 Answer
Jan 3, 2017

Answer:

  1. Get the #x#-intercept by setting #y=0# and solving for #x#.
  2. Get the #y#-intercept by setting #x=0# and solving for #y#.

Graph these two points, and connect them.

Explanation:

An intercept is simply a point where a line (or any function) crosses an axis. At such a point, the non-axis coordinate is 0. (e.g. all points on the #x#-axis have no displacement in the #y# direction, so the #y#-coordinate is #0#.)

We can use this to help us get two points that are on the line, graph those two points, and finally connect them by drawing a line through them.

The #x#-intercept occurs when #y=0#. To find this intercept, we just need to plug in 0 for #y# in our line equation:

#color(white)(=>)x+y=5 " @ "y=0#
#=>x+0=5#
#=>xcolor(white)+color(white)0=5#

So, when #y=0#, we have #x=5#, meaning #(5,0)# is a point on our line, and it is the #x#-intercept.

Similarly, to find the #y#-intercept, we let #x=0# and solve for #y#:

#color(white)(=>)x+y=5 " @ "x=0#
#=>0+y=5#
#=>color(white)x color(white)+ y=5#

So when #x=0#, #y=5#, and thus #(0,5)# is our #y#-intercept.

Finally, we graph these two points, and connect them with a line:

graph{(x+y-5)((x-5)^2+y^2-0.06)(x^2+(y-5)^2-0.06)=0 [-8.325, 11.68, -2.3, 7.7]}

Bonus:

For a linear equation of the form #x+y=k# for some number #k#, the equation is saying "two numbers sum to #k#". From this, it is easy to see that when either number is 0, the other one must be #k#, so that the sum will be #k#. This means the intercepts will always be #(0,k)# and #(k,0)#.