How do you graph #2x-1=y# using the intercepts?

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Answer 1 · 2017-11-02T21:45:46

See a solution process below:

First, solve for the #x# and #y#-intercepts and plot these points:

x-intercept - set y = 0 and solve for x:

#2x - 1 = 0#

#2x - 1 + color(red)(1) = 0 + color(red)(1)#

#2x - 0 = 1#

#2x = 1#

#(2x)/color(red)(2) = 1/color(red)(2)#

#x = 1/2# or #(1/2, 0)#

y-intercept - set x = 0 and solve for y:

#(2 xx 0) - 1 = y#

#0 - 1 = y#

#-1 = y#

#y = -1# or #(0, -1)#

We can next plot the two points on the coordinate plane:

graph{(x^2+(y+1)^2-0.025)((x-(1/2))^2+y^2-0.025)=0 [-10, 10, -5, 5]}

Now, we can draw a straight line through the two points to graph the line:

graph{(2x-1-y)(x^2+(y+1)^2-0.025)((x-(1/2))^2+y^2-0.025)=0 [-10, 10, -5, 5]}