How do you graph #f(x)=2abs(x-1)-3#?

1 Answer
Jul 23, 2015

Answer:

Plot the point where #(x-1) = 0# and one point for each of #(x-1) > 0# and #(x-1) < 0#; connect the initial point through each of the other 2 points as 2 line segments.

Explanation:

Given #f(x) = 2abs(x-1) -3#

When #x= 1#
#color(white)("XXXX")##f(1) = -3#
#(1,-3)# is a point common to both line segments (#x<=1# and #x>=1#)

Picking #x=3# as an arbitrary point with #x > 1#
#color(white)("XXXX")##f(3) = 2abs(3-1) -3 = 1#
#(3,1)# is a point on the line segment for #x >1#

Picking #x=0# as an arbitrary point with #x < 1#
#color(white)("XXXX")##f(0) = 2abs(0-1) -3 =-1#
#(0,-1)# is a point on the line segment for #x < 1#
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