How do you graph, find the zeros, intercepts, domain and range of #f(x)=1/3abs(2x-1)#?

1 Answer
Feb 20, 2018

Answer:

domain # {x, x in(-oo,+oo)}#
range #color(white)("d") {y, y in[0,+oo)}#

See explanation.

Explanation:

This is of graph type #vv#.

The part #|2x-1|# is always positive so #1/3|2x-1|# is also always
positive. Thus #y>=0#

#color(blue)("Determin the vertex")#

Set #y=0=1/3|2x-1|#

Multiply both sides by #3/1#

#y=0=|2x-1|#

#y=0=|0|#

So #2x-1=0=>x=1/2#

So the the vertex of #vv# is at #(x,y)->(1/2,0)#

Thus the vertex is on the x-axis. There is no plot below the x-axis.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the y-intercept")#

Set #x=0# giving:

#y=1/3|0-1|#

#y=1/3xx(+1) = 1/3#

#y_("intercept")->(x,y)=(0,1/3)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Plot two graphs but cit them off so that there is no line below the x-axis

Line 1 #y=1/3(2x-1)# where #x<=0#
Line 2 #y=1/3(2x-1)# where #x>=0#

Tony B