How do you graph, find the zeros, intercepts, domain and range of f(x)=1/3abs(2x-1)f(x)=13|2x1|?

1 Answer
Feb 20, 2018

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

See explanation.

Explanation:

This is of graph type vv.

The part |2x-1||2x1| is always positive so 1/3|2x-1|13|2x1| is also always
positive. Thus y>=0y0

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

Set y=0=1/3|2x-1|y=0=13|2x1|

Multiply both sides by 3/131

y=0=|2x-1|y=0=|2x1|

y=0=|0|y=0=|0|

So 2x-1=0=>x=1/22x1=0x=12

So the the vertex of vv is at (x,y)->(1/2,0)(x,y)(12,0)

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

Set x=0x=0 giving:

y=1/3|0-1|y=13|01|

y=1/3xx(+1) = 1/3y=13×(+1)=13

y_("intercept")->(x,y)=(0,1/3)yintercept(x,y)=(0,13)
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Plot two graphs but cit them off so that there is no line below the x-axis

Line 1 y=1/3(2x-1)y=13(2x1) where x<=0x0
Line 2 y=1/3(2x-1)y=13(2x1) where x>=0x0

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