How do you graph #f(x)= -1/4|x-2|+2#?

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Answer 1 · 2016-12-22T02:13:19

Socratic graph is inserted. See the explanation.

As #8 - 4y = abs(x-2) >= 0, y <=2#.

Let me introduce the inverse operator #abs(...)^(-1)# for abs =

#abs(...)#.

If g(y) = #abs(f(x))#, then

f(x) =

#(abs)^(-1)(g(y)) = g(y)#, for #f(x) >= 0# and

                          = - g(y), for #f(x) <= 0#.

Here, f(x) = x - 2 and g(y) = 8 - 4y. And so,

#x -2 = 8 - 4y# or #x = 10 - 4y#, for #x-2 >= 0#

#x - 2= 4y - 8# or #x = 4y - 6#, for #x - 2 <= 0#.

The given equation is the combined equation, for these separate

piecewise equations

See the graph.

The vertex #(2, 2)# is the zenith of this pair.

graph{y+1/4|x-2|-2=0 [-20, 20, -10, 10]}