How do you write # y= |x-5| -4# as a piecewise function?

2 Answers
Oct 20, 2017

Start with the piecewise definition of the absolute value function:

#|f(x)| = {(f(x); f(x) >= 0),(-f(x); f(x) < 0):}#

Substitute #x - 5# for #f(x)#

#|x - 5| = {(x - 5; x - 5 >= 0),(-(x - 5); x - 5 < 0):}#

Simplify the inequalities:

#|x - 5| = {(x - 5; x >= 5),(-(x - 5); x < 5):}#

Distribute the -1:

#|x - 5| = {(x - 5; x >= 5),(5-x; x < 5):}#

Make two separate equation of the original:

#y = {((x-5)-4;x>=5),((5-x)-4; x < 5):}#

Simplify:

#y = {(x-9;x>=5),(1-x; x < 5):}#

Oct 20, 2017

Answer:

#y=-x+1# for #x<=5#
#y=x-9# for #x>5#

Explanation:

First let's look at the graph or the original function #y=abs(x-5)#

desmos.com

The left half of the graph where #x<=5# has a slope of #m=-1# and a y intercept of #(0,1)#.

Using the slope intercept form of a line #y=mx+b# where #m=-1# and #b=1# gives the equation is #y=-1x+1# or #y=-x+1#

The right half of the graph where #x>=5# has a slope of #m=1# and passes through the point #(5,-4)#, which is the vertex of the graph as shown in the original equation.

Using the slope intercept form of the line #y-y_1=m(x-x_1)# where #m=1# and #(x_1,y_1) = (5,-4)# gives the equation

#y- -4 = 1 (x-5)#
#y+4= x-5#
#y=x-9# for #x<=5#

Written in piecewise form

#y=-x+1# for #x<=5#
#y=x-9# for #x>5#

Note that I arbitrarily assigned #x=5# to the first equation.