How do you graph #y = - abs(x-3) + 5#?

1 Answer
Grégory V.
Jul 6, 2015

graph{-abs(x-3)+5 [-7.27, 10.93, -2.696, 6.404]}

#y = x+2 , if x<=3#
#y = -x+8, if x>3#

Explanation:

Recall : The "absolute value" function is defined as :

#abs(x)= x, if x>=0#
#abs(x) =-x, if x<0#
# #
Graph of #abs(x)# :
graph{abs(x) [-, 10, -5, 5]}

Here, we have #abs(x-3)#. Using the definition of #abs(x)# we can write :
# #
#abs(x-3) = x-3, if x>=3#
#abs(x-3) = -(x-3) = -x+3, if x<3#
# #
Your function doesn't contain #abs(x-3)# but #-abs(x-3)#, we have to adapt what I wrote in the last paragraph.

#color(red)-abs(x-3) = -x+3, if x>=3#
#color(red)-abs(x-3) = x-3, if x<3#
# #
Graph of #-abs(x-3)# :
graph{-abs(x-3) [-10, 10, -5, 5]}
# #
Now we have to add 5 to #-abs(x-3)# without regard of the value of #x#
# #
Then :

#-abs(x-3)+5 = -x+3+5 = -x+8, if x>=3#
#-abs(x-3) = x-3+5 = x+2, if x<3#
# #
# #
Note : I don't know how to use multi-line equations with socratic, then sorry for heavy notations.

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