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.

Related questions
See all questions in Graphs of Absolute Value Equations
Impact of this question
1577 views around the world
You can reuse this answer
Creative Commons License