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

1 Answer
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.