How do you graph #y=1/3abs(x-3)+4#?

1 Answer

Answer:

Use the definition of the absolute value function:
#|a| ={(a" for " a >=0),(-a" for "a<0):}#

Explanation:

Applying the definition to the given equation:

#y = {(1/3(x - 3) + 4" for "x -3 >=0),(1/3(3-x)+4" for "x - 3 < 0):}#

Simplify the inequalities:

#y = {(1/3(x - 3) + 4" for "x >=3),(1/3(3-x)+4" for "x < 3):}#

Distribute the #1/3#

#y = {(x/3 - 1 + 4" for "x >=3),(1-x/3+4" for "x < 3):}#

Combine the constants:

#y = {(x/3 + 3" for "x >=3),(-x/3+5" for "x < 3):}#

Check for continuity by evaluating both at x = 3:

#y = {(3/3 + 3" for "x >=3),(-3/3+5" for "x < 3):}#

In both cases, #y = 4#; this checks.

You graph

#y = {(x/3 + 3" for "x >=3),(-x/3+5" for "x < 3):}#

using two rays, originating a the point #(3,4)#

A point for the ray on the right side is #(9,6)#

A point for the ray on the left side is #(-3,6)#

graph{1/3abs(x-3)+4 [-23.65, 27.66, -2.88, 22.78]}