Question

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

Answer 1

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]}