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

Mar 16, 2017

Use the definition of the absolute value function:
$| a | = \left\{\begin{matrix}a \text{ for " a >=0 \\ -a" for } a < 0\end{matrix}\right.$

Explanation:

Applying the definition to the given equation:

$y = \left\{\begin{matrix}\frac{1}{3} \left(x - 3\right) + 4 \text{ for "x -3 >=0 \\ 1/3(3-x)+4" for } x - 3 < 0\end{matrix}\right.$

Simplify the inequalities:

$y = \left\{\begin{matrix}\frac{1}{3} \left(x - 3\right) + 4 \text{ for "x >=3 \\ 1/3(3-x)+4" for } x < 3\end{matrix}\right.$

Distribute the $\frac{1}{3}$

$y = \left\{\begin{matrix}\frac{x}{3} - 1 + 4 \text{ for "x >=3 \\ 1-x/3+4" for } x < 3\end{matrix}\right.$

Combine the constants:

$y = \left\{\begin{matrix}\frac{x}{3} + 3 \text{ for "x >=3 \\ -x/3+5" for } x < 3\end{matrix}\right.$

Check for continuity by evaluating both at x = 3:

$y = \left\{\begin{matrix}\frac{3}{3} + 3 \text{ for "x >=3 \\ -3/3+5" for } x < 3\end{matrix}\right.$

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

You graph

$y = \left\{\begin{matrix}\frac{x}{3} + 3 \text{ for "x >=3 \\ -x/3+5" for } x < 3\end{matrix}\right.$

using two rays, originating a the point $\left(3 , 4\right)$

A point for the ray on the right side is $\left(9 , 6\right)$

A point for the ray on the left side is $\left(- 3 , 6\right)$

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