# Where is the vertex for y= |x/3-4 |?

Oct 26, 2014

By pulling $\frac{1}{3}$ out of the absolute value sign, we have

$y = \frac{1}{3} | x - 12 |$.

Remember that $y = | x |$ has a vertex at $\left(0 , 0\right)$.

Since vertical shrinkage does not affect the vertex, $y = \frac{1}{3} | x |$ still has a vertex at $\left(0 , 0\right)$.

By shifting right by 12 units, $y = \frac{1}{3} | x - 12 |$ has a vertex at $\left(12 , 0\right)$.

Hence, the vertex is at $\left(12 , 0\right)$.

I hope that this was helpful.