Where is the vertex for $y = ∣ ∣ \frac{x}{3} - 4 ∣ ∣$ $y= |x/3-4 |$ ?

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Answer 1 · 2014-10-26T00:25:26

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

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

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

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

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

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

I hope that this was helpful.

Where is the vertex for y=∣∣x3−4∣∣y= |x/3-4 |?