which functions have a vertex with a x - value of 0? select three options.\n□f(x)=|x|\n□f(x)=|x| +…

Explanation:

Step1: Recall vertex - form of absolute - value function

The general form of an absolute - value function is $y=a|x - h|+k$, and its vertex is at the point $(h,k)$.

Step2: Analyze $f(x)=|x|$

For $f(x)=|x|$, which can be written as $f(x)=1|x - 0|+0$. Here, $h = 0$ and $k = 0$, so the vertex is $(0,0)$.

Step3: Analyze $f(x)=|x|+3$

For $f(x)=|x|+3$, which is $f(x)=1|x - 0|+3$. Here, $h = 0$ and $k = 3$, so the vertex is $(0,3)$.

Step4: Analyze $f(x)=|x + 3|$

For $f(x)=|x + 3|$, which can be written as $f(x)=1|x-(-3)|+0$. Here, $h=-3$ and $k = 0$, so the vertex is $(-3,0)$.

Step5: Analyze $f(x)=|x|-6$

For $f(x)=|x|-6$, which is $f(x)=1|x - 0|-6$. Here, $h = 0$ and $k=-6$, so the vertex is $(0,-6)$.

Step6: Analyze $f(x)=|x + 3|-6$

For $f(x)=|x + 3|-6$, which is $f(x)=1|x-(-3)|-6$. Here, $h=-3$ and $k=-6$, so the vertex is $(-3,-6)$.

Answer:

$f(x)=|x|$, $f(x)=|x|+3$, $f(x)=|x|-6$