Answer:
The range of f(x) is [tex]f(x)\geq -2[/tex] and option D is correct.
Step-by-step explanation:
The given function is
[tex]f(x)+2=\frac{1}{6}|x-3|[/tex]
It can be written as
[tex]f(x)=\frac{1}{6}|x-3|-2[/tex] .... (1)
The function is in the form of
[tex]g(x)=a|x-h|+k[/tex] ....(2)
Where, a is scale factor and (h,k) is vertex of the graph.
On comparing (1) and (2), we get
[tex]a=\frac{1}{6}[/tex]
[tex]h=3[/tex]
[tex]k=-2[/tex]
Therefore the vertex of f(x) is (3,-2). Option A is incorrect.
The value of a is [tex]\frac{1}{6}[/tex]. So, the graph compressed vertically. The value of a is positive, therefore the graph of f(x) opens upward.
We know the absolute value is always greater than or equal to 0.
[tex]|x-3|\geq 0[/tex]
[tex]\frac{1}{6}|x-3|\geq \frac{1}{6}(0)[/tex]
[tex]\frac{1}{6}|x-3|\geq 0[/tex]
[tex]\frac{1}{6}|x-3|-2\geq 0-2[/tex]
[tex]f(x)\geq -2[/tex]
Therefore the range of f(x) is [tex]f(x)\geq -2[/tex] and option D is correct.
