Question

Which statement is true about f(x)+2=1/6|x-3|?

A. The graph of f(x) has a vertex of (-3,2).
B. The graph of f(x) is horizontally compressed.
C. The graph of f(x) opens downward.
D. The graph of f(x) has range of f(x)_>-2.

Answer 1

It’s D i just took the test

Answer 2

The range of f(x) is
`f(x)\geq -2` and option D is correct.

Explanation:

The given function is

`f(x)+2=(1)/(6)|x-3|`

It can be written as

`f(x)=(1)/(6)|x-3|-2` .... (1)

The function is in the form of

`g(x)=a|x-h|+k` ....(2)

Where, a is scale factor and (h,k) is vertex of the graph.

On comparing (1) and (2), we get

`a=(1)/(6)`

`h=3`

`k=-2`

Therefore the vertex of f(x) is (3,-2). Option A is incorrect.

The value of a is
`(1)/(6)`. 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.

`|x-3|\geq 0`

`(1)/(6)|x-3|\geq (1)/(6)(0)`

`(1)/(6)|x-3|\geq 0`

`(1)/(6)|x-3|-2\geq 0-2`

`f(x)\geq -2`

Therefore the range of f(x) is
`f(x)\geq -2` and option D is correct.