Question

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

Answer 1

Answer:

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

Step-by-step explanation:

The given function is

$f(x)+2=\frac{1}{6}|x-3|$

It can be written as

$f(x)=\frac{1}{6}|x-3|-2$          .... (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

$a=\frac{1}{6}$

$h=3$

$k=-2$

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

The value of a is  $\frac{1}{6}$. 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.

$|x-3|\geq 0$

$\frac{1}{6}|x-3|\geq \frac{1}{6}(0)$

$\frac{1}{6}|x-3|\geq 0$

$\frac{1}{6}|x-3|-2\geq 0-2$

$f(x)\geq -2$

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

Answer 2

It’s D i just took the test