Question

36. A function g is described below

Answer 1

Answer: F

Step-by-step explanation:

A square is always positive:

$(x-3)^2 \geq 0\\\\g(x)=(x-3)^2+1 \geq 1\\\\answer\ F$

Answer 2

Answer:

F.

Step-by-step explanation:

The expression (x - 3)^2 is always >= 0.

The minimum value of g occurs when x = 3 :- that is (3-3)^2 + 1 = 1.

The range of g(x) is all real numbers >=  1.