Question

determine the range of ( f(x) = sqrt{x + 5} - 12 ).
options: ( y geq -12 ), ( y geq 12 ), ( y geq -5 ), ( y geq 5 )

Explanation:

Step1: Analyze the square root function

The square root function \( \sqrt{x + 5} \) has a range of \( y\geq0 \) because the square root of a number is always non - negative (for real numbers, the square root of a non - negative number \( a\) is a non - negative number \( b\) such that \( b^2=a\)). So, \( \sqrt{x + 5}\geq0 \) for all \( x\) in the domain of the function (where \( x+5\geq0\), i.e., \( x\geq - 5\)).

Step2: Analyze the transformed function

We have the function \( f(x)=\sqrt{x + 5}-12 \). If we start with \( \sqrt{x + 5}\geq0 \), and we subtract 12 from both sides of the inequality, we get \( \sqrt{x + 5}-12\geq0 - 12\).
Simplifying the right - hand side of the inequality, we have \( \sqrt{x + 5}-12\geq - 12\). Since \( f(x)=\sqrt{x + 5}-12\), this means that \( f(x)\geq - 12\), or in terms of the range of the function, \( y\geq - 12\).

Answer:

\( y\geq - 12 \) (corresponding to the option: \( y\geq - 12 \))