Answer :
To determine which statement best describes the function [tex]\( f(x) = k^{-\sqrt{x+2}} \)[/tex] where [tex]\( k \)[/tex] is a constant such that [tex]\( k > 1 \)[/tex] and [tex]\( x \geq -2 \)[/tex], let's analyze the behavior of the function step-by-step.
1. Understanding the Function:
[tex]\[ f(x) = k^{-\sqrt{x+2}} \][/tex]
Here, [tex]\( k \)[/tex] is a positive constant greater than 1, and we are considering [tex]\( x \geq -2 \)[/tex].
2. Analyzing the Exponent:
For [tex]\( x \geq -2 \)[/tex], [tex]\( \sqrt{x+2} \)[/tex] will always yield non-negative values (since the square root of a non-negative number is also non-negative).
3. Behavior of [tex]\( \sqrt{x+2} \)[/tex]:
As [tex]\( x \)[/tex] increases from [tex]\(-2\)[/tex], the value of [tex]\( \sqrt{x+2} \)[/tex] also increases. For example:
- When [tex]\( x = -2 \)[/tex], [tex]\( \sqrt{x+2} = 0 \)[/tex].
- When [tex]\( x = 0 \)[/tex], [tex]\( \sqrt{x+2} = \sqrt{2} \)[/tex].
- When [tex]\( x \)[/tex] becomes very large, [tex]\( \sqrt{x+2} \)[/tex] also becomes very large.
4. Impact on [tex]\( k^{-\sqrt{x+2}} \)[/tex]:
Given that [tex]\( k > 1 \)[/tex], any positive exponent will result in a number greater than 1. When that exponent is negative, we have a fraction between 0 and 1. Specifically:
- As [tex]\( x \)[/tex] increases, [tex]\( \sqrt{x+2} \)[/tex] becomes larger.
- The exponent [tex]\( -\sqrt{x+2} \)[/tex] becomes more negative, making [tex]\( k^{-\sqrt{x+2}} \)[/tex] a smaller positive number.
5. Exponential Decay:
The function [tex]\( f(x) = k^{-\sqrt{x+2}} \)[/tex] will exhibit exponential decay because as [tex]\( x \)[/tex] increases, [tex]\( -\sqrt{x+2} \)[/tex] decreases (becomes more negative), and [tex]\( k \)[/tex] to the power of this more negative exponent yields smaller and smaller positive values.
In conclusion, the function [tex]\( f(x) = k^{-\sqrt{x+2}} \)[/tex] with [tex]\( k > 1 \)[/tex] and [tex]\( x \geq -2 \)[/tex] is best described by statement A:
A. It is an exponential decay function.
1. Understanding the Function:
[tex]\[ f(x) = k^{-\sqrt{x+2}} \][/tex]
Here, [tex]\( k \)[/tex] is a positive constant greater than 1, and we are considering [tex]\( x \geq -2 \)[/tex].
2. Analyzing the Exponent:
For [tex]\( x \geq -2 \)[/tex], [tex]\( \sqrt{x+2} \)[/tex] will always yield non-negative values (since the square root of a non-negative number is also non-negative).
3. Behavior of [tex]\( \sqrt{x+2} \)[/tex]:
As [tex]\( x \)[/tex] increases from [tex]\(-2\)[/tex], the value of [tex]\( \sqrt{x+2} \)[/tex] also increases. For example:
- When [tex]\( x = -2 \)[/tex], [tex]\( \sqrt{x+2} = 0 \)[/tex].
- When [tex]\( x = 0 \)[/tex], [tex]\( \sqrt{x+2} = \sqrt{2} \)[/tex].
- When [tex]\( x \)[/tex] becomes very large, [tex]\( \sqrt{x+2} \)[/tex] also becomes very large.
4. Impact on [tex]\( k^{-\sqrt{x+2}} \)[/tex]:
Given that [tex]\( k > 1 \)[/tex], any positive exponent will result in a number greater than 1. When that exponent is negative, we have a fraction between 0 and 1. Specifically:
- As [tex]\( x \)[/tex] increases, [tex]\( \sqrt{x+2} \)[/tex] becomes larger.
- The exponent [tex]\( -\sqrt{x+2} \)[/tex] becomes more negative, making [tex]\( k^{-\sqrt{x+2}} \)[/tex] a smaller positive number.
5. Exponential Decay:
The function [tex]\( f(x) = k^{-\sqrt{x+2}} \)[/tex] will exhibit exponential decay because as [tex]\( x \)[/tex] increases, [tex]\( -\sqrt{x+2} \)[/tex] decreases (becomes more negative), and [tex]\( k \)[/tex] to the power of this more negative exponent yields smaller and smaller positive values.
In conclusion, the function [tex]\( f(x) = k^{-\sqrt{x+2}} \)[/tex] with [tex]\( k > 1 \)[/tex] and [tex]\( x \geq -2 \)[/tex] is best described by statement A:
A. It is an exponential decay function.
