the statements below describe the exponential function $f(x)$.- the graph of $f(x)$ has been shifted 2 units…

the statements below describe the exponential function $f(x)$.- the graph of $f(x)$ has been shifted 2 units to the left of its parent function- the range of $f(x)$ is $y < - 5$- the y-intercept of the graph of $f(x)$ is $(0, - 9)$which of the following equations could represent $f(x)$?$square$ $f(x) = - 2^{x+2} - 5$$square$ $f(x) = - 9 cdot 2^{x-2} - 5$$square$ $f(x) = 2^{x+2} - 5$$square$ $f(x) = 2^{x-2} - 5$

the statements below describe the exponential function $f(x)$.- the graph of $f(x)$ has been shifted 2 units to the left of its parent function- the range of $f(x)$ is $y < - 5$- the y-intercept of the graph of $f(x)$ is $(0, - 9)$which of the following equations could represent $f(x)$?$square$ $f(x) = - 2^{x+2} - 5$$square$ $f(x) = - 9 cdot 2^{x-2} - 5$$square$ $f(x) = 2^{x+2} - 5$$square$ $f(x) = 2^{x-2} - 5$

Answer

Explanation:

Step1: Analyze range condition

The range (y < -5) means the function has a vertical shift down by 5, and the exponential term is negative (so it approaches 0 from below, making (f(x)) approach -5 from below). So the general form is (f(x) = -a \cdot 2^{x-h} - 5) where (a>0).

Step2: Analyze left shift condition

A left shift of 2 units means (x) is replaced with (x+2) in the parent function (2^x), so the exponent becomes (x+2) (for left shift: (f(x) \to f(x+h)) for (h>0) left shift).

Step3: Verify y-intercept condition

The y-intercept is at (x=0), (f(0)=-9). Test the first option: Substitute (x=0) into (f(x) = -2^{x+2} - 5): (f(0) = -2^{0+2} - 5 = -4 - 5 = -9), which matches the given y-intercept. Check the range: Since (2^{x+2} > 0), then (-2^{x+2} < 0), so (f(x) = -2^{x+2} -5 < -5), which matches. The left shift: The parent function is (2^x), shifted left 2 units gives (2^{x+2}), reflected over x-axis gives (-2^{x+2}), shifted down 5 gives (-2^{x+2}-5), which matches the shift condition.

Answer:

A. (f(x) = - 2^{x+2} - 5)