Question

use the table of values to write the exponential function.
$f(x) = \boxed{quad} \boxed{quad}^x$
the table is:

$x$	$f(x)$
-1	2
0	0.5
1	0.125
2	0.03125

the dropdown options for the first box are -2, 0, 0.25, 0.5

Explanation:

Step1: Recall exponential function form

The general form of an exponential function is \( f(x) = a \cdot b^x \), where \( a \) is the initial value (when \( x = 0 \)) and \( b \) is the base.

Step2: Find the initial value \( a \)

When \( x = 0 \), \( f(0)=0.5 \). Substituting into \( f(x)=a\cdot b^x \), we get \( 0.5 = a\cdot b^0 \). Since \( b^0 = 1 \), then \( a = 0.5 \).

Step3: Find the base \( b \)

Use another point, say \( x = 1 \), \( f(1)=0.125 \). We know \( a = 0.5 \), so substitute into \( f(x)=a\cdot b^x \): \( 0.125 = 0.5\cdot b^1 \). Solve for \( b \): \( b=\frac{0.125}{0.5}=0.25 \)? Wait, no, wait. Wait, let's check \( x=-1 \): \( f(-1)=2 \), \( a = 0.5 \), so \( 2 = 0.5\cdot b^{-1} \). Then \( b^{-1}=\frac{2}{0.5}=4 \), so \( b=\frac{1}{4}=0.25 \)? Wait, no, wait when \( x = -1 \), \( f(-1)=2 \), \( a = 0.5 \), so \( 2 = 0.5 \cdot b^{-1} \implies b^{-1}=4 \implies b=\frac{1}{4}=0.25 \)? Wait, but when \( x = 1 \), \( f(1)=0.125 \), \( 0.5 \cdot (0.25)^1 = 0.125 \), which matches. When \( x=-2 \), \( f(-2)=8 \), \( 0.5 \cdot (0.25)^{-2}=0.5 \cdot 16 = 8 \), which matches. When \( x=-1 \), \( 0.5 \cdot (0.25)^{-1}=0.5 \cdot 4 = 2 \), which matches. When \( x=0 \), \( 0.5 \cdot (0.25)^0 = 0.5 \cdot 1 = 0.5 \), which matches. So the function is \( f(x)=0.5 \cdot (0.25)^x \)? Wait, no, wait the first dropdown is for \( a \), the second for \( b \). Wait, the first dropdown options are -2, 0, 0.25, 0.5. We found \( a = 0.5 \), so the first box is 0.5, the second box is 0.25? Wait, no, wait the general form is \( f(x)=a\cdot b^x \). We found \( a = 0.5 \), \( b = 0.25 \)? Wait, but let's check the ratio between consecutive terms. From \( x=-2 \) to \( x=-1 \): \( 8 \) to \( 2 \), ratio is \( 2/8 = 0.25 \). From \( x=-1 \) to \( x=0 \): \( 2 \) to \( 0.5 \), ratio is \( 0.5/2 = 0.25 \). From \( x=0 \) to \( x=1 \): \( 0.5 \) to \( 0.125 \), ratio is \( 0.125/0.5 = 0.25 \). So the common ratio (base) is 0.25, and the initial value (when \( x=0 \)) is 0.5. So the function is \( f(x)=0.5 \cdot (0.25)^x \). So the first dropdown (the coefficient) is 0.5, the second (the base) is 0.25? Wait, but the first dropdown options are -2, 0, 0.25, 0.5. So the first box is 0.5, the second box is 0.25? Wait, no, wait maybe I mixed up. Wait, let's re-express the exponential function. Let's take two points. Let's use \( x=0 \), \( f(0)=0.5 \), so \( a = 0.5 \). Then, when \( x=1 \), \( f(1)=0.125 \), so \( 0.125 = 0.5 \cdot b^1 \implies b = 0.125 / 0.5 = 0.25 \). So yes, \( f(x)=0.5 \cdot (0.25)^x \). So the first dropdown (the coefficient) is 0.5, the second (the base) is 0.25. Wait, but the first dropdown options are -2, 0, 0.25, 0.5. So the first box is 0.5, the second box is 0.25? Wait, no, maybe the first dropdown is the base? Wait, no, the general form is \( f(x)=a \cdot b^x \), so \( a \) is the coefficient, \( b \) is the base. So from the table, when \( x=0 \), \( f(0)=0.5 \), so \( a=0.5 \). Then, the ratio between \( f(x+1) \) and \( f(x) \) is \( 0.25 \), so \( b=0.25 \). So the function is \( f(x)=0.5 \cdot (0.25)^x \). So the first dropdown (the first box) is 0.5, the second box is 0.25. Wait, but the first dropdown options are -2, 0, 0.25, 0.5. So the first box is 0.5, the second box is 0.25? Wait, maybe I made a mistake. Let's check \( x=-1 \): \( f(-1)=2 \). If \( f(x)=0.5 \cdot (0.25)^x \), then \( f(-1)=0.5 \cdot (0.25)^{-1}=0.5 \cdot 4 = 2 \), which is correct. \( x=-2 \): \( 0.5 \cdot (0.25)^{-2}=0.5 \cdot 16 = 8 \), correct. \( x=0 \): 0.5, correct. \( x=1 \): 0.5 * 0.25 = 0.125, correct. \( x=2 \): 0.5 * (0.25)^2 = 0.5 * 0.0625 = 0.03125, correct. So yes, the function is \( f(x)=0.5 \cdot (0.25)^x \). So the first dropdown (the coefficient) is 0.5, the second (the base) is 0.25. Wait, but the first dropdown options are -2, 0, 0.25, 0.5. So the first box is 0.5, the second box is 0.25. Wait, maybe the first dropdown is the base? No, because when \( x=0 \), the base to the 0 power is 1, so the coefficient is the value at \( x=0 \), which is 0.5. So the first box is 0.5, the second box is 0.25.

Answer:

The first dropdown (coefficient) is \( 0.5 \), the second dropdown (base) is \( 0.25 \). So \( f(x) = \boldsymbol{0.5} \cdot \boldsymbol{0.25}^x \)