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analyzing exponential decay graphs analyze the graph of the exponential…

Question

analyzing exponential decay graphs
analyze the graph of the exponential decay function.
the initial value is
the base, or rate o is
the domain is
graph has points (-1, 3), (0, 1), (1, 1/3), (2, 1/9)

Explanation:

Step1: Find Initial Value

The initial value of an exponential function \( y = ab^x \) is the value when \( x = 0 \). From the graph, when \( x = 0 \), \( y = 1 \)? Wait, no, wait. Wait the point \( (0,1) \) is on the graph? Wait the graph has points \( (-1, 3) \), \( (0,1) \), \( (1, \frac{1}{3}) \), \( (2, \frac{1}{9}) \). So the general form of exponential decay is \( y = ab^x \). Let's use \( x = 0 \): \( y = a \cdot b^0 = a \). So when \( x = 0 \), \( y = 1 \)? Wait no, the point \( (0,1) \) is on the graph? Wait the graph shows \( (0,1) \)? Wait the grid: at \( x = 0 \), the y - coordinate is 1? Wait the red dot at \( (0,1) \)? Wait the point \( (-1, 3) \), \( (0,1) \), \( (1, 1/3) \), \( (2, 1/9) \). So for \( x = 0 \), \( y = 1 \). Wait but the dropdown has options 0, 1/3, 1, 3. Wait maybe I made a mistake. Wait let's check the function. Let's take two points. Let's take \( x=-1 \), \( y = 3 \); \( x = 0 \), \( y =? \); \( x = 1 \), \( y = 1/3 \); \( x = 2 \), \( y = 1/9 \). Let's assume the function is \( y = ab^x \). When \( x = 0 \), \( y = a \cdot b^0 = a \). If \( x = 1 \), \( y = ab^1 = ab \); \( x = 2 \), \( y = ab^2 \). From \( x = 1 \) to \( x = 2 \), \( y \) goes from \( 1/3 \) to \( 1/9 \), so the ratio is \( (1/9)/(1/3)=1/3 \), so \( b = 1/3 \). Then when \( x = 1 \), \( y = ab = 1/3 \), and \( b = 1/3 \), so \( a \cdot (1/3)=1/3 \), so \( a = 1 \). Wait but when \( x=-1 \), \( y = ab^{-1}=a/b \). If \( a = 1 \) and \( b = 1/3 \), then \( y = 1/(1/3)=3 \), which matches the point \( (-1, 3) \). So the function is \( y = 1 \cdot (1/3)^x \). So the initial value (when \( x = 0 \)) is \( a = 1 \)? Wait no, the initial value is \( a \), which is 1? But the dropdown has 0, 1/3, 1, 3. Wait maybe the initial value is the value when \( x = 0 \), which is 1. Wait but let's check the graph again. The point \( (0,1) \) is on the graph. So the initial value (the y - intercept) is 1. Wait but the dropdown options include 1. Wait maybe I was wrong. Wait let's re - evaluate. Wait the function is \( y = 3 \cdot (1/3)^{x + 1}\)? Let's check \( x=-1 \): \( 3\cdot(1/3)^{0}=3\cdot1 = 3 \), correct. \( x = 0 \): \( 3\cdot(1/3)^{1}=3\cdot(1/3)=1 \), correct. \( x = 1 \): \( 3\cdot(1/3)^{2}=3\cdot(1/9)=1/3 \), correct. \( x = 2 \): \( 3\cdot(1/3)^{3}=3\cdot(1/27)=1/9 \), correct. So the function can also be written as \( y = 3\cdot(1/3)^{x + 1}=3\cdot(1/3)^x\cdot(1/3)^{-1}=3\cdot3\cdot(1/3)^x = 9\cdot(1/3)^x \)? No, that's not right. Wait \( (1/3)^{-1}=3 \), so \( 3\cdot(1/3)^{x + 1}=3\cdot(1/3)^x\cdot(1/3)= (1/3)^x \cdot1 \). Wait no, \( 3\times(1/3)=1 \), so \( 3\cdot(1/3)^{x + 1}=(1/3)^x \). Wait that's the same as \( y=(1/3)^x \). Wait when \( x=-1 \), \( (1/3)^{-1}=3 \), correct. \( x = 0 \), \( (1/3)^0 = 1 \), correct. \( x = 1 \), \( (1/3)^1=1/3 \), correct. So the function is \( y=(1/3)^x \)? Wait no, \( (1/3)^{-1}=3 \), which matches \( (-1,3) \). So the initial value (when \( x = 0 \)) is \( y=(1/3)^0 = 1 \). So the initial value is 1.

Step2: Find the Base

The base of the exponential function. Looking at the points: from \( x = 0 \) to \( x = 1 \), \( y \) goes from 1 to \( 1/3 \), so the ratio is \( (1/3)/1 = 1/3 \). From \( x = 1 \) to \( x = 2 \), \( y \) goes from \( 1/3 \) to \( 1/9 \), ratio is \( (1/9)/(1/3)=1/3 \). So the base \( b = 1/3 \).

Step3: Find the Domain

The domain of an exponential function \( y = b^x \) (or any exponential function) is all real numbers, because we can plug in any real number for \( x \). So the domain is all real numbers (\( (-\infty, \infty) \)).

Wait but the first dropdown is f…

Answer:

Initial value: 1; Base: \( \frac{1}{3} \); Domain: All real numbers (\( (-\infty, \infty) \))

(Assuming the first dropdown is for initial value, the second for base, and the third for domain. So the initial value is 1, the base is \( 1/3 \), and the domain is all real numbers.)