Answer :
To determine which set of ordered pairs could be generated by an exponential function, let's analyze each given set and see whether they match the general form of an exponential function.
An exponential function typically has the form \( y = ab^x \), where \( a \) is the initial value (the \( y \)-intercept) and \( b \) is the base of the exponential.
Option 1: \(\left(-1,-\frac{1}{2}\right),(0,0),\left(1, \frac{1}{2}\right),(2,1)\)
- At \( x = 0 \), \( y = 0 \).
Since exponential functions never pass through \((0,0)\) unless they are zero everywhere, this option cannot represent an exponential function.
Option 2: \((-1,-1),(0,0),(1,1),(2,8)\)
- At \( x = 0 \), \( y = 0 \).
Again, exponential functions do not pass through \((0,0)\) unless they are zero everywhere, so this option cannot represent an exponential function.
Option 3: \(\left(-1, \frac{1}{2}\right),(0,1),(1,2),(2,4)\)
Let's test if these pairs fit an exponential function \( y = ab^x \).
- At \( x = 0 \), \( y = 1 \); hence \( a = 1 \).
- Now, we use other points to find \( b \).
Using \( (1, 2) \):
[tex]\[ 2 = 1 \cdot b^1 \Rightarrow b = 2 \][/tex]
Using \( (2, 4) \):
[tex]\[ 4 = 1 \cdot 2^2 \Rightarrow 4 = 4 \][/tex]
This works.
Using \( (-1, \frac{1}{2} )\):
[tex]\[ \frac{1}{2} = 1 \cdot 2^{-1} \Rightarrow \frac{1}{2} = \frac{1}{2} \][/tex]
This also works.
This set of ordered pairs fits the exponential function \( y = 2^x \).
Option 4: \((-1,1),(0,0),(1,1),(2,4)\)
- At \( x = 0 \), \( y = 0 \).
Since exponential functions do not pass through \((0,0)\) unless they are zero everywhere, this set cannot represent an exponential function.
Therefore, the set of ordered pairs [tex]\(\left(-1, \frac{1}{2}\right),(0,1),(1,2),(2,4)\)[/tex] can be generated by an exponential function.
An exponential function typically has the form \( y = ab^x \), where \( a \) is the initial value (the \( y \)-intercept) and \( b \) is the base of the exponential.
Option 1: \(\left(-1,-\frac{1}{2}\right),(0,0),\left(1, \frac{1}{2}\right),(2,1)\)
- At \( x = 0 \), \( y = 0 \).
Since exponential functions never pass through \((0,0)\) unless they are zero everywhere, this option cannot represent an exponential function.
Option 2: \((-1,-1),(0,0),(1,1),(2,8)\)
- At \( x = 0 \), \( y = 0 \).
Again, exponential functions do not pass through \((0,0)\) unless they are zero everywhere, so this option cannot represent an exponential function.
Option 3: \(\left(-1, \frac{1}{2}\right),(0,1),(1,2),(2,4)\)
Let's test if these pairs fit an exponential function \( y = ab^x \).
- At \( x = 0 \), \( y = 1 \); hence \( a = 1 \).
- Now, we use other points to find \( b \).
Using \( (1, 2) \):
[tex]\[ 2 = 1 \cdot b^1 \Rightarrow b = 2 \][/tex]
Using \( (2, 4) \):
[tex]\[ 4 = 1 \cdot 2^2 \Rightarrow 4 = 4 \][/tex]
This works.
Using \( (-1, \frac{1}{2} )\):
[tex]\[ \frac{1}{2} = 1 \cdot 2^{-1} \Rightarrow \frac{1}{2} = \frac{1}{2} \][/tex]
This also works.
This set of ordered pairs fits the exponential function \( y = 2^x \).
Option 4: \((-1,1),(0,0),(1,1),(2,4)\)
- At \( x = 0 \), \( y = 0 \).
Since exponential functions do not pass through \((0,0)\) unless they are zero everywhere, this set cannot represent an exponential function.
Therefore, the set of ordered pairs [tex]\(\left(-1, \frac{1}{2}\right),(0,1),(1,2),(2,4)\)[/tex] can be generated by an exponential function.
