Answer :
To determine which set of ordered pairs could be generated by an exponential function, we consider the definition of an exponential function. An exponential function takes the form [tex]\( y = k \cdot a^x \)[/tex], where [tex]\( k \)[/tex] and [tex]\( a \)[/tex] are constants, [tex]\( k \neq 0 \)[/tex], and [tex]\( a > 0 \)[/tex].
Let's analyze each set of ordered pairs to see if they follow this form:
1. [tex]\(\left(-1,-\frac{1}{2}\right), (0,0), \left(1, \frac{1}{2}\right), (2,1)\)[/tex]
- At [tex]\( x = -1 \)[/tex], [tex]\( y = -\frac{1}{2} \)[/tex]
- At [tex]\( x = 0 \)[/tex], [tex]\( y = 0 \)[/tex]
- At [tex]\( x = 1 \)[/tex], [tex]\( y = \frac{1}{2} \)[/tex]
- At [tex]\( x = 2 \)[/tex], [tex]\( y = 1 \)[/tex]
Here, the y-values don't follow a clear pattern that fits the form [tex]\( y = k \cdot a^x \)[/tex]. For an exponential function, the change in y-values should be multiplicative, but there is an inconsistency here, particularly with [tex]\( y = 0 \)[/tex] when [tex]\( x = 0 \)[/tex]. This set cannot be generated by an exponential function.
2. [tex]\((-1,-1), (0,0), (1,1), (2,8)\)[/tex]
- At [tex]\( x = -1 \)[/tex], [tex]\( y = -1 \)[/tex]
- At [tex]\( x = 0 \)[/tex], [tex]\( y = 0 \)[/tex]
- At [tex]\( x = 1 \)[/tex], [tex]\( y = 1 \)[/tex]
- At [tex]\( x = 2 \)[/tex], [tex]\( y = 8 \)[/tex]
This set does not appear to follow an exponential pattern either because the y-values do not reflect multiplicative consistency that an exponential function requires. The y-values do not scale correctly according to the exponential form for an [tex]\( a \)[/tex] and [tex]\( k \)[/tex].
3. [tex]\(\left(-1, \frac{1}{2}\right), (0,1), (1,2), (2,4)\)[/tex]
- At [tex]\( x = -1 \)[/tex], [tex]\( y = \frac{1}{2} \)[/tex]
- At [tex]\( x = 0 \)[/tex], [tex]\( y = 1 \)[/tex]
- At [tex]\( x = 1 \)[/tex], [tex]\( y = 2 \)[/tex]
- At [tex]\( x = 2 \)[/tex], [tex]\( y = 4 \)[/tex]
To check if these points fit the form [tex]\( y = k \cdot a^x \)[/tex]:
- For [tex]\( x = 0 \)[/tex], [tex]\( y = 1 \)[/tex]: [tex]\( 1 = k \cdot a^0 \Rightarrow 1 = k \cdot 1 \Rightarrow k = 1 \)[/tex]
- For [tex]\( x = 1 \)[/tex], [tex]\( y = 2 \)[/tex]: [tex]\( 2 = 1 \cdot a \Rightarrow a = 2 \)[/tex]
- For [tex]\( x = -1 \)[/tex], [tex]\( y = \frac{1}{2} \)[/tex]: [tex]\( \frac{1}{2} = 1 \cdot 2^{-1} \Rightarrow \frac{1}{2} = \frac{1}{2} \)[/tex]
- For [tex]\( x = 2 \)[/tex], [tex]\( y = 4 \)[/tex]: [tex]\( 4 = 1 \cdot 2^2 \Rightarrow 4 = 4 \)[/tex]
These values fit the form [tex]\( y = 2^x \)[/tex], making this set capable of being generated by an exponential function.
4. [tex]\((-1,1), (0,0), (1,1), (2,4)\)[/tex]
- At [tex]\( x = -1 \)[/tex], [tex]\( y = 1 \)[/tex]
- At [tex]\( x = 0 \)[/tex], [tex]\( y = 0 \)[/tex]
- At [tex]\( x = 1 \)[/tex], [tex]\( y = 1 \)[/tex]
- At [tex]\( x = 2 \)[/tex], [tex]\( y = 4 \)[/tex]
The presence of [tex]\( y = 0 \)[/tex] for [tex]\( x = 0 \)[/tex] disrupts the multiplicative pattern typical of exponential functions. This set also does not follow an exponential pattern.
As analyzed:
The only set of ordered pairs that could be generated by an exponential function is:
[tex]\(\left(-1, \frac{1}{2}\right), (0,1), (1,2), (2,4)\)[/tex]
So, the answer is the 3rd set of ordered pairs.
Let's analyze each set of ordered pairs to see if they follow this form:
1. [tex]\(\left(-1,-\frac{1}{2}\right), (0,0), \left(1, \frac{1}{2}\right), (2,1)\)[/tex]
- At [tex]\( x = -1 \)[/tex], [tex]\( y = -\frac{1}{2} \)[/tex]
- At [tex]\( x = 0 \)[/tex], [tex]\( y = 0 \)[/tex]
- At [tex]\( x = 1 \)[/tex], [tex]\( y = \frac{1}{2} \)[/tex]
- At [tex]\( x = 2 \)[/tex], [tex]\( y = 1 \)[/tex]
Here, the y-values don't follow a clear pattern that fits the form [tex]\( y = k \cdot a^x \)[/tex]. For an exponential function, the change in y-values should be multiplicative, but there is an inconsistency here, particularly with [tex]\( y = 0 \)[/tex] when [tex]\( x = 0 \)[/tex]. This set cannot be generated by an exponential function.
2. [tex]\((-1,-1), (0,0), (1,1), (2,8)\)[/tex]
- At [tex]\( x = -1 \)[/tex], [tex]\( y = -1 \)[/tex]
- At [tex]\( x = 0 \)[/tex], [tex]\( y = 0 \)[/tex]
- At [tex]\( x = 1 \)[/tex], [tex]\( y = 1 \)[/tex]
- At [tex]\( x = 2 \)[/tex], [tex]\( y = 8 \)[/tex]
This set does not appear to follow an exponential pattern either because the y-values do not reflect multiplicative consistency that an exponential function requires. The y-values do not scale correctly according to the exponential form for an [tex]\( a \)[/tex] and [tex]\( k \)[/tex].
3. [tex]\(\left(-1, \frac{1}{2}\right), (0,1), (1,2), (2,4)\)[/tex]
- At [tex]\( x = -1 \)[/tex], [tex]\( y = \frac{1}{2} \)[/tex]
- At [tex]\( x = 0 \)[/tex], [tex]\( y = 1 \)[/tex]
- At [tex]\( x = 1 \)[/tex], [tex]\( y = 2 \)[/tex]
- At [tex]\( x = 2 \)[/tex], [tex]\( y = 4 \)[/tex]
To check if these points fit the form [tex]\( y = k \cdot a^x \)[/tex]:
- For [tex]\( x = 0 \)[/tex], [tex]\( y = 1 \)[/tex]: [tex]\( 1 = k \cdot a^0 \Rightarrow 1 = k \cdot 1 \Rightarrow k = 1 \)[/tex]
- For [tex]\( x = 1 \)[/tex], [tex]\( y = 2 \)[/tex]: [tex]\( 2 = 1 \cdot a \Rightarrow a = 2 \)[/tex]
- For [tex]\( x = -1 \)[/tex], [tex]\( y = \frac{1}{2} \)[/tex]: [tex]\( \frac{1}{2} = 1 \cdot 2^{-1} \Rightarrow \frac{1}{2} = \frac{1}{2} \)[/tex]
- For [tex]\( x = 2 \)[/tex], [tex]\( y = 4 \)[/tex]: [tex]\( 4 = 1 \cdot 2^2 \Rightarrow 4 = 4 \)[/tex]
These values fit the form [tex]\( y = 2^x \)[/tex], making this set capable of being generated by an exponential function.
4. [tex]\((-1,1), (0,0), (1,1), (2,4)\)[/tex]
- At [tex]\( x = -1 \)[/tex], [tex]\( y = 1 \)[/tex]
- At [tex]\( x = 0 \)[/tex], [tex]\( y = 0 \)[/tex]
- At [tex]\( x = 1 \)[/tex], [tex]\( y = 1 \)[/tex]
- At [tex]\( x = 2 \)[/tex], [tex]\( y = 4 \)[/tex]
The presence of [tex]\( y = 0 \)[/tex] for [tex]\( x = 0 \)[/tex] disrupts the multiplicative pattern typical of exponential functions. This set also does not follow an exponential pattern.
As analyzed:
The only set of ordered pairs that could be generated by an exponential function is:
[tex]\(\left(-1, \frac{1}{2}\right), (0,1), (1,2), (2,4)\)[/tex]
So, the answer is the 3rd set of ordered pairs.
