Whether you're a student or a professional, IDNLearn.com has answers for everyone. Find reliable solutions to your questions quickly and easily with help from our experienced experts.
Sagot :
To determine which set of ordered pairs could be generated by an exponential function, we'll analyze each set and see if they fit the form of an exponential function, [tex]\( y = a \cdot b^x \)[/tex].
1. Set 1: [tex]\((1,1), \left(2, \frac{1}{2}\right), \left(3, \frac{1}{3}\right), \left(4, \frac{1}{4}\right)\)[/tex]
Let's check if this set can be represented by an exponential function:
- For [tex]\( x = 1 \)[/tex], [tex]\( y = 1 \)[/tex]
- For [tex]\( x = 2 \)[/tex], [tex]\( y = \frac{1}{2} \)[/tex]
- For [tex]\( x = 3 \)[/tex], [tex]\( y = \frac{1}{3} \)[/tex]
- For [tex]\( x = 4 \)[/tex], [tex]\( y = \frac{1}{4} \)[/tex]
Notice that the pattern in the [tex]\( y \)[/tex]-values is not exponential (it looks more like a harmonic sequence), so this set does not match an exponential function.
2. Set 2: [tex]\((1,1), \left(2, \frac{1}{4}\right), \left(3, \frac{1}{9}\right), \left(4, \frac{1}{16}\right)\)[/tex]
Let's check this set for an exponential pattern:
- For [tex]\( x = 1 \)[/tex], [tex]\( y = 1 \)[/tex]
- For [tex]\( x = 2 \)[/tex], [tex]\( y = \frac{1}{4} = \left(\frac{1}{2}\right)^2 \)[/tex]
- For [tex]\( x = 3 \)[/tex], [tex]\( y = \frac{1}{9} = \left(\frac{1}{3}\right)^2 \)[/tex]
- For [tex]\( x = 4 \)[/tex], [tex]\( y = \frac{1}{16} = \left(\frac{1}{4}\right)^2 \)[/tex]
We can see that [tex]\( y = \left(\frac{1}{x}\right)^2 \)[/tex], which implies that [tex]\( y = x^{-2} \)[/tex], matching the form of an exponential function [tex]\( y = a \cdot b^x \)[/tex] with [tex]\( a = 1 \)[/tex] and [tex]\( b = x^{-2} \)[/tex].
3. Set 3: [tex]\(\left(1, \frac{1}{2}\right), \left(2, \frac{1}{4}\right), \left(2, \frac{1}{8}\right), \left(4, \frac{1}{16}\right)\)[/tex]
This set has a repeated [tex]\( x \)[/tex]-value:
- For [tex]\( x = 1 \)[/tex], [tex]\( y = \frac{1}{2} \)[/tex]
- For [tex]\( x = 2 \)[/tex], [tex]\( y = \frac{1}{4} \)[/tex] and another [tex]\( y = \frac{1}{8} \)[/tex]
- For [tex]\( x = 4 \)[/tex], [tex]\( y = \frac{1}{16} \)[/tex]
The presence of repeated and conflicting [tex]\(y \)[/tex]-values for [tex]\( x = 2 \)[/tex] makes this inconsistent. Therefore, this set cannot represent an exponential function.
4. Set 4: [tex]\(\left(1, \frac{1}{2}\right), \left(2, \frac{1}{4}\right), \left(3, \frac{1}{6}\right), \left(4, \frac{1}{8}\right)\)[/tex]
Analyzing the [tex]\( y \)[/tex]-values for an exponential function:
- For [tex]\( x = 1 \)[/tex], [tex]\( y = \frac{1}{2} \)[/tex]
- For [tex]\( x = 2 \)[/tex], [tex]\( y = \frac{1}{4} \)[/tex]
- For [tex]\( x = 3 \)[/tex], [tex]\( y = \frac{1}{6} \)[/tex]
- For [tex]\( x = 4 \)[/tex], [tex]\( y = \frac{1}{8} \)[/tex]
There's no clear exponential pattern here: for consistent exponential behavior, we'd expect a constant ratio between consecutive [tex]\( y \)[/tex]-values, which is not the case here ([tex]\( \frac{1/4}{1/2} \neq \frac{1/6}{1/4} \)[/tex]).
From this analysis, the set of ordered pairs that fits the pattern of an exponential function is:
Set 2: [tex]\((1,1), \left(2, \frac{1}{4}\right), \left(3, \frac{1}{9}\right), \left(4, \frac{1}{16}\right)\)[/tex]
1. Set 1: [tex]\((1,1), \left(2, \frac{1}{2}\right), \left(3, \frac{1}{3}\right), \left(4, \frac{1}{4}\right)\)[/tex]
Let's check if this set can be represented by an exponential function:
- For [tex]\( x = 1 \)[/tex], [tex]\( y = 1 \)[/tex]
- For [tex]\( x = 2 \)[/tex], [tex]\( y = \frac{1}{2} \)[/tex]
- For [tex]\( x = 3 \)[/tex], [tex]\( y = \frac{1}{3} \)[/tex]
- For [tex]\( x = 4 \)[/tex], [tex]\( y = \frac{1}{4} \)[/tex]
Notice that the pattern in the [tex]\( y \)[/tex]-values is not exponential (it looks more like a harmonic sequence), so this set does not match an exponential function.
2. Set 2: [tex]\((1,1), \left(2, \frac{1}{4}\right), \left(3, \frac{1}{9}\right), \left(4, \frac{1}{16}\right)\)[/tex]
Let's check this set for an exponential pattern:
- For [tex]\( x = 1 \)[/tex], [tex]\( y = 1 \)[/tex]
- For [tex]\( x = 2 \)[/tex], [tex]\( y = \frac{1}{4} = \left(\frac{1}{2}\right)^2 \)[/tex]
- For [tex]\( x = 3 \)[/tex], [tex]\( y = \frac{1}{9} = \left(\frac{1}{3}\right)^2 \)[/tex]
- For [tex]\( x = 4 \)[/tex], [tex]\( y = \frac{1}{16} = \left(\frac{1}{4}\right)^2 \)[/tex]
We can see that [tex]\( y = \left(\frac{1}{x}\right)^2 \)[/tex], which implies that [tex]\( y = x^{-2} \)[/tex], matching the form of an exponential function [tex]\( y = a \cdot b^x \)[/tex] with [tex]\( a = 1 \)[/tex] and [tex]\( b = x^{-2} \)[/tex].
3. Set 3: [tex]\(\left(1, \frac{1}{2}\right), \left(2, \frac{1}{4}\right), \left(2, \frac{1}{8}\right), \left(4, \frac{1}{16}\right)\)[/tex]
This set has a repeated [tex]\( x \)[/tex]-value:
- For [tex]\( x = 1 \)[/tex], [tex]\( y = \frac{1}{2} \)[/tex]
- For [tex]\( x = 2 \)[/tex], [tex]\( y = \frac{1}{4} \)[/tex] and another [tex]\( y = \frac{1}{8} \)[/tex]
- For [tex]\( x = 4 \)[/tex], [tex]\( y = \frac{1}{16} \)[/tex]
The presence of repeated and conflicting [tex]\(y \)[/tex]-values for [tex]\( x = 2 \)[/tex] makes this inconsistent. Therefore, this set cannot represent an exponential function.
4. Set 4: [tex]\(\left(1, \frac{1}{2}\right), \left(2, \frac{1}{4}\right), \left(3, \frac{1}{6}\right), \left(4, \frac{1}{8}\right)\)[/tex]
Analyzing the [tex]\( y \)[/tex]-values for an exponential function:
- For [tex]\( x = 1 \)[/tex], [tex]\( y = \frac{1}{2} \)[/tex]
- For [tex]\( x = 2 \)[/tex], [tex]\( y = \frac{1}{4} \)[/tex]
- For [tex]\( x = 3 \)[/tex], [tex]\( y = \frac{1}{6} \)[/tex]
- For [tex]\( x = 4 \)[/tex], [tex]\( y = \frac{1}{8} \)[/tex]
There's no clear exponential pattern here: for consistent exponential behavior, we'd expect a constant ratio between consecutive [tex]\( y \)[/tex]-values, which is not the case here ([tex]\( \frac{1/4}{1/2} \neq \frac{1/6}{1/4} \)[/tex]).
From this analysis, the set of ordered pairs that fits the pattern of an exponential function is:
Set 2: [tex]\((1,1), \left(2, \frac{1}{4}\right), \left(3, \frac{1}{9}\right), \left(4, \frac{1}{16}\right)\)[/tex]
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Accurate answers are just a click away at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.
