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Which set of ordered pairs could be generated by an exponential function?

A. \((1,1),\left(2, \frac{1}{2}\right),\left(3, \frac{1}{3}\right),\left(4, \frac{1}{4}\right)\)

B. \((1,1),\left(2, \frac{1}{4}\right),\left(3, \frac{1}{9}\right),\left(4, \frac{1}{16}\right)\)

C. \(\left(1, \frac{1}{2}\right),\left(2, \frac{1}{4}\right),\left(3, \frac{1}{8}\right),\left(4, \frac{1}{16}\right)\)

D. [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]

Sagot :

To determine which set of ordered pairs could be generated by an exponential function, let's check each set of points. An exponential function has the form \(y = a \cdot b^x\) for constants \(a\) and \(b\). Below each set will be analyzed to see if it follows this form.

### Set 1:
[tex]\[ (1, 1), \left(2, \frac{1}{2}\right), \left(3, \frac{1}{3}\right), \left(4, \frac{1}{4}\right) \][/tex]
For this set to follow the form \(y = a \cdot b^x\), the ratio \(\frac{y_{i+1}}{y_i}\) must be constant.

- \(\frac{\frac{1}{2}}{1} = \frac{1}{2}\)
- \(\frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}\)
- \(\frac{\frac{1}{4}}{\frac{1}{3}} = \frac{3}{4}\)

These ratios are not constant, so this set does not represent an exponential function.

### Set 2:
[tex]\[ (1, 1), \left(2, \frac{1}{4}\right), \left(3, \frac{1}{9}\right), \left(4, \frac{1}{16}\right) \][/tex]
Again, we check the ratios:

- \(\frac{\frac{1}{4}}{1} = \frac{1}{4}\)
- \(\frac{\frac{1}{9}}{\frac{1}{4}} = \frac{4}{9}\)
- \(\frac{\frac{1}{16}}{\frac{1}{9}} = \frac{9}{16}\)

These ratios are not constant either, so this set does not represent an exponential function.

### Set 3:
[tex]\[ \left(1, \frac{1}{2}\right), \left(2, \frac{1}{4}\right), \left(3, \frac{1}{8}\right), \left(4, \frac{1}{16}\right) \][/tex]
Let's check the ratios for this set:

- \(\frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}\)
- \(\frac{\frac{1}{8}}{\frac{1}{4}} = \frac{1}{2}\)
- \(\frac{\frac{1}{16}}{\frac{1}{8}} = \frac{1}{2}\)

The ratio is constant (\(\frac{1}{2}\)), indicating that each successive term is obtained by multiplying the previous term by \(\frac{1}{2}\). Therefore, this set follows the form \(y = a \cdot b^x\) with \(a = \frac{1}{2}\) and \(b = \frac{1}{2}\), indicating it is exponential.

### 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]
Let's check the ratios:

- \(\frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}\)
- \(\frac{\frac{1}{6}}{\frac{1}{4}} = \frac{2}{3}\)
- \(\frac{\frac{1}{8}}{\frac{1}{6}} = \frac{3}{4}\)

These ratios are not constant, indicating this set does not represent an exponential function.

### Conclusion:
The set of ordered pairs that could be generated by an exponential function is:

[tex]\[ \left(1, \frac{1}{2}\right), \left(2, \frac{1}{4}\right), \left(3, \frac{1}{8}\right), \left(4, \frac{1}{16}\right) \][/tex]

Thus, the correct answer is:
[tex]\[ \left( 3 \right) \][/tex]