Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
To determine which set of ordered pairs could be generated by an exponential function, we need to analyze each set to see if they fit the general form of an exponential function [tex]\( y = a \cdot b^x \)[/tex], where [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are constants.
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 see if we can derive a consistent set of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] that match these points:
For point [tex]\((1, 1)\)[/tex]:
[tex]\[ 1 = a \cdot b^1 \][/tex]
[tex]\[ a \cdot b = 1 \][/tex]
For point [tex]\((2, \frac{1}{2})\)[/tex]:
[tex]\[ \frac{1}{2} = a \cdot b^2 \][/tex]
Using [tex]\( a \cdot b = 1 \)[/tex] to solve for [tex]\( a \)[/tex]:
[tex]\[ a = \frac{1}{b} \][/tex]
Substitute [tex]\( a \)[/tex] into second equation:
[tex]\[ \frac{1}{2} = \frac{1}{b} \cdot b^2 \][/tex]
[tex]\[ \frac{1}{2} = b \][/tex]
Substituting [tex]\( b = \frac{1}{2} \)[/tex] in [tex]\( a \cdot b = 1 \)[/tex]:
[tex]\[ a \cdot \frac{1}{2} = 1 \][/tex]
[tex]\[ a = 2 \][/tex]
Now let's check the other points:
For point [tex]\((3, \frac{1}{3})\)[/tex]:
[tex]\[ \frac{1}{3} = 2 \cdot \left(\frac{1}{2}\right)^3 \][/tex]
[tex]\[ \frac{1}{3} = \frac{2}{8} = \frac{1}{4} \][/tex]
This contradicts, so Set 1 is not exponential.
Set 2: [tex]\((1, 1), \left(2, \frac{1}{4}\right), \left(3, \frac{1}{9}\right), \left(4, \frac{1}{16}\right)\)[/tex]
For point [tex]\((1, 1)\)[/tex]:
[tex]\[ 1 = a \cdot b^1 \][/tex]
[tex]\[ a \cdot b = 1 \][/tex]
For point [tex]\((2, \frac{1}{4})\)[/tex]:
[tex]\[ \frac{1}{4} = a \cdot b^2 \][/tex]
Using [tex]\( a \cdot b = 1 \)[/tex]:
[tex]\[ a = \frac{1}{b} \][/tex]
Substitute:
[tex]\[ \frac{1}{4} = \frac{1}{b} \cdot b^2 \][/tex]
[tex]\[ \frac{1}{4} = b \][/tex]
Now [tex]\( a \)[/tex]:
[tex]\[ a \cdot \frac{1}{4} = 1 \][/tex]
[tex]\[ a = 4 \][/tex]
Check other points:
For point [tex]\((3, \frac{1}{9})\)[/tex]:
[tex]\[ \frac{1}{9} = 4 \cdot \left(\frac{1}{4}\right)^3 \][/tex]
[tex]\[ \frac{1}{9} = 4 \cdot \frac{1}{64} = \frac{1}{16} \][/tex]
This contradicts, so Set 2 is not exponential.
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]
For point [tex]\((1, \frac{1}{2})\)[/tex]:
[tex]\[ \frac{1}{2} = a \cdot b^1 \][/tex]
[tex]\[ a \cdot b = \frac{1}{2} \][/tex]
For point [tex]\((2, \frac{1}{4})\)[/tex]:
[tex]\[ \frac{1}{4} = a \cdot b^2 \][/tex]
Using [tex]\( a \cdot b = \frac{1}{2} \)[/tex]:
[tex]\[ a = \frac{1}{2b} \][/tex]
Substitute:
[tex]\[ \frac{1}{4} = \frac{1}{2b} \cdot b^2 \][/tex]
[tex]\[ \frac{1}{4} = \frac{b}{2} \][/tex]
[tex]\[ b = \frac{1}{2} \][/tex]
Now [tex]\( a \)[/tex]:
[tex]\[ a \cdot \frac{1}{2} = \frac{1}{2} \][/tex]
[tex]\[ a = 1 \][/tex]
Check other points:
For point [tex]\((3, \frac{1}{8})\)[/tex]:
[tex]\[ \frac{1}{8} = 1 \cdot \left(\frac{1}{2}\right)^3 \][/tex]
[tex]\[ \frac{1}{8} = \frac{1}{8} \][/tex]
For point [tex]\((4, \frac{1}{16})\)[/tex]:
[tex]\[ \frac{1}{16} = 1 \cdot \left(\frac{1}{2}\right)^4 \][/tex]
[tex]\[ \frac{1}{16} = \frac{1}{16} \][/tex]
All points fit, so Set 3 is generated by an exponential function.
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]
Using similar steps:
For point [tex]\((1, \frac{1}{2})\)[/tex]:
[tex]\[ \frac{1}{2} = a \cdot b \][/tex]
For point [tex]\((2, \frac{1}{4})\)[/tex]:
[tex]\[ \frac{1}{4} = a \cdot b^2 \][/tex]
Using [tex]\( a \cdot b = \frac{1}{2} \)[/tex]:
[tex]\[ a = \frac{1}{2b} \][/tex]
Substitute:
[tex]\[ \frac{1}{4} = \frac{1}{2b} \cdot b^2 \][/tex]
[tex]\[ \frac{1}{4} = \frac{b}{2} \][/tex]
[tex]\[ b = \frac{1}{2} \][/tex]
Now [tex]\( a \)[/tex]:
[tex]\[ a \cdot \frac{1}{2} = \frac{1}{2} \][/tex]
[tex]\[ a = 1 \][/tex]
Check other points:
For point [tex]\((3, \frac{1}{6})\)[/tex]:
[tex]\[ \frac{1}{6} = 1 \cdot \left(\frac{1}{2}\right)^3 \][/tex]
[tex]\[ \frac{1}{6} \neq \frac{1}{8} \][/tex]
This contradicts, so Set 4 is not exponential.
Thus, the correct set of ordered pairs that could be generated by an exponential function is 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].
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 see if we can derive a consistent set of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] that match these points:
For point [tex]\((1, 1)\)[/tex]:
[tex]\[ 1 = a \cdot b^1 \][/tex]
[tex]\[ a \cdot b = 1 \][/tex]
For point [tex]\((2, \frac{1}{2})\)[/tex]:
[tex]\[ \frac{1}{2} = a \cdot b^2 \][/tex]
Using [tex]\( a \cdot b = 1 \)[/tex] to solve for [tex]\( a \)[/tex]:
[tex]\[ a = \frac{1}{b} \][/tex]
Substitute [tex]\( a \)[/tex] into second equation:
[tex]\[ \frac{1}{2} = \frac{1}{b} \cdot b^2 \][/tex]
[tex]\[ \frac{1}{2} = b \][/tex]
Substituting [tex]\( b = \frac{1}{2} \)[/tex] in [tex]\( a \cdot b = 1 \)[/tex]:
[tex]\[ a \cdot \frac{1}{2} = 1 \][/tex]
[tex]\[ a = 2 \][/tex]
Now let's check the other points:
For point [tex]\((3, \frac{1}{3})\)[/tex]:
[tex]\[ \frac{1}{3} = 2 \cdot \left(\frac{1}{2}\right)^3 \][/tex]
[tex]\[ \frac{1}{3} = \frac{2}{8} = \frac{1}{4} \][/tex]
This contradicts, so Set 1 is not exponential.
Set 2: [tex]\((1, 1), \left(2, \frac{1}{4}\right), \left(3, \frac{1}{9}\right), \left(4, \frac{1}{16}\right)\)[/tex]
For point [tex]\((1, 1)\)[/tex]:
[tex]\[ 1 = a \cdot b^1 \][/tex]
[tex]\[ a \cdot b = 1 \][/tex]
For point [tex]\((2, \frac{1}{4})\)[/tex]:
[tex]\[ \frac{1}{4} = a \cdot b^2 \][/tex]
Using [tex]\( a \cdot b = 1 \)[/tex]:
[tex]\[ a = \frac{1}{b} \][/tex]
Substitute:
[tex]\[ \frac{1}{4} = \frac{1}{b} \cdot b^2 \][/tex]
[tex]\[ \frac{1}{4} = b \][/tex]
Now [tex]\( a \)[/tex]:
[tex]\[ a \cdot \frac{1}{4} = 1 \][/tex]
[tex]\[ a = 4 \][/tex]
Check other points:
For point [tex]\((3, \frac{1}{9})\)[/tex]:
[tex]\[ \frac{1}{9} = 4 \cdot \left(\frac{1}{4}\right)^3 \][/tex]
[tex]\[ \frac{1}{9} = 4 \cdot \frac{1}{64} = \frac{1}{16} \][/tex]
This contradicts, so Set 2 is not exponential.
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]
For point [tex]\((1, \frac{1}{2})\)[/tex]:
[tex]\[ \frac{1}{2} = a \cdot b^1 \][/tex]
[tex]\[ a \cdot b = \frac{1}{2} \][/tex]
For point [tex]\((2, \frac{1}{4})\)[/tex]:
[tex]\[ \frac{1}{4} = a \cdot b^2 \][/tex]
Using [tex]\( a \cdot b = \frac{1}{2} \)[/tex]:
[tex]\[ a = \frac{1}{2b} \][/tex]
Substitute:
[tex]\[ \frac{1}{4} = \frac{1}{2b} \cdot b^2 \][/tex]
[tex]\[ \frac{1}{4} = \frac{b}{2} \][/tex]
[tex]\[ b = \frac{1}{2} \][/tex]
Now [tex]\( a \)[/tex]:
[tex]\[ a \cdot \frac{1}{2} = \frac{1}{2} \][/tex]
[tex]\[ a = 1 \][/tex]
Check other points:
For point [tex]\((3, \frac{1}{8})\)[/tex]:
[tex]\[ \frac{1}{8} = 1 \cdot \left(\frac{1}{2}\right)^3 \][/tex]
[tex]\[ \frac{1}{8} = \frac{1}{8} \][/tex]
For point [tex]\((4, \frac{1}{16})\)[/tex]:
[tex]\[ \frac{1}{16} = 1 \cdot \left(\frac{1}{2}\right)^4 \][/tex]
[tex]\[ \frac{1}{16} = \frac{1}{16} \][/tex]
All points fit, so Set 3 is generated by an exponential function.
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]
Using similar steps:
For point [tex]\((1, \frac{1}{2})\)[/tex]:
[tex]\[ \frac{1}{2} = a \cdot b \][/tex]
For point [tex]\((2, \frac{1}{4})\)[/tex]:
[tex]\[ \frac{1}{4} = a \cdot b^2 \][/tex]
Using [tex]\( a \cdot b = \frac{1}{2} \)[/tex]:
[tex]\[ a = \frac{1}{2b} \][/tex]
Substitute:
[tex]\[ \frac{1}{4} = \frac{1}{2b} \cdot b^2 \][/tex]
[tex]\[ \frac{1}{4} = \frac{b}{2} \][/tex]
[tex]\[ b = \frac{1}{2} \][/tex]
Now [tex]\( a \)[/tex]:
[tex]\[ a \cdot \frac{1}{2} = \frac{1}{2} \][/tex]
[tex]\[ a = 1 \][/tex]
Check other points:
For point [tex]\((3, \frac{1}{6})\)[/tex]:
[tex]\[ \frac{1}{6} = 1 \cdot \left(\frac{1}{2}\right)^3 \][/tex]
[tex]\[ \frac{1}{6} \neq \frac{1}{8} \][/tex]
This contradicts, so Set 4 is not exponential.
Thus, the correct set of ordered pairs that could be generated by an exponential function is 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].
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
