Question

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

Answer 1

The general expression for the exponential function is $f(x)=a\cdot b^x,$ where $a\neq 0,\ b>0,\ b\neq 1.$

Note that the graph of each exponential function passes through the point (0,a), because $f(0)=a\cdot b^0=a.$ Since $a\neq 0,$ then options A and  C are false.

First point in options B and D is (0,1), then

$f(0)=1\Rightarrow a=1$

and you get the expression $f(x)=b^x$ for the exponential function.

Points (0,1), (1,3), (2,9) and (3,27) represent the powers of 3. In this case $f(x)=3^x.$

Points (0,1), (1,2), (2,5) and (3,10) couldn't generate any exponential function, because if $b^1=2,$ then $b^2=2^2=4\neq 5.$

Answer: correct choice is D

Answer 2

Answer:

Option 4

(0, 1), (1, 3), (2, 9), (3, 27) set of ordered pairs could be generated by an exponential function $f(x)=3^x$.

Step-by-step explanation:

Given : Set of ordered pairs could be generated by an exponential function

(0, 0), (1, 1), (2, 8), (3, 27)  

(0, 1), (1, 2), (2, 5), (3, 10)

(0, 0), (1, 3), (2, 6), (3, 9)

(0, 1), (1, 3), (2, 9), (3, 27)

To find : Which set?

Solution :

The general form of an exponential function is $f(x)=ab^x$

where,

a is the initial amount  $a\neq 0$

b is the rate of change $b>0 , b\neq 1$

The graph of each exponential function passes through the point (0,a),

As $f(0)=a\cdot b^0=a$

and $a\neq 0$

Therefore, Option 1 and 3 are false.

Now, in Option 2 and 4 first point is (0,1)

i.e, $f(0)=a\cdot b^0\Rightarrow a=1$

Then, The exponential function form is

$f(x)=b^x$

In Option 2,

Point (1,2)

$2=b^1\Rightarrow, b=2$

Point (2,5)

$f(x)=2^2\Rightarrow, f(x)=4$

which is not correct.

Option 2 is false.

Now, In Option 4      

Point (1,3)

$3=b^1\Rightarrow, b=3$

Point (2,9)

$f(x)=3^2\Rightarrow, f(x)=9$ satisfying

Point (3,27)

$f(x)=3^3\Rightarrow, f(x)=27$ satisfying

Option 4 is true.

Therefore, The exponential form is $f(x)=3^x$

Hence, option 4 is correct.

(0, 1), (1, 3), (2, 9), (3, 27) set of ordered pairs could be generated by an exponential function $f(x)=3^x$.