Exercise 40 Page 968

We are given a table that represents an exponential function.

x	f(x)
- 2	12
-1	6
0	3
1	1.5

We want to write an appropriate equation for this data. To do so, let's recall the form for an exponential function. f(x)=ab^xTo find the values of a and b, we will use two of the ordered pairs given in the table. For simplicity, we will use ( 0, 3) and ( - 1, 6). Let's start by substituting 0 for x and 3 for f(x).

f(x)=ab^x

SubstituteII

x= 0, f(x)= 3

3=ab^0

▼

Solve for a

ExponentZero

a^0=1

3=a(1)

IdPropMult

Identity Property of Multiplication

3=a

RearrangeEqn

Rearrange equation

a=3

Now that we know that a= 3, we can partially write the equation. f(x)= 3b^x To find the value of b, we will substitute - 1 for x and 6 for f(x) in the above equation.

f(x)=3b^x

SubstituteII

x= - 1, f(x)= 6

6=3b^(- 1)

▼

Solve for b

NegOneExponentToFrac

a^(- 1)=1/a

6=3/b

MultEqn

LHS * b=RHS* b

6b=3

DivEqn

.LHS /6.=.RHS /6.

b= 3/6

ReduceFrac

a/b=.a /3./.b /3.

b= 1/2

Now that we know that a= 3 and b= 12, we can write the full equation that models the data in the given table.

f(x) = ab^x

SubstituteII

a= 3, b= 1/2

f(x) = 3 ( 1/2)^x

PowQuot

(a/b)^m=a^m/b^m

f(x)=3 (1^x/2^x )

BaseOne

1^a=1

f(x)=3 (1/2^x )

FracToNegExponent

1/a^m=a^(- m)

f(x) = 3(2^(- x))

This corresponds to option A.