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Core Connections Integrated I, 2013

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Core Connections Integrated I, 2013 View details

2. Section 8.2

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Exercise 112 Page 479

a In an exponential function

f (x) = a b^{x},

the variable

a

is the function's initial value. This is the function's value when

x = 0 .

Examining the table, we see that the corresponding

y -

value when

x = 0

is not given. However, we do know two points,

(- 1, 3)

and

(1, 75) .

With these points we can create two equations.

Equation (I): Equation (II): 3 = a b^{- 1} 75 = a b^{1}

By combining these equations we get a system of equations, which we can solve with the Substitution Method.

{3 = a b^{- 1} 75 = a b^{1} (I) (II)

$(II):$ $a^{1} = a$

{3 = a b^{- 1} 75 = a b

▼

(I):

Solve for

a

NegOneExponentToFrac

$(I):$ $a^{- 1} = \frac{1}{a}$

{3 = a \cdot \frac{1}{b} 75 = a b

$(I):$ $LHS \cdot b = RHS \cdot b$

{3 b = a 75 = a b

$(I):$ Rearrange equation

{a = 3 b 75 = a b

▼

(II):

Solve by substitution

$(II):$ $a = 3 b$

{a = 3 b 75 = (3 b) b

$(II):$ Multiply

{a = 3 b 75 = 3 b^{2}

$(II):$ Rearrange equation

{a = 3 b 3 b^{2} = 75

$(II):$ $LHS / 3 = RHS / 3$

{a = 3 b b^{2} = 25

$(II):$ $LHS = RHS$

{a = 3 b b = \pm 5

$(II):$ $b > 0$

{a = 3 b b = 5

When we know the value of

b

we can substitute this into the first equation to solve for

a .

{a = 3 b b = 5

$(I):$ $b = 5$

{a = 3 (5) b = 5

$(I):$ Multiply

{a = 15 b = 5

Now we can write the complete function.

f (x) = 15 (5)^{x}

If we substitute

0,

2,

and

3

instead of

x

in the function, we can calculate their corresponding

y -

values.

$x$	$15 (5)^{x}$	$f (x)$
$0$	$15 (5)^{0}$	$15$
$2$	$15 (5)^{2}$	$375$
$3$	$15 (5)^{3}$	$1875$

Now we can complete the table.

b Like in Part A, the table does not show the initial value of the function. However, we know two points —

(2, 96.64)

and

(3, 77.312) .

Also, notice that these are consecutive terms, which means the quotient of

77.312

and

96.64

will give us the multiplier,

b .

b = \frac{7 7 . 3 1 2}{9 6 . 6 4} \Leftrightarrow b = 0.8

So far we can write the equation as

f (x) = a (0.8)^{x} .

To find the value of

a

we have to substitute either of the known points in the equation and solve for

a .

f (x) = a (0.8)^{x}

$x = 2$ , $f (x) = 96.64$

96.64 = a (0.8)^{2}

▼

Solve for

a

Calculate power

96.64 = a \cdot 0.64

Rearrange equation

a \cdot 0.64 = 96.64

$LHS / 0.64 = RHS / 0.64$

a = 151

Now we can write the complete function.

f (x) = 151 (0.8)^{x}

If we substitute

0,

1,

and

4

instead of

x

in the function, we can calculate their corresponding

y -

values.

$x$	$151 (0.8)^{x}$	$f (x)$
$0$	$151 (0.8)^{0}$	$151$
$1$	$151 (0.8)^{1}$	$120.8$
$4$	$151 (0.8)^{4}$	$61.8496$

Now we can complete the table.

Subchapter links

8.2.1 Curve Fitting p.471-472

8.2.2 Curved Best-Fit Models p.478-480

8.2.3 Solving a System of Exponential Functions Graphically p.483-484