Exercise 123 Page 484

Practice makes perfect

a Using the function

f (x),

we want to evaluate for the given value,

f (- 1) .

To do this we need to substitute

- 1

for

x

in each instance of the

x -

variable and simplify.

f (x) = 3 (2)^{x}

Substitute

$x = - 1$

f (- 1) = 3 (2)^{- 1}

NegOneExponentToFrac

$a^{- 1} = \frac{1}{a}$

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

MoveLeftFacToNumOne

$a \cdot \frac{1}{b} = \frac{a}{b}$

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

b Using the function

f (x),

we want to evaluate for the given value,

f (0) .

To do this, we need to substitute

0

for

x

in each instance of the

x -

variable and simplify.

f (x) = 3 (2)^{x}

Substitute

$x = 0$

f (0) = 3 (2)^{0}

ExponentZero

$a^{0} = 1$

f (0) = 3 (1)

IdPropMult

Identity Property of Multiplication

f (0) = 3

c Using the function

f (x),

we want to evaluate for the given value,

f (1) .

To do this, we need to substitute

1

for

x

in each instance of the

x -

variable and simplify.

f (x) = 3 (2)^{x}

Substitute

$x = 1$

f (1) = 3 (2)^{1}

ExponentOne

$a^{1} = a$

f (1) = 3 (2)

Multiply

f (1) = 6

d Using the function

f (x),

we want to find the value of

x,

that gives

f (x) = 12 .

To do this, we need to substitute

12

for

f (x)

and solve for

x .

f (x) = 3 (2)^{x}

Substitute

$f (x) = 12$

12 = 3 (2)^{x}

DivEqn

$LHS / 3 = RHS / 3$

4 = 2^{x}

RearrangeEqn

Rearrange equation

2^{x} = 4

Rewrite

Rewrite $4$ as $2^{2}$

2^{x} = 2^{2}

Now, notice that

2

is the value of

x,

that satisfies this equation. It means that

x = 2

gives

f (x) = 12 .

e The given function is written in the function notation. For simplicity let's rewrite it using the

y -

variable.

f (x) = 3 (2)^{x} \Rightarrow y = 3 (2)^{x}

To determine the points where the graph crosses the

x -

and

y -

axis, we need to substitute

0

for one variable, solve, then repeat for the other variable.

Finding the $x -$ intercept

Think of the point where the graph of an equation crosses the

x -

axis. The

y -

value of that

(x, y)

coordinate pair is

0,

and the

x -

value is the

x -

intercept. To find the

x -

intercept of the equation, we should substitute

0

for

y

and solve for

x .

y = 3 (2)^{x}

Substitute

$y = 0$

0 = 3 (2)^{x}

DivEqn

$LHS / 3 = RHS / 3$

0 = 2^{x}

There is no

x

such that

2

raised to the power of

x

will have the value of

0 .

Thus, our substitution resulted in contradiction. It means that the given function does not have an

x -

intercept.

Finding the $y -$ intercept

Let's use the same concept to find the

y -

intercept. Consider the point where the graph of the equation crosses the

y -

axis. The

x

-value of the

(x, y)

coordinate pair at the

y -

intercept is

0 .

Therefore, substituting

0

for

x

will give us the

y -

intercept.

y = 3 (2)^{x}

Substitute

$x = 0$

y = 3 (2)^{0}

ExponentZero

$a^{0} = 1$

y = 3 (1)

IdPropMult

Identity Property of Multiplication

y = 3

y -

intercept of

3

means that the graph passes through the

y -

axis at the point

(0, 3) .

Subchapter links

8.2.1 Curve Fitting p.471-472

100

101

102

8.2.2 Curved Best-Fit Models p.478-480

108

109

110

111

112

113

114

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

119

120

121

122

123

124

125

Exercise 123 Page 484

Hints

Check the answer

Practice exercises

Asses your skill

Finding the $x -$ intercept

Finding the $y -$ intercept

Exercise 123 Page 484

Hints

Check the answer

Practice exercises

Asses your skill

Finding the x-intercept

Finding the y-intercept

Finding the $x -$ intercept

Finding the $y -$ intercept