Exponential Functions Notes and Worksheets - Lindsay Bowden - Free Printable

Let’s solve each problem step by step.

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Problem 2: Use the graph to find characteristics of f(x) = (1/2)^x + 4

We are given a graph and the function:
f(x) = (1/2)^x + 4

This is an exponential decay function because the base (1/2) is between 0 and 1. The “+4” shifts the whole graph up by 4 units.

Let’s find each characteristic:

- Growth or Decay?
Since the base is 1/2 (< 1), it’s decay.

- Asymptote at:
The horizontal asymptote for y = a·b^x + k is y = k. Here, k = 4 → so asymptote is y = 4

- Domain:
Exponential functions are defined for all real numbers → all real numbers or (-∞, ∞)

- Range:
Because it decays toward y=4 but never reaches it, and goes up to infinity as x → -∞, range is y > 4 or (4, ∞)

- Y-intercept:
Plug in x = 0:
f(0) = (1/2)^0 + 4 = 1 + 4 = 5 → point (0, 5)

✔ So for Problem 2:
- growth or decay? → decay
- asymptote at → y = 4
- domain → all real numbers
- range → y > 4
- y-intercept → (0, 5)

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Problem 3: Use key table to graph f(x) = 4^(x−1) − 3

First, let’s fill in the table using the function f(x) = 4^(x−1) − 3

Compute for each x:

- x = -2:
f(-2) = 4^(-2−1) − 3 = 4^(-3) − 3 = 1/(4^3) − 3 = 1/64 − 3 ≈ -2.984 → we can write as fraction: -191/64 or decimal ≈ -2.98

But maybe better to keep exact fractions unless told otherwise.

Actually, let’s compute exactly:

x = -2: 4^(-3) = 1/64 → 1/64 - 3 = 1/64 - 192/64 = -191/64

x = -1: 4^(-2) = 1/16 → 1/16 - 3 = 1/16 - 48/16 = -47/16

x = 0: 4^(-1) = 1/4 → 1/4 - 3 = 1/4 - 12/4 = -11/4

x = 1: 4^(0) = 1 → 1 - 3 = -2

x = 2: 4^(1) = 4 → 4 - 3 = 1

x = 3: 4^(2) = 16 → 16 - 3 = 13

So table:

| x | f(x) |
|-----|-------------|
| -2 | -191/64 |
| -1 | -47/16 |
| 0 | -11/4 |
| 1 | -2 |
| 2 | 1 |
| 3 | 13 |

Now, characteristics:

Function: f(x) = 4^(x−1) − 3

Base = 4 (>1) → this is growth

Asymptote: Horizontal asymptote is y = -3 (because of the -3 shift)

Domain: All real numbers → (-∞, ∞)

Range: Since it grows from near y=-3 upward → y > -3

Y-intercept: We already computed x=0 → f(0) = -11/4 → point (0, -11/4)

✔ So for Problem 3:
- growth or decay? → growth
- asymptote at → y = -3
- domain → all real numbers
- range → y > -3
- y-intercept → (0, -11/4)

*(Note: You’d plot the points from the table on the grid to sketch the curve — starts low on left, rises sharply right, approaches y=-3 on left side.)*

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Problem 4: Use key table to graph f(x) = 2·(2/3)^x + 3

Fill in table:

f(x) = 2·(2/3)^x + 3

x = -2:
(2/3)^(-2) = (3/2)^2 = 9/4 → 2*(9/4) + 3 = 18/4 + 3 = 9/2 + 3 = 4.5 + 3 = 7.5 or 15/2

x = -1:
(2/3)^(-1) = 3/2 → 2*(3/2) + 3 = 3 + 3 = 6

x = 0:
(2/3)^0 = 1 → 2*1 + 3 = 5

x = 1:
(2/3)^1 = 2/3 → 2*(2/3) + 3 = 4/3 + 3 = 4/3 + 9/3 = 13/3 ≈ 4.333

x = 2:
(2/3)^2 = 4/9 → 2*(4/9) + 3 = 8/9 + 27/9 = 35/9 ≈ 3.888...

Table:

| x | f(x) |
|-----|---------|
| -2 | 15/2 |
| -1 | 6 |
| 0 | 5 |
| 1 | 13/3 |
| 2 | 35/9 |

Characteristics:

Function: f(x) = 2·(2/3)^x + 3

Base = 2/3 (<1) → decay

Asymptote: y = 3 (from the +3)

Domain: all real numbers

Range: since it decays toward y=3 from above → y > 3

Y-intercept: x=0 → f(0)=5 → (0,5)

✔ So for Problem 4:
- growth or decay? → decay
- asymptote at → y = 3
- domain → all real numbers
- range → y > 3
- y-intercept → (0, 5)

*(Graph: Plot points — high on left, decreases toward y=3 on right.)*

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Final Answer:

Problem 2:
- growth or decay? → decay
- asymptote at → y = 4
- domain → all real numbers
- range → y > 4
- y-intercept → (0, 5)

Problem 3 Table:
x | f(x)
--|------
-2 | -191/64
-1 | -47/16
0 | -11/4
1 | -2
2 | 1
3 | 13

Characteristics:
- growth or decay? → growth
- asymptote at → y = -3
- domain → all real numbers
- range → y > -3
- y-intercept → (0, -11/4)

Problem 4 Table:
x | f(x)
--|------
-2 | 15/2
-1 | 6
0 | 5
1 | 13/3
2 | 35/9

Characteristics:
- growth or decay? → decay
- asymptote at → y = 3
- domain → all real numbers
- range → y > 3
- y-intercept → (0, 5)