This worksheet helps students practice graphing exponential functions and identifying their key characteristics.

Let's solve each problem step by step, analyzing the exponential functions and their characteristics.

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Problem 1: $ f(x) = 2(3)^x $

We are given the function and a graph. Let's analyze it.

#### Step-by-step Analysis:

- Growth or Decay?
The base is $ 3 > 1 $, so this is an exponential growth function.

- Asymptote:
Exponential functions of the form $ f(x) = a(b)^x $ have a horizontal asymptote at $ y = 0 $ when there’s no vertical shift. Here, there is no vertical shift added, so the asymptote is at $ \boxed{y = 0} $.

- Domain:
All real numbers: $ \boxed{(-\infty, \infty)} $

- Range:
Since the function grows exponentially and approaches 0 but never reaches it (from above), the range is $ \boxed{(0, \infty)} $

- Y-intercept:
Plug in $ x = 0 $:
$ f(0) = 2(3)^0 = 2(1) = 2 $
So, y-intercept is $ \boxed{(0, 2)} $

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✔ Answer Summary for Problem 1:
- Growth
- Asymptote at $ y = 0 $
- Domain: $ (-\infty, \infty) $
- Range: $ (0, \infty) $
- Y-intercept: $ (0, 2) $

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Problem 2: $ f(x) = 3\left(\frac{1}{2}\right)^x + 4 $

This is an exponential function with a base between 0 and 1 → decay.

#### Step-by-step Analysis:

- Growth or Decay?
Base $ \frac{1}{2} < 1 $ → Exponential decay

- Asymptote:
The function has a vertical shift of +4, so the horizontal asymptote is at $ \boxed{y = 4} $

- Domain:
All real numbers: $ \boxed{(-\infty, \infty)} $

- Range:
Since $ \left(\frac{1}{2}\right)^x > 0 $, multiplying by 3 gives positive values, then adding 4 → $ f(x) > 4 $. So, range is $ \boxed{(4, \infty)} $

- Y-intercept:
Plug in $ x = 0 $:
$ f(0) = 3\left(\frac{1}{2}\right)^0 + 4 = 3(1) + 4 = 7 $
So, y-intercept is $ \boxed{(0, 7)} $

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✔ Answer Summary for Problem 2:
- Decay
- Asymptote at $ y = 4 $
- Domain: $ (-\infty, \infty) $
- Range: $ (4, \infty) $
- Y-intercept: $ (0, 7) $

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Problem 3: $ f(x) = 4^{x-1} - 3 $

We need to:
1. Fill in the table
2. Graph the function
3. Find characteristics

#### Step 1: Evaluate $ f(x) $ for given x-values

| $ x $ | $ f(x) = 4^{x-1} - 3 $ |
|--------|--------------------------|
| -2 | $ 4^{-3} - 3 = \frac{1}{64} - 3 = -2.984375 $ ≈ $ -2.98 $ |
| -1 | $ 4^{-2} - 3 = \frac{1}{16} - 3 = -2.9375 $ ≈ $ -2.94 $ |
| 0 | $ 4^{-1} - 3 = \frac{1}{4} - 3 = -2.75 $ |
| 1 | $ 4^{0} - 3 = 1 - 3 = -2 $ |
| 2 | $ 4^{1} - 3 = 4 - 3 = 1 $ |
| 3 | $ 4^{2} - 3 = 16 - 3 = 13 $ |

So the table becomes:

| $ x $ | $ f(x) $ |
|--------|-----------|
| -2 | -2.98 |
| -1 | -2.94 |
| 0 | -2.75 |
| 1 | -2 |
| 2 | 1 |
| 3 | 13 |

#### Step 2: Characteristics

- Growth or Decay?
Base is $ 4 > 1 $ → Exponential growth

- Asymptote:
As $ x \to -\infty $, $ 4^{x-1} \to 0 $, so $ f(x) \to -3 $
→ Horizontal asymptote at $ \boxed{y = -3} $

- Domain: $ (-\infty, \infty) $

- Range: Since $ 4^{x-1} > 0 $, $ f(x) > -3 $ → $ \boxed{(-3, \infty)} $

- Y-intercept:
At $ x = 0 $, $ f(0) = 4^{-1} - 3 = \frac{1}{4} - 3 = -2.75 $
→ $ \boxed{(0, -2.75)} $

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✔ Answer Summary for Problem 3:
- Growth
- Asymptote at $ y = -3 $
- Domain: $ (-\infty, \infty) $
- Range: $ (-3, \infty) $
- Y-intercept: $ (0, -2.75) $

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Problem 4: $ f(x) = 2\left(\frac{1}{2}\right)^x + 3 $

Again, we’ll:
1. Complete the table
2. Graph
3. Analyze

#### Step 1: Evaluate $ f(x) $

| $ x $ | $ f(x) = 2\left(\frac{1}{2}\right)^x + 3 $ |
|--------|---------------------------------------------|
| -2 | $ 2\left(\frac{1}{2}\right)^{-2} + 3 = 2(4) + 3 = 8 + 3 = 11 $ |
| -1 | $ 2\left(\frac{1}{2}\right)^{-1} + 3 = 2(2) + 3 = 4 + 3 = 7 $ |
| 0 | $ 2(1) + 3 = 2 + 3 = 5 $ |
| 1 | $ 2\left(\frac{1}{2}\right) + 3 = 1 + 3 = 4 $ |
| 2 | $ 2\left(\frac{1}{4}\right) + 3 = 0.5 + 3 = 3.5 $ |
| 3 | $ 2\left(\frac{1}{8}\right) + 3 = 0.25 + 3 = 3.25 $ |

Table:

| $ x $ | $ f(x) $ |
|--------|-----------|
| -2 | 11 |
| -1 | 7 |
| 0 | 5 |
| 1 | 4 |
| 2 | 3.5 |
| 3 | 3.25 |

#### Step 2: Characteristics

- Growth or Decay?
Base $ \frac{1}{2} < 1 $ → Exponential decay

- Asymptote:
As $ x \to \infty $, $ \left(\frac{1}{2}\right)^x \to 0 $, so $ f(x) \to 3 $
→ Horizontal asymptote at $ \boxed{y = 3} $

- Domain: $ (-\infty, \infty) $

- Range: Since $ \left(\frac{1}{2}\right)^x > 0 $, $ f(x) > 3 $ → $ \boxed{(3, \infty)} $

- Y-intercept:
$ f(0) = 2(1) + 3 = 5 $ → $ \boxed{(0, 5)} $

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✔ Answer Summary for Problem 4:
- Decay
- Asymptote at $ y = 3 $
- Domain: $ (-\infty, \infty) $
- Range: $ (3, \infty) $
- Y-intercept: $ (0, 5) $

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✔ Final Answers Summary:

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Problem 1:
- Growth
- Asymptote: $ y = 0 $
- Domain: $ (-\infty, \infty) $
- Range: $ (0, \infty) $
- Y-intercept: $ (0, 2) $

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Problem 2:
- Decay
- Asymptote: $ y = 4 $
- Domain: $ (-\infty, \infty) $
- Range: $ (4, \infty) $
- Y-intercept: $ (0, 7) $

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Problem 3:
- Growth
- Asymptote: $ y = -3 $
- Domain: $ (-\infty, \infty) $
- Range: $ (-3, \infty) $
- Y-intercept: $ (0, -2.75) $
- Table:
$$
\begin{array}{c|c}
x & f(x) \\
\hline
-2 & -2.98 \\
-1 & -2.94 \\
0 & -2.75 \\
1 & -2 \\
2 & 1 \\
3 & 13 \\
\end{array}
$$

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Problem 4:
- Decay
- Asymptote: $ y = 3 $
- Domain: $ (-\infty, \infty) $
- Range: $ (3, \infty) $
- Y-intercept: $ (0, 5) $
- Table:
$$
\begin{array}{c|c}
x & f(x) \\
\hline
-2 & 11 \\
-1 & 7 \\
0 & 5 \\
1 & 4 \\
2 & 3.5 \\
3 & 3.25 \\
\end{array}
$$

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Let me know if you'd like help graphing these!