Exponential Functions Notes and Worksheets - Lindsay Bowden - Free Printable

Problem Analysis and Solution

The worksheet involves analyzing exponential functions and their graphs. Let's solve each part step by step.

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#### Problem 1:
Function: \( f(x) = 2(3)^x \)

- Graph Characteristics:
- The graph is provided, showing an exponential growth curve.

1. Growth or Decay?
- Since the base \( 3 > 1 \), the function represents exponential growth.

2. Asymptote at:
- Exponential functions of the form \( a(b)^x \) have a horizontal asymptote at \( y = 0 \) (the x-axis).
- Asymptote: \( y = 0 \).

3. Domain:
- The domain of an exponential function is all real numbers.
- Domain: \( (-\infty, \infty) \).

4. Range:
- The range of \( f(x) = 2(3)^x \) is all positive values because \( 2(3)^x > 0 \) for all \( x \).
- Range: \( (0, \infty) \).

5. y-intercept:
- The y-intercept occurs when \( x = 0 \):
\[
f(0) = 2(3)^0 = 2(1) = 2
\]
- y-intercept: \( (0, 2) \).

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

- Graph Characteristics:
- The graph is provided, showing an exponential decay curve shifted upward.

1. Growth or Decay?
- Since the base \( \frac{1}{2} < 1 \), the function represents exponential decay.

2. Asymptote at:
- The function is of the form \( a(b)^x + c \), where \( c = 4 \). The horizontal asymptote is at \( y = c \).
- Asymptote: \( y = 4 \).

3. Domain:
- The domain of an exponential function is all real numbers.
- Domain: \( (-\infty, \infty) \).

4. Range:
- The range of \( f(x) = 3\left(\frac{1}{2}\right)^x + 4 \) is all values greater than 4 because \( 3\left(\frac{1}{2}\right)^x > 0 \) for all \( x \), and adding 4 shifts the range up.
- Range: \( (4, \infty) \).

5. y-intercept:
- The y-intercept occurs when \( x = 0 \):
\[
f(0) = 3\left(\frac{1}{2}\right)^0 + 4 = 3(1) + 4 = 7
\]
- y-intercept: \( (0, 7) \).

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#### Problem 3:
Function: \( f(x) = 4^{x-1} - 3 \)

1. Complete the x-y table:
- Calculate \( f(x) \) for each given \( x \)-value:
\[
f(x) = 4^{x-1} - 3
\]
- For \( x = -2 \):
\[
f(-2) = 4^{-2-1} - 3 = 4^{-3} - 3 = \frac{1}{4^3} - 3 = \frac{1}{64} - 3 = -\frac{191}{64}
\]
- For \( x = -1 \):
\[
f(-1) = 4^{-1-1} - 3 = 4^{-2} - 3 = \frac{1}{4^2} - 3 = \frac{1}{16} - 3 = -\frac{47}{16}
\]
- For \( x = 0 \):
\[
f(0) = 4^{0-1} - 3 = 4^{-1} - 3 = \frac{1}{4} - 3 = -\frac{11}{4}
\]
- For \( x = 1 \):
\[
f(1) = 4^{1-1} - 3 = 4^0 - 3 = 1 - 3 = -2
\]
- For \( x = 2 \):
\[
f(2) = 4^{2-1} - 3 = 4^1 - 3 = 4 - 3 = 1
\]
- For \( x = 3 \):
\[
f(3) = 4^{3-1} - 3 = 4^2 - 3 = 16 - 3 = 13
\]

- Completed Table:
\[
\begin{array}{|c|c|}
\hline
x & f(x) \\
\hline
-2 & -\frac{191}{64} \\
-1 & -\frac{47}{16} \\
0 & -\frac{11}{4} \\
1 & -2 \\
2 & 1 \\
3 & 13 \\
\hline
\end{array}
\]

2. Graph the function:
- Plot the points from the table and sketch the curve.

3. Characteristics:
- Growth or Decay?
- Since the base \( 4 > 1 \), the function represents exponential growth.
- Asymptote at:
- The function is of the form \( a(b)^{x-h} + k \), where \( k = -3 \). The horizontal asymptote is at \( y = k \).
- Asymptote: \( y = -3 \).
- Domain:
- The domain of an exponential function is all real numbers.
- Domain: \( (-\infty, \infty) \).
- Range:
- The range of \( f(x) = 4^{x-1} - 3 \) is all values greater than \(-3\) because \( 4^{x-1} > 0 \) for all \( x \), and subtracting 3 shifts the range down.
- Range: \( (-3, \infty) \).
- y-intercept:
- The y-intercept occurs when \( x = 0 \):
\[
f(0) = 4^{0-1} - 3 = 4^{-1} - 3 = \frac{1}{4} - 3 = -\frac{11}{4}
\]
- y-intercept: \( (0, -\frac{11}{4}) \).

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

1. Complete the x-y table:
- Calculate \( f(x) \) for each given \( x \)-value:
\[
f(x) = 2\left(\frac{1}{2}\right)^x + 3
\]
- For \( x = -2 \):
\[
f(-2) = 2\left(\frac{1}{2}\right)^{-2} + 3 = 2(4) + 3 = 8 + 3 = 11
\]
- For \( x = -1 \):
\[
f(-1) = 2\left(\frac{1}{2}\right)^{-1} + 3 = 2(2) + 3 = 4 + 3 = 7
\]
- For \( x = 0 \):
\[
f(0) = 2\left(\frac{1}{2}\right)^0 + 3 = 2(1) + 3 = 2 + 3 = 5
\]
- For \( x = 1 \):
\[
f(1) = 2\left(\frac{1}{2}\right)^1 + 3 = 2\left(\frac{1}{2}\right) + 3 = 1 + 3 = 4
\]
- For \( x = 2 \):
\[
f(2) = 2\left(\frac{1}{2}\right)^2 + 3 = 2\left(\frac{1}{4}\right) + 3 = \frac{1}{2} + 3 = \frac{7}{2}
\]
- For \( x = 3 \):
\[
f(3) = 2\left(\frac{1}{2}\right)^3 + 3 = 2\left(\frac{1}{8}\right) + 3 = \frac{1}{4} + 3 = \frac{13}{4}
\]

- Completed Table:
\[
\begin{array}{|c|c|}
\hline
x & f(x) \\
\hline
-2 & 11 \\
-1 & 7 \\
0 & 5 \\
1 & 4 \\
2 & \frac{7}{2} \\
3 & \frac{13}{4} \\
\hline
\end{array}
\]

2. Graph the function:
- Plot the points from the table and sketch the curve.

3. Characteristics:
- Growth or Decay?
- Since the base \( \frac{1}{2} < 1 \), the function represents exponential decay.
- Asymptote at:
- The function is of the form \( a(b)^x + c \), where \( c = 3 \). The horizontal asymptote is at \( y = c \).
- Asymptote: \( y = 3 \).
- Domain:
- The domain of an exponential function is all real numbers.
- Domain: \( (-\infty, \infty) \).
- Range:
- The range of \( f(x) = 2\left(\frac{1}{2}\right)^x + 3 \) is all values greater than 3 because \( 2\left(\frac{1}{2}\right)^x > 0 \) for all \( x \), and adding 3 shifts the range up.
- Range: \( (3, \infty) \).
- y-intercept:
- The y-intercept occurs when \( x = 0 \):
\[
f(0) = 2\left(\frac{1}{2}\right)^0 + 3 = 2(1) + 3 = 2 + 3 = 5
\]
- y-intercept: \( (0, 5) \).

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Final Answers:

1.
- Growth or decay: growth
- Asymptote: \( y = 0 \)
- Domain: \( (-\infty, \infty) \)
- Range: \( (0, \infty) \)
- y-intercept: \( (0, 2) \)

2.
- Growth or decay: decay
- Asymptote: \( y = 4 \)
- Domain: \( (-\infty, \infty) \)
- Range: \( (4, \infty) \)
- y-intercept: \( (0, 7) \)

3.
- Growth or decay: growth
- Asymptote: \( y = -3 \)
- Domain: \( (-\infty, \infty) \)
- Range: \( (-3, \infty) \)
- y-intercept: \( (0, -\frac{11}{4}) \)

4.
- Growth or decay: decay
- Asymptote: \( y = 3 \)
- Domain: \( (-\infty, \infty) \)
- Range: \( (3, \infty) \)
- y-intercept: \( (0, 5) \)

\boxed{\text{See detailed solutions above}}