Unit 7: Exponential and Logarithmic Functions - Graphing Logarithmic Functions

Overview of Unit 7: Exponential and Logarithmic Functions

The materials provided comprise a comprehensive homework assignment titled "Homework 5: Graphing Logarithmic Functions," which is part of a larger curriculum unit on Exponential and Logarithmic Functions. This specific document was developed by Gina Wilson of "All Things Algebra" in the year 2015. The assignment is structured as a two-page document designed to guide students through the process of graphing various logarithmic equations and identifying their critical algebraic and geometric characteristics. For each function presented, students are required to determine the domain, the range, the end behavior as $x$ approaches specific limits, the x-intercept of the graph, and the equation of the vertical asymptote.

Fundamental Concepts in Graphing Logarithmic Functions

When graphing logarithmic functions of the general form $f(x) = a \times \text{log}_{b}(x - h) + k$, several mathematical principles must be applied to identify the function's characteristics. The domain of a logarithmic function is restricted because the argument of the logarithm, $(x - h)$, must always be greater than zero. This leads to a domain defined by $x > h$. Unlike the domain, the range of a logarithmic function is typically all real numbers, expressed as $(-\text{∞}, \text{∞})$. Every logarithmic function of this type possesses a vertical asymptote, which is the line $x = h$. The x-intercept is discovered by setting the function value $f(x)$ to zero and solving the resulting equation for $x$. End behavior for these functions is analyzed by observing the limit of $f(x)$ as $x$ approaches the vertical asymptote from the right and as $x$ approaches positive infinity. For growth models where the base $b > 1$, the function increases; for decay models where the base $0 < b < 1$, the function decreases.

Comprehensive Analysis of Problems 1 and 2

Problem 1 introduces the base-level function $f(x) = \text{log}_{3}(x)$. In this instance, the base $b$ is $3$, and there are no horizontal or vertical shifts ($h = 0$, $k = 0$). The domain is defined as $(0, \text{∞})$ and the range is $(-\text{∞}, \text{∞})$. The vertical asymptote is located at the y-axis, represented by the equation $x = 0$. Regarding end behavior, as $x \rightarrow \text{∞}$, the function value $f(x) \rightarrow \text{∞}$. Conversely, as $x \rightarrow 0$, the function value $f(x) \rightarrow -\text{∞}$. The x-intercept is located at $(1, 0)$.

Problem 2 presents the function $f(x) = \text{log}_{\frac{1}{2}}(x) + 3$. This function utilizes a fractional base of $\frac{1}{2}$, indicating a logarithmic decay shape, and incorporates a vertical shift upward by $3$ units ($k = 3$). The domain remains $(0, \text{∞})$ because there is no horizontal shift, and the range is $(-\text{∞}, \text{∞})$. The vertical asymptote remains at $x = 0$. For the end behavior, as $x \rightarrow \text{∞}$, the function value $f(x) \rightarrow -\text{∞}$ due to the fractional base. As $x \rightarrow 0$, the function value $f(x) \rightarrow \text{∞}$. To find the x-intercept, one solves $0 = \text{log}_{\frac{1}{2}}(x) + 3$, which simplifies to $-3 = \text{log}_{\frac{1}{2}}(x)$, or $(\frac{1}{2})^{-3} = x$, resulting in an intercept at $x = 8$.

Detailed Breakdown of Horizontal and Vertical Translations (Problems 3 and 4)

Problem 3 features the function $f(x) = \text{log}_{4}(x + 5)$. Here, a horizontal shift of $5$ units to the left is introduced ($h = -5$). This shift moves the vertical asymptote to the line $x = -5$. Consequently, the domain is restricted to $(-5, \text{∞})$, and the range continues to be $(-\text{∞}, \text{∞})$. The end behavior indicates that as $x \rightarrow \text{∞}$, $f(x) \rightarrow \text{∞}$, and as $x \rightarrow -5$, $f(x) \rightarrow -\text{∞}$. The x-intercept occurs where $\text{log}_{4}(x + 5) = 0$, meaning $x + 5 = 4^{0} = 1$, which yields $x = -4$.

Problem 4 involves the function $f(x) = \text{log}_{2}(x + 8) - 4$. This function undergoes both a horizontal shift of $8$ units left ($h = -8$) and a vertical shift of $4$ units down ($k = -4$). The domain is $(-8, \text{∞})$ and the range is $(-\text{∞}, \text{∞})$. Its vertical asymptote is situated at $x = -8$. The end behavior follows the growth pattern: as $x \rightarrow \text{∞}$, $f(x) \rightarrow \text{∞}$, and as $x \rightarrow -8$, $f(x) \rightarrow -\text{∞}$. To calculate the x-intercept, solve $0 = \text{log}_{2}(x + 8) - 4$, which leads to $2^{4} = x + 8$. Since $16 = x + 8$, the x-intercept is located at $x = 8$.

Analysis of Decay Functions and Shifts (Problems 5 and 6)

Problem 5 explores $f(x) = \text{log}_{\frac{1}{3}}(x - 4)$. This function involves a base of $\frac{1}{3}$ and a horizontal shift of $4$ units to the right ($h = 4$). The vertical asymptote is shifted to $x = 4$, making the domain $(4, \text{∞})$ and the range $(-\text{∞}, \text{∞})$. Because the base is less than one, the end behavior shows that as $x \rightarrow \text{∞}$, $f(x) \rightarrow -\text{∞}$, and as $x \rightarrow 4$, $f(x) \rightarrow \text{∞}$. The x-intercept is found by setting $x - 4 = (\frac{1}{3})^{0} = 1$, resulting in $x = 5$.

Problem 6 presents the function $f(x) = \text{log}_{5}(x - 2) - 1$. The transformations include a rightward horizontal shift of $2$ units ($h = 2$) and a downward vertical shift of $1$ unit ($k = -1$). The vertical asymptote is defined by the equation $x = 2$. This results in a domain of $(2, \text{∞})$ and a range of $(-\text{∞}, \text{∞})$. The end behavior is characterized by $f(x) \rightarrow \text{∞}$ as $x \rightarrow \text{∞}$ and $f(x) \rightarrow -\text{∞}$ as $x \rightarrow 2$. The x-intercept is calculated via the equation $0 = \text{log}_{5}(x - 2) - 1$, which gives $5^{1} = x - 2$, meaning the intercept is at $x = 7$.

Complex Transformations in Problem 7 and Assignment Directions

Problem 7 requires the student to graph and identify the characteristics for $f(x) = \text{log}_{\frac{1}{4}}(x - 1) + 1$. This function combines a fractional base of $\frac{1}{4}$, a horizontal shift of $1$ unit to the right ($h = 1$), and an upward vertical shift of $1$ unit ($k = 1$). The vertical asymptote is identified as $x = 1$. The domain is subsequently $(1, \text{∞})$ while the range remains $(-\text{∞}, \text{∞})$. The end behavior for this logarithmic decay function is as follows: as $x \rightarrow \text{∞}$, $f(x) \rightarrow -\text{∞}$, and as $x \rightarrow 1$, $f(x) \rightarrow \text{∞}$. Determining the x-intercept involves solving $0 = \text{log}_{\frac{1}{4}}(x - 1) + 1$, resulting in $-1 = \text{log}_{\frac{1}{4}}(x - 1)$. This equates to $(\frac{1}{4})^{-1} = x - 1$, which is $4 = x - 1$, placing the x-intercept at $x = 5$.

The overall directions provided in the transcript emphasize that the student must graph each of these seven functions and explicitly list all identifying characteristics. The worksheet concludes with the footer information identifying Gina Wilson and the "All Things Algebra" copyright for 2015, used as study material for Unit 7 mathematical standards.