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Chapter 4: Problem 32

Graph the function. $$y=\ln (3 x+5)$$

Short Answer

Expert verified
The graph of the function \(y = \ln(3x + 5)\) starts from the vertical asymptote at \(x = -5/3\) and increases without bound as \(x\) increases. It intercepts the y-axis at \(y = \ln(5)\). There are no x-intercepts. The domain of the function is \(-5/3 < x < \infty\) and the range is \(-\infty < y < \infty\).

Step by step solution

01

Identify Domain

Before graphing, it's important to understand the domain of the function. The natural log function, \(ln(x)\), is defined for \(x>0\). Given the function \(y = \ln(3x + 5)\), the argument of the ln function is \(3x+5\), and it should be greater than 0, i.e, \(3x+5 > 0\). Solving for \(x\) gives \(x > -5/3\). Hence, the domain of \(y = \ln(3x + 5)\) is \(-5/3 < x < \infty\).
02

Identify Intercepts

When \(x = -5/3\), y becomes undefined as there is a vertical asymptote at \(x = -5/3\). Hence, there are no x-intercepts. The y-intercept occurs when \(x = 0\), and we substitute \(x = 0\) into the function we get: \(y = \ln(3(0) + 5) = \ln(5)\). Hence, the y-intercept is \(y = \ln(5)\).
03

Identify Range

The range of a logarithmic function is the set of all real numbers. So the range of \(y = \ln(3x + 5)\) is \(-\infty < y < \infty\).
04

Sketch the Graph

Drawing on the above information, the graph can be sketched. The graph will start from the vertical asymptote at \(x = -5/3\) and increase without bound as \(x\) increases. It intercepts the y-axis at \(y = \ln(5)\) and there are no x-intercepts.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Domain of a function
To understand any function, figuring out its domain is essential. The **domain** of a function is the set of all possible input values (x-values) that make the function work. For the natural logarithmic function, written as \(y = \ln(3x + 5)\), the expression inside the parentheses, called the **argument**, must always be positive. This makes sense because the logarithm of a number is only defined for positive values.

For our function \(3x + 5 > 0\). If we solve this inequality:
  • Subtract 5 from both sides to get \(3x > -5\).
  • Divide by 3, leading to \(x > -\frac{5}{3}\).
Therefore, the domain of the function is \(-\frac{5}{3} < x < \infty\). This tells us that the function only works for x-values greater than \(-\frac{5}{3}\).
Vertical asymptote
A vital feature of many functions, especially logarithmic ones, is the **vertical asymptote**. This is a vertical line that the graph of the function approaches but never actually touches or crosses. For the function \(y = \ln(3x + 5)\), there is a vertical asymptote at \(x = -\frac{5}{3}\).

This asymptote occurs because at \(x = -\frac{5}{3}\), the argument of the natural log becomes zero \((3x + 5 = 0)\). The natural logarithm of zero is undefined, so the graph cannot exist at this point, creating a vertical barrier. As x approaches \(-\frac{5}{3}\) from the right, the function value decreases rapidly towards negative infinity.
Natural logarithm
The **natural logarithm**, denoted as \(\ln(x)\), is a special mathematical function. It represents the logarithm to the base of \(e\), where \(e\approx 2.718\). Natural logs are used frequently in mathematics because they have unique properties that make complex operations simpler.

For the function \(y = \ln(3x + 5)\), the presence of \(3x + 5\) shifts and stretches the basic \(\ln(x)\) graph. When graphed, larger x-values make \(3x + 5\) significantly larger, increasing y, whereas as x nears the vertical asymptote, y plummets. This stretching feature influences how steep the graph is.
Intercepts in graphing
When graphing, it鈥檚 helpful to locate intercepts. Intercepts are points where the graph crosses the axes. For \(y = \ln(3x + 5)\):
  • **X-Intercept:** This is the point where the graph crosses the x-axis \((y = 0)\). Since an ln function is undefined when the argument is zero \(3x+5 = 0\), meaning it approaches but never touches, there鈥檚 no x-intercept.
  • **Y-Intercept:** Set \(x = 0\) to find the y-intercept. Substitute 0 in the function: \(y = \ln(3(0) + 5) = \ln(5)\). Therefore, the y-intercept is \(y = \ln(5)\). This means the graph crosses the y-axis at this point, providing a fixed reference point for the graph.
Understanding intercepts helps in constructing and interpreting graphs by indicating fixed points on the axes.

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