Problem 11 - Precalculus: Functions and Graphs

Question

Sketch the graph of \(f\) $$f(x)=\frac{4 x-1}{2 x+3}$$

Step-by-Step Solution

Answer

Vertical asymptote at \(x = -\frac{3}{2}\), horizontal asymptote \(y = 2\); intercepts are \((0, -\frac{1}{3})\) and \(\left(\frac{1}{4}, 0\right)\).

Step 1: Identify the Function Type

The given function \( f(x) = \frac{4x - 1}{2x + 3} \) is a rational function. A rational function is a ratio of two polynomials.

Step 2: Find the Vertical Asymptote

A vertical asymptote occurs where the denominator is zero (and the numerator is not zero). Setting \(2x + 3 = 0\) and solving for \(x\), we get \(x = -\frac{3}{2}\). So, there is a vertical asymptote at \(x = -\frac{3}{2}\).

Step 3: Find the Horizontal Asymptote

Rational functions have horizontal asymptotes determined by the degrees of the polynomials in the numerator and denominator. Since both the numerator \((4x - 1)\) and denominator \((2x + 3)\) are of degree 1, the horizontal asymptote is \(y = \frac{4}{2} = 2\).

Step 4: Find the Intercepts

To find the y-intercept, set \(x = 0\): \(f(0) = \frac{4(0) - 1}{2(0) + 3} = -\frac{1}{3}\). Thus, the y-intercept is \((0, -\frac{1}{3})\). To find the x-intercept, set \(f(x) = 0\). Solve \(4x - 1 = 0\) which gives \(x = \frac{1}{4}\). So, the x-intercept is \((\frac{1}{4}, 0)\).

Step 5: Sketch the Graph

Plot the vertical asymptote as a dashed line at \(x = -\frac{3}{2}\) and the horizontal asymptote at \(y = 2\). Mark the intercepts \((0, -\frac{1}{3})\) and \((\frac{1}{4}, 0)\) on the graph. The curve approaches the vertical asymptote going to positive or negative infinity and approaches the horizontal asymptote as \(x\) goes to infinity or negative infinity.

Key Concepts

Understanding Vertical Asymptotes in Rational Functions

When dealing with rational functions, understanding vertical asymptotes is crucial. A vertical asymptote is a vertical line on a graph where the function becomes unbounded, meaning it moves towards infinity or negative infinity.
To find a vertical asymptote in a rational function, you need to identify where the denominator equals zero, while the numerator is not zero. This is because division by zero is undefined, thus creating a break in the graph.
Let's look at the function given by: \[f(x) = \frac{4x - 1}{2x + 3}\] For this function, we find the vertical asymptote by setting the denominator to zero: \[2x + 3 = 0\] Solving for \(x\), we find that \(x = -\frac{3}{2}\). This indicates a vertical asymptote at \(x = -\frac{3}{2}\), represented by a dashed vertical line on the graph. The graph of the function will approach this line but never actually touch or cross it.

Exploring Horizontal Asymptotes

Horizontal asymptotes show how the function behaves as the input \(x\) becomes very large, either positively or negatively. For rational functions, horizontal asymptotes depend on the degrees of the polynomials in the numerator and denominator.
To determine the horizontal asymptote, compare the highest degree terms from each polynomial.For the given function:\[f(x) = \frac{4x - 1}{2x + 3}\]both the numerator \((4x - 1)\) and the denominator \((2x + 3)\) are linear polynomials (each of degree 1). In such cases, the horizontal asymptote is given by the ratio of the leading coefficients, leading to: \[y = \frac{4}{2} = 2\] This implies that as \(x\) approaches infinity or negative infinity, the value of \(f(x)\) approaches 2. On the graph, this is shown as a dashed horizontal line at \(y = 2\), indicating that the function will get closer and closer to this value.

Calculating Intercepts of Rational Functions

Intercepts are the points where a graph crosses the axes, and they play a significant role in defining the overall shape of the graph. For any given function, you can find both x- and y-intercepts.
**Finding the y-intercept**
To find the y-intercept, set \(x = 0\) and solve for \(f(x)\). In our function: \[f(0) = \frac{4(0) - 1}{2(0) + 3} = -\frac{1}{3}\] Thus, the y-intercept is \((0, -\frac{1}{3})\), indicating the point where the graph crosses the y-axis.**Finding the x-intercept**
To calculate the x-intercept, set the entire function equal to zero: \[f(x) = \frac{4x - 1}{2x + 3} = 0\] This condition equals identifying where the numerator is zero. Solving \(4x - 1 = 0\) gives us:\[4x = 1\] \[x = \frac{1}{4}\] Therefore, the x-intercept is \((\frac{1}{4}, 0)\), showing where the graph crosses the x-axis.

Together, these intercepts help map the initial path of the graph before it is influenced by the asymptotes.