Problem 13 - Algebra and Trigonometry with Analytic Geometry

Question

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

Step-by-Step Solution

Answer

The graph has a hole at (2, 1), a vertical asymptote at \(x=-\frac{3}{2}\), horizontal asymptote at \(y=2\), and intercepts at \((\frac{1}{4}, 0)\) and \((0, -\frac{1}{3})\).

Step 1: Identify Domain and Simplify

First, determine the domain of the function by identifying where the denominator is zero. For \(f(x)\), the denominator is \((2x+3)(x-2)\). This will be zero when \(x=-\frac{3}{2}\) or \(x=2\). The function is undefined at these points. The \((x-2)\) in both the numerator and denominator cancels out, simplifying \(f(x)\) to \(\frac{4x-1}{2x+3}\) for \(x eq 2\).

Step 2: Determine Asymptotes

Identify vertical and horizontal asymptotes. The vertical asymptote is at \(x=-\frac{3}{2}\) because the denominator \(2x+3=0\) at this point and does not cancel. The simplified function, \(\frac{4x-1}{2x+3}\), approaches a horizontal asymptote of \(y=2\) as \(x \to \infty\) because the degrees of the polynomial in the numerator and denominator are equal. The leading coefficients ratio is \(\frac{4}{2}=2\).

Step 3: Find Intercepts

To find the y-intercept, set \(x=0\). The function becomes \(\frac{4(0)-1}{2(0)+3}=-\frac{1}{3}\). The y-intercept is thus \( (0, -\frac{1}{3}) \). For the x-intercept, set the simplified function's numerator equal to zero: \(4x-1=0\), yielding \(x=\frac{1}{4}\). Thus, the x-intercept is \(\left(\frac{1}{4}, 0\right)\).

Step 4: Address Hole in the Graph

Since \(x=2\) canceled during simplification, \(f(x)\) has a hole there, not a vertical asymptote. At \(x=2\), substitute back into \(\frac{4x-1}{2x+3}\) to find the y-value of the hole: \(\frac{4(2)-1}{2(2)+3} = \frac{7}{7} = 1\). The hole is located at the point \((2, 1)\).

Step 5: Sketch the Graph

Start by plotting the x and y-intercepts at \(\left(\frac{1}{4}, 0\right)\) and \( (0, -\frac{1}{3}) \). Plot the vertical asymptote at \(x = -\frac{3}{2}\) and the horizontal asymptote at \(y=2\). Mark the hole at \((2, 1)\). The curve should approach these asymptotes, pass through the intercepts, and reflect the existence of the hole.

Key Concepts

Domain of a Function

The domain of a function refers to all possible input values (x-values) for which the function is defined. For rational functions like \(f(x) = \frac{(4x-1)(x-2)}{(2x+3)(x-2)}\), we must be cautious of any values of \(x\) that make the denominator zero as the function would be undefined at those points.

In this exercise, the denominator is \((2x+3)(x-2)\). Setting each factor to zero gives possible discontinuities: \(x = -\frac{3}{2}\) and \(x = 2\). Therefore, the domain of this function consists of all real numbers except \(x = -\frac{3}{2}\) and \(x = 2\).

It's important to note that any overlapping factors between the numerator and denominator, such as \((x-2)\), can be canceled out. After the cancellation, these provide more insight into the behavior of the graph at those points, which we'll discuss later.

Asymptotes

Asymptotes are lines that the graph of a function approaches but never actually touches. In rational functions, there are typically vertical and horizontal asymptotes, though slant asymptotes can also occur if certain conditions are met.

For the function \(f(x) = \frac{(4x-1)}{(2x+3)}\), after canceling out the \((x-2)\) term, we need to find the asymptotes to understand its behavior.

• Vertical Asymptotes: These occur where the denominator equals zero and the factor does not cancel with any part of the numerator. Here, the function has a vertical asymptote at \(x = -\frac{3}{2}\) because \(2x + 3 = 0\) at this point.



• Horizontal Asymptotes: These are found by comparing the degrees of the polynomial in the numerator and denominator. If they are the same, as in this case, the horizontal asymptote is the ratio of the leading coefficients. Hence, we have a horizontal asymptote at \(y = 2\), derived from \(\frac{4}{2} = 2\).

Understanding these asymptotes helps us sketch the rational function by guiding the end behavior of the graph.

Intercepts

Intercepts are specific points where the graph crosses the axes. Identifying intercepts provides fixed points that help to accurately plot the graph.

**Y-intercept**:
The y-intercept is where the curve hits the y-axis, which occurs when \(x = 0\). Substituting \(x = 0\) into our function \(f(x) = \frac{4x-1}{2x+3}\) results in \(\frac{4(0)-1}{2(0)+3} = -\frac{1}{3}\). Thus, the y-intercept is at the point \((0, -\frac{1}{3})\).

**X-intercept**:
The x-intercept occurs where the function crosses the x-axis, meaning \(f(x) = 0\). This requires setting the numerator equal to zero while keeping the denominator non-zero: \(4x - 1 = 0\) leads to \(x = \frac{1}{4}\). Thus, the x-intercept is at \(\left(\frac{1}{4}, 0\right)\).

Plotting these intercepts forms the basis to sketching the curve, acting like anchor points that the curve passes through, satisfying the initial conditions of the function.

Holes in Graphs

In some rational functions, certain points make the expression undefined due to a zero in the denominator that also cancels with a zero in the numerator. These points are known as `holes` in the graph.

In our function \(f(x) = \frac{(4x-1)(x-2)}{(2x+3)(x-2)}\), the factor \((x-2)\) cancels out, but it still implies the presence of a hole at \(x = 2\).

Even though \(x=2\) is no longer an asymptote because of this cancellation, it is crucial to recognize it as a hole. To find the y-value where this hole occurs, substitute \(x = 2\) back into the simplified form of the function \(f(x) = \frac{4x-1}{2x+3}\):
The calculation \(\frac{4(2)-1}{2(2)+3} = 1\) shows that the hole is actually at \((2, 1)\).

Recognizing and plotting holes ensures that we fully capture the behavior of the function, providing a more accurate and comprehensive graph.