Problem 47 - Precalculus : Building Concepts and Connections

Question

Sketch a graph of the rational function. Indicate any vertical and horizontal asymptote(s) and all intercepts. $$f(x)=\frac{x-1}{2 x^{2}-5 x-3}$$

Step-by-Step Solution

Answer

For the function \(f(x) = \frac{x-1}{2x^{2}-5x-3}\), the vertical asymptotes are at \(x = 1.5\) and \(x = -1\), the horizontal asymptote is at \(y = 0\), the x-intercept is at \(x=1\), and the y-intercept is at (0,-1/3).

Step 1: Find the Vertical Asymptote

The vertical asymptote of a rational function can be found by setting the denominator equal to zero and solving for x. In this case, we have \(2x^{2} - 5x - 3 = 0\). Solving this quadratic equation, we get \(x = 1.5, x = -1\).

Step 2: Determine the Horizontal Asymptote

The horizontal asymptote of a rational function can be determined by looking at the degree of the polynomial in the numerator and the denominator. If the degree of the denominator is larger than the degree of the numerator, then the horizontal asymptote is \(y = 0\). In this case, since the degree of the denominator (which is 2) is greater than the degree of the numerator (which is 1), the horizontal asymptote for this function is \(y = 0\).

Step 3: Find the x-intercept

The x-intercept of the function can be found by setting the function equal to zero and solving for x. Therefore, \(0 = \frac{x-1}{2x^{2}-5x-3}\). Multiplying both sides by \(2x^{2}-5x-3\) to eliminate the denominator, we get 0 = \(x-1\) Thus, \(x=1\) is the x-intercept.

Step 4: Identify the y-intercept

The y-intercept of the function can be found by setting x equal to zero in the function. Thus, \(f(x=0) = \frac{0 - 1}{2 \cdot 0^{2} - 5 \cdot 0 - 3} = \frac{1}/{3}\). Hence, the y-intercept is at (0, -1/3).

Step 5: Sketch the Graph

To sketch the graph, first draw the vertical and horizontal asymptotes as dashed lines on your graph. Then, plot the x intercept and y intercept on the graph. Remember that the graph will approach the asymptotes but never cross them. With these points and the asymptotes, draw the curve of the function.

Key Concepts

Vertical Asymptote

In rational functions, vertical asymptotes occur where the denominator of the function equals zero, while the numerator is non-zero. They represent values that the function cannot take, causing the graph to shoot up to positive or negative infinity. To find the vertical asymptotes for the function \(f(x)=\frac{x-1}{2x^2-5x-3}\), we set the denominator to zero: \(2x^2 - 5x - 3 = 0\). Solving this quadratic equation involves factoring or using the quadratic formula. Here, it factors to give two solutions: \(x = 1.5\) and \(x = -1\). These values are the vertical asymptotes.

When sketching the graph, vertical asymptotes are especially important as they define boundary lines that the function approaches but never actually reaches. You can draw these boundaries as dashed lines to visualize these invisible barriers on your graph.

Horizontal Asymptote

Horizontal asymptotes describe the behavior of the graph as \(x\) approaches infinity or negative infinity. They illustrate where the graph 'levels off'. For rational functions, the horizontal asymptote depends on the degrees of the numerator and the denominator polynomials. If the degree of the numerator is less than the degree of the denominator, as in this function where \(f(x)=\frac{x-1}{2x^2-5x-3}\), the horizontal asymptote is \(y = 0\).

This indicates that as \(x\) grows larger and larger, the value of \(f(x)\) will get closer and closer to zero. When graphing, this horizontal line is drawn parallel to the x-axis, showing the end behavior. Remember, the graph may cross this horizontal line, unlike a vertical asymptote, but ultimately it will tend towards this boundary.

Intercepts

Intercepts are essential points where the graph intersects the axes, providing key insights into the graph's behavior. Let's explore both the x-intercept and y-intercept for \(f(x)=\frac{x-1}{2x^2-5x-3}\).

**Finding the x-intercept:** Set the function equal to zero and solve \(\frac{x-1}{2x^2-5x-3} = 0\). Upon solving, you find that \(x=1\) is the x-intercept. This tells you that at \(x=1\), the graph crosses the x-axis.

**Finding the y-intercept:** Replace \(x\) with zero in the function and solve for \(f(0)\). Using \(f(0)=\frac{0-1}{0-0-3}\), results in \(f(0)=-\frac{1}{3}\). Thus, at the point \((0, -\frac{1}{3})\), the graph crosses the y-axis.

Both x- and y-intercepts are valuable for plotting the graph accurately, as they are easy reference points that contribute to shaping the overall curve.