Problem 43 - College Algebra and Calculus: An Applied Approach

Question

Sketch the graph of the rational function. To aid in sketching the graphs, check for intercepts, symmetry, vertical asymptotes, and horizontal asymptotes. $$C(x)=\frac{5+2 x}{1+x}$$

Step-by-Step Solution

Answer

No x-intercept exists. The y-intercept is at (0,5). The function is neither even nor odd. Vertical asymptote is at \(x=-1\), and horizontal asymptote is at \(y=2\).

Step 1: Find the Intercepts

The x-intercept is found by setting \(C(x) = 0\) and solving for \(x\). For the y-intercept, set \(x = 0\). For the x-intercept, we get the equation:\(0 = \frac{5+2x}{1+x}\)We can realize that no solution exists for the above equation. Hence, there's no x-intercept.For the y-intercept, \(C(0) = \frac{5+2*0}{1+0} = 5\) So, the y-intercept is at (0,5).

Step 2: Check for Symmetry

For a function \(f(x)\) to be even, \(f(-x) = f(x)\). For it to be odd, \(f(-x) = -f(x)\). Calculate \(C(-x) = \frac{5-2x}{1-x}\). This doesn't equate to \(C(x)\) or \(-C(x)\), so the function is neither even nor odd, and hence has no symmetry.

Step 3: Find the Vertical and Horizontal Asymptotes

Vertical asymptotes are found when the denominator equals 0. Thus, when \(1+x = 0\), we get \(x = -1\) as the vertical asymptote.For the horizontal asymptote of a rational function \(\frac{f(x)}{g(x)}\), if the degree of the numerator is equal to the degree of the denominator, it is \(y = \frac{a}{b}\), where a and b are the coefficients of the highest degrees of f(x) and g(x). Here, the degrees are equal, so we have \(y = \frac{2}{1}\) or \(y = 2\) as the horizontal asymptote.

Step 4: Sketch the Graph

With these key features, the sketch of the function can be made. The graph has a vertical asymptote at \(x=-1\), a horizontal asymptote at \(y=2\), a y-intercept at (0,5), and no x-intercept. This gives us enough information to sketch the accurate curve of the function.

Key Concepts

X-Intercepts

Understanding the x-intercepts of a graph is crucial as it represents where the graph crosses the x-axis. In essence, these are the points where the value of the function is zero. For a rational function like the one presented, finding the x-intercept involves setting the entire function equal to zero, effectively the numerator equal to zero, since a fraction is only zero when its numerator is zero.

In our example, the equation to find the x-intercept is set as \(0 = \frac{5 + 2x}{1 + x}\). As observed from the calculation, this particular function does not have a real solution for the x-intercept, indicated by the fact that the equation does not have a valid solution where the numerator can equal zero. This piece of information implies that the graph does not touch or cross the x-axis at any point.

Y-Intercepts

When we refer to the y-intercept, we're talking about the point where the graph meets the y-axis. This is found simply by evaluating the function when \(x\) is zero. It's like asking the question, 'What's the output of the function when there's no input?'

In this problem, the y-intercept is found by plugging zero into our function, resulting in \(C(0) = \frac{5 + 2(0)}{1 + 0} = 5\). Therefore, the y-intercept is at the coordinate \((0, 5)\), meaning this is the starting point of the graph on the y-axis. Pinpointing the y-intercept gives us a clear reference point for beginning the graph.

Vertical Asymptotes

A vertical asymptote occurs at a specific \(x\)-value where a function moves towards infinity or negative infinity. It essentially represents the boundaries of a function's graph. For rational functions, vertical asymptotes are found when the denominator approaches zero, since division by zero is undefined.

In our function \(C(x) = \frac{5 + 2x}{1 + x}\), we set the denominator \((1 + x)\) to zero and solve for \(x\) to find the vertical asymptote. We discover that \(x = -1\) is the value at which the graph of our function cannot exist, illustrating that at this point, the graph will have a line that extends vertically in both directions towards infinity, but never actually touches \(x = -1\).

Horizontal Asymptotes

Horizontal asymptotes guide us in understanding the behavior of a function as \(x\) approaches infinity or negative infinity. For rational functions, if the degrees of the numerator and the denominator polynomials are the same, the horizontal asymptote can be found by simply dividing the leading coefficients.

Our example function \(C(x)\) has the same degree in the numerator and the denominator, which leads to a horizontal asymptote determined by the ratio of the leading coefficients. In this case, it's the ratio of 2 to 1, resulting in a horizontal asymptote at \(y = 2\). This means as \(x\) increases or decreases without bound, the function approaches but never actually reaches the asymptote line \(y = 2\), shaping the overall graph.