Problem 46 - 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. $$h(x)=\frac{1}{x-3}+1$$

Step-by-Step Solution

Answer

The graph of the function \( h(x)=\frac{1}{x-3}+1 \) has a vertical asymptote at \( x=3 \) and a horizontal asymptote at \( y=1 \). The function has no x-intercept and the y-intercept is \( h(0) = -\frac{1}{3}+1 \). The function is not symmetric.

Step 1: Identify the intercepts

Firstly, find where the function crosses the x-axis and y-axis, these are the x and y intercepts. Set \( h(x) \) to 0 and solve for x to get the x-intercept. Set \( x \) to 0 and solve for \( h(x) \) to get the y-intercept.

Step 2: Test for symmetry

A function is symmetrical if \( f(-x) = f(x) \) for even symmetry, or \( f(-x) = -f(x) \) for odd symmetry. Substitute \( -x \) into the function to test for symmetry.

Step 3: Find vertical asymptotes

Vertical asymptotes occur when the denominator of the rational function is zero. Set \( x-3 \) to 0 and solve for x to find the vertical asymptotes.

Step 4: Check horizontal asymptotes

Horizontal asymptotes are found by calculating the limit of the function as \( x \) tends to infinity or minus infinity.

Step 5: Sketch the graph

Compile all the information found in the first four steps: plot the intercepts, plot the asymptotes as dotted lines, plot the point of symmetry if any. After that, you will be able to sketch a curve that fits all of these constraints.

Key Concepts

Rational Function Intercepts

Understanding the intercepts of a rational function is a crucial step in graphing its curve. The x-intercept occurs where the function crosses the x-axis, which implies the output value, or the value of the function is zero. To find it for the rational function
\( h(x) = \frac{1}{x-3} + 1 \),
you set \( h(x) \) to zero and solve for \( x \). However, for this function, it's clear that there is no real value of \( x \) that will make \( h(x) \) equal to zero, so it does not have an x-intercept.

Conversely, the y-intercept is found by plugging in zero for \( x \) in the function, giving us
\( h(0) = \frac{1}{0-3} + 1 = -\frac{1}{3} + 1 \),
which simplifies to \( \frac{2}{3} \). Therefore, the y-intercept of this rational function is at the point \( (0, \frac{2}{3}) \). Understanding and plotting these intercepts helps us to start shaping the graph of a rational function.

Vertical Asymptotes

Vertical asymptotes of rational functions are straight lines that the function approaches but never crosses or touches. They correspond to the values of \( x \) that make the function undefined, typically where the denominator equals zero. In our function,
\( h(x) = \frac{1}{x-3} + 1 \),
the denominator \( x-3 \) is zero when \( x = 3 \). This means there is a vertical asymptote at \( x = 3 \), which is a line that the graph approaches near that point but does not intersect. Vertical asymptotes are an important feature that significantly influences the shape and direction of the graph around those values, illustrating the behavior of the function as values get large positive or negative near the asymptote.

Horizontal Asymptotes

Horizontal asymptotes indicate the value that the output of a function approaches as the input either increases or decreases without bounds. In simpler terms, it shows where the graph of the function goes when \( x \) heads towards infinity or negative infinity. For the given function,
\( h(x) = \frac{1}{x-3} + 1 \),
you calculate the horizontal asymptote by taking the limit of \( h(x) \) as \( x \) approaches infinity. Since the degree of the polynomial in the numerator (0, as there's no \( x \) in the numerator) is less than the degree of the polynomial in the denominator (1, from \( x-3 \)), the horizontal asymptote will be the horizontal line where the highest degrees' coefficients' ratio is for large magnitudes of \( x \), which in this case is \( y = 1 \). This line is a guide for the end behavior of the function and suggests that as \( x \) gets very large or very negative, the graph flattens out approaching the line \( y = 1 \).

Function Symmetry

The analysis of symmetry in a function can provide insights into the behavior of its graph. A function can have even symmetry, which means it is symmetrical about the y-axis, or odd symmetry, where it is symmetrical about the origin. To determine the symmetry of the rational function
\( h(x) = \frac{1}{x-3} + 1 \),
we replace \( x \) with \( -x \) and compare it to the original function. If \( h(-x) = h(x) \), the function has even symmetry. If \( h(-x) = -h(x) \), it has odd symmetry. For our function, substitution gives us
\( h(-x) = \frac{1}{-x-3} + 1 \),
which is not equal to \( h(x) \) nor is it the negation of \( h(x) \). Thus, the function does not exhibit even or odd symmetry. However, in cases where symmetry is present, it simplifies the graphing task, as you would only need to plot half of the graph before reflecting it across the axis or the origin.