Graphing Rational Functions: Videos & Practice Problems

Understanding how to graph rational functions involves recognizing the components of the function and applying transformations to create the desired graph. A fundamental example is the function \( f(x) = \frac{1}{x} \), which serves as a base for applying transformations to graph more complex rational functions.

Transformations can be categorized mainly into reflections and shifts. Reflections occur when a negative sign is present in the function. A negative outside the function indicates a reflection over the x-axis, while a negative inside the function reflects over the y-axis. Shifts are determined by the parameters \( h \) and \( k \) in the function \( g(x) = \frac{1}{x - h} + k \). Here, \( h \) represents a horizontal shift, moving the graph left or right, and \( k \) represents a vertical shift, moving the graph up or down.

To graph a transformed function, start by identifying the vertical and horizontal asymptotes. For the function \( g(x) = \frac{1}{x - 3} + 1 \), the vertical asymptote is found at \( x = h = 3 \) and the horizontal asymptote at \( y = k = 1 \). These asymptotes are plotted as dashed lines on the graph, indicating that the function approaches but never touches these lines.

Next, check for reflections. If there are no negative signs in the function, as in this case, no reflections are necessary. The next step involves shifting reference points based on the transformations. For example, the reference points \( (1, 1) \) and \( (-1, -1) \) from the base function are shifted 3 units to the right and 1 unit up, resulting in new points \( (4, 2) \) and \( (2, 0) \).

Finally, sketch the curves of the function, ensuring they approach the asymptotes correctly. The resulting graph will resemble the original \( \frac{1}{x} \) graph but will be repositioned according to the transformations applied.

When determining the domain and range of the function, consider the asymptotes. The domain is expressed in set notation as \( (-\infty, 3) \cup (3, \infty) \), indicating that the function is undefined at \( x = 3 \). Similarly, the range is \( (-\infty, 1) \cup (1, \infty) \), showing that the function does not take the value \( y = 1 \). Both domain and range exclude the asymptote values, hence the use of parentheses.

By mastering these transformations and understanding how to manipulate the graph of \( \frac{1}{x} \), students can effectively graph more complex rational functions and analyze their properties.

To graph the function \( g(x) = -\frac{1}{(x - 2)^2} \), we start by recognizing that this is a transformation of the basic function \( \frac{1}{x^2} \). The process involves several key steps to identify asymptotes, reflections, and shifts.

First, we determine the vertical asymptote, which occurs at \( x = h \). In this function, \( h \) is 2, so we plot a vertical dashed line at \( x = 2 \). Next, we identify the horizontal asymptote, represented by \( y = k \). Since there is no vertical shift in this case, \( k \) is 0, placing the horizontal asymptote at \( y = 0 \), the same as the original function.

Next, we check for reflections. The negative sign in front of the function indicates a reflection over the x-axis. This means that points on the graph will be mirrored across the x-axis. For example, if we take the point \( (1, -1) \), its reflection will be \( (1, 1) \). Similarly, the point \( (-1, -1) \) will reflect to \( (-1, 1) \).

After reflecting the points, we apply the horizontal and vertical shifts defined by \( h \) and \( k \). Here, \( h = 2 \) and \( k = 0 \) means we shift the points 2 units to the right. Thus, the point \( (1, 1) \) shifts to \( (3, 1) \), and \( (-1, 1) \) shifts to \( (1, 1) \).

Now, we sketch the curve, ensuring it approaches the asymptotes. The shape of the graph remains similar to that of \( \frac{1}{x^2} \), but it is reflected and shifted. The resulting graph will approach the vertical asymptote at \( x = 2 \) and the horizontal asymptote at \( y = 0 \).

Finally, we identify the domain and range of the function. The domain is all real numbers except for the vertical asymptote, which can be expressed as \( (-\infty, 2) \cup (2, \infty) \). The range, since the graph is entirely below the x-axis, is \( (-\infty, 0) \).

In summary, the function \( g(x) = -\frac{1}{(x - 2)^2} \) has a vertical asymptote at \( x = 2 \), a horizontal asymptote at \( y = 0 \), and reflects over the x-axis, resulting in a domain of \( (-\infty, 2) \cup (2, \infty) \) and a range of \( (-\infty, 0) \).

Graphing a rational function involves several systematic steps to ensure accuracy. Let's consider the function \( f(x) = \frac{2x - 3}{x - 1} \) as an example to illustrate the process.

First, we need to determine the domain of the function. The domain is found by setting the denominator equal to zero. For our function, \( x - 1 = 0 \) leads to \( x = 1 \). Therefore, the domain is all real numbers except \( x = 1 \).

Next, we check for any holes in the graph by identifying common factors in the numerator and denominator. In this case, there are no common factors, so we can skip this step.

To find the x-intercepts, we set the numerator equal to zero: \( 2x - 3 = 0 \). Solving this gives \( 2x = 3 \) or \( x = \frac{3}{2} \) (1.5). Since the factor \( 2x - 3 \) occurs once, the multiplicity is 1, indicating that the graph will cross the x-axis at this point.

Next, we find the y-intercept by evaluating \( f(0) \): \( f(0) = \frac{2(0) - 3}{0 - 1} = \frac{-3}{-1} = 3 \). Thus, the y-intercept is at \( (0, 3) \).

Now, we identify the vertical asymptote by setting the denominator to zero, which we already determined gives \( x = 1 \). This asymptote is represented by a dashed line at \( x = 1 \).

To find the horizontal asymptote, we compare the degrees of the numerator and denominator. Both have a degree of 1, so we divide the leading coefficients: \( \frac{2}{1} = 2 \). Therefore, the horizontal asymptote is at \( y = 2 \).

With the intercepts and asymptotes established, we can now create intervals based on these points. The intervals are: \( (-\infty, 1) \), \( (1, \frac{3}{2}) \), and \( (\frac{3}{2}, \infty) \).

Next, we select test points from each interval to determine the behavior of the function:

• For the interval \( (-\infty, 1) \), choose \( x = -2 \): \( f(-2) = \frac{2(-2) - 3}{-2 - 1} = \frac{-4 - 3}{-3} = \frac{-7}{-3} = \frac{7}{3} \approx 2.33 \).
• For the interval \( (1, \frac{3}{2}) \), choose \( x = \frac{5}{4} \): \( f\left(\frac{5}{4}\right) = \frac{2\left(\frac{5}{4}\right) - 3}{\frac{5}{4} - 1} = \frac{\frac{10}{4} - 3}{\frac{5}{4} - \frac{4}{4}} = \frac{\frac{10}{4} - \frac{12}{4}}{\frac{1}{4}} = \frac{-\frac{2}{4}}{\frac{1}{4}} = -2 \).
• For the interval \( (\frac{3}{2}, \infty) \), choose \( x = 3 \): \( f(3) = \frac{2(3) - 3}{3 - 1} = \frac{6 - 3}{2} = \frac{3}{2} = 1.5 \).

Now we have the points: \( (-2, \frac{7}{3}) \), \( \left(\frac{5}{4}, -2\right) \), and \( (3, \frac{3}{2}) \). We can plot these points on the graph along with the intercepts and asymptotes.

Finally, we connect the points while ensuring the graph approaches the asymptotes. The graph will cross the x-axis at \( \frac{3}{2} \) and will approach the vertical asymptote at \( x = 1 \) and the horizontal asymptote at \( y = 2 \). This process results in a complete sketch of the rational function.

By following these steps, you can graph any rational function accurately, ensuring you account for all critical components such as intercepts, asymptotes, and the overall behavior of the function across its domain.

To graph the rational function \( f(x) = \frac{2x^2}{x^2 - 1} \), we begin by determining the domain. The numerator \( 2x^2 \) does not factor further, while the denominator factors as \( (x + 1)(x - 1) \). Setting the denominator equal to zero, we find the restrictions: \( x \neq -1 \) and \( x \neq 1 \). Thus, the domain of the function excludes these values.

Next, we check for holes in the function by identifying any common factors between the numerator and denominator. Since there are no common factors, the function is already in its simplest form, indicating there are no holes.

To find the x-intercepts, we set the numerator equal to zero: \( 2x^2 = 0 \). Solving this gives \( x = 0 \). The multiplicity of this intercept is 2 (since it comes from \( 2x^2 \)), indicating that the graph will touch the x-axis at this point but not cross it.

For the y-intercept, we evaluate \( f(0) \): \( f(0) = \frac{2(0)^2}{(0)^2 - 1} = 0 \). This confirms that the y-intercept is also at the origin, \( (0, 0) \).

Next, we identify vertical asymptotes by setting the denominator equal to zero, which gives us the same restrictions as the domain: vertical asymptotes occur at \( x = -1 \) and \( x = 1 \).

To find the horizontal asymptote, we compare the degrees of the numerator and denominator. Both have a degree of 2, so we divide the leading coefficients: \( \frac{2}{1} = 2 \). Therefore, the horizontal asymptote is at \( y = 2 \).

Now, we divide the graph into intervals based on the asymptotes and intercepts: \( (-\infty, -1) \), \( (-1, 0) \), \( (0, 1) \), and \( (1, \infty) \). We will select test points from each interval to determine the behavior of the function.

For the interval \( (-\infty, -1) \), we choose \( x = -3 \): \[f(-3) = \frac{2(-3)^2}{(-3)^2 - 1} = \frac{18}{8} = 2.25.\]This gives us the point \( (-3, 2.25) \).

In the interval \( (-1, 0) \), we select \( x = -\frac{1}{2} \):\[f\left(-\frac{1}{2}\right) = \frac{2\left(-\frac{1}{2}\right)^2}{\left(-\frac{1}{2}\right)^2 - 1} = \frac{2 \cdot \frac{1}{4}}{\frac{1}{4} - 1} = \frac{\frac{1}{2}}{-\frac{3}{4}} = -\frac{2}{3} \approx -0.67.\]This gives us the point \( \left(-\frac{1}{2}, -\frac{2}{3}\right) \).

For the interval \( (0, 1) \), we again use \( x = \frac{1}{2} \):\[f\left(\frac{1}{2}\right) = -\frac{2}{3} \text{ (same as above)}.\]This gives us the point \( \left(\frac{1}{2}, -\frac{2}{3}\right) \).

Finally, in the interval \( (1, \infty) \), we choose \( x = 2 \):\[f(2) = \frac{2(2)^2}{(2)^2 - 1} = \frac{8}{3} \approx 2.67.\]This gives us the point \( (2, 2.67) \).

With all points plotted, we can now sketch the graph. The function approaches the vertical asymptotes at \( x = -1 \) and \( x = 1 \) and the horizontal asymptote at \( y = 2 \). The graph will touch the x-axis at the origin and will not cross it due to the even multiplicity of the x-intercept. The final graph will reflect these behaviors, connecting the plotted points and approaching the asymptotes as described.