I will be the first to admit that I'm not too fond of graphing functions. There are a lot of steps involved, and sometimes, I get overwhelmed by all the information necessary to start graphing. That's what motivated me to write this post.\n\n\n\nWhen graphing a rational function, it is necessary to find various components before we start graphing the function itself. When graphing a rational function, there are six necessary components. These are the horizontal asymptotes, oblique asymptotes, vertical asymptotes, x-intercepts, y-intercepts, and various points in between. \n\n\n\nLet's break it down to understand better where each part fits into the grander scheme.\n\n\n\nAs we find each component for graphing, we'll walk through one example.\n\n\n\nWhere to begin - horizontal asymptotes.\n\n\n\nThe easiest place to start involves no factoring. Finding the horizontal asymptote involves us looking over the function and noting a few things. \n\n\n\nWe have to look for three conditions when determining the horizontal asymptote for a rational function. In all cases, we are looking at the highest power in the numerator and the highest power in the denominator. Once we locate these, we can find the horizontal asymptote.\n\n\n\nIf the power in the numerator and denominator are equal, then the horizontal asymptote is equal to the coefficient of the numerator divided by the coefficient of the denominator. Pretty easy. \n\n\n\nIn the example below, we see that the numerator and the denominator share the same highest power. We then take the coefficients of each and express them as a fraction. In this case, the horizontal asymptote becomes y=1\/3.\n\n\n\n\n\n\n\nLet's look at the following rational function:\n\n\n\n\n\n\n\nIn this example, the power in the numerator and the denominator are equal. Since both are equal, we take both coefficients and place them as a fraction. In this case, our horizontal asymptote would be at y=1\/1 or y=1.\n\n\n\nNow let's move on to the vertical asymptote.\n\n\n\nCheck for an oblique asymptote.\n\n\n\nNot every rational function will have every asymptote, but it's still important to check. \n\n\n\nAn oblique asymptote pops up when the degree in the numerator is one degree higher than the degree in the denominator. \n\n\n\n\n\n\n\nOur example finds that this rational function does not have an oblique asymptote. \n\n\n\n\n\n\n\nWhen we check out the degree in the numerator and denominator, we find that they are equal. Therefore, this rational function does not have an oblique asymptote.\n\n\n\nIf the numerator and denominator degrees met the requirements, we would then move on to find the actual oblique asymptote. As noted before, it is found when you divide the denominator into numerator via long division. \n\n\n\nOnce you go through long division, the result should be in the form of y=ax+b. This equation represents the oblique asymptote for your rational function.\n\n\n\nNow let's try to find the vertical asymptotes.\n\n\n\nThe vertical asymptote will occur where the rational function's denominator equals zero once the function has been reduced. \n\n\n\n\n\n\n\nIn our example, the rational function has already been reduced. We have to set the denominator equal to zero and solve. \n\n\n\n\n\n\n\nIn this case, we have x-4=0. The vertical asymptote occurs at x=4. This point shows where the rational function is undefined.\n\n\n\nEasy enough, right? Now let's move on to finding the x and y-intercepts of our function.\n\n\n\nNow let's find the intercepts.\n\n\n\nThe x and y-intercepts are where our function either crosses the y-axis or the x-axis. Let's start with the x-intercept.\n\n\n\nThe x-intercept will occur where the numerator is equal to zero. When the numerator equals zero, our whole function takes on that zero value. When the denominator is equal to zero, the whole function is undefined.\n\n\n\n\n\n\n\nIn this function, the numerator is equal to x+4. To find the spot where the x-intercept lies, we have to set the numerator equal to zero. In this case, we have x+4=0. Once we solve this equation, we get x=-4. Our x-intercept is at (-4,0).\n\n\n\nTo find the y-intercept, we must replace all x's with zeros in this equation and solve. When we do that with our rational function, we get -4\/4 or -1.\n\n\n\nTherefore, our y-intercept is at (0,-1). \n\n\n\nNow let's pick a few points.\n\n\n\nWe have to find some extra (random) points to fill out the graph to complete the picture. Once we know where some of the asymptotes and intercepts occur, we can start to pick random x and y values near the more interesting points on the graph.\n\n\n\n\n\n\n\nThe points that I picked occur around the asymptotes or in areas where no points show up. I was curious to see what was going on with other areas around the graph. \n\n\n\nOnce I picked random points, I started to see the shape of the graph. I mainly concentrated on areas with no intercepts and only asymptotes.\n\n\n\nLet's look at all of this on a graph.\n\n\n\nHere is the completed graph with all of the previous components labeled. As you can see, the area to the left of the vertical asymptote is where most of the intercepts showed up. \n\n\n\nMost of the random points that I picked lay to the right of the vertical asymptote.\n\n\n\n\n\n\n\nHere is the first example (removable discontinuity included).\n\n\n\nHere is the first example we'll walkthrough.\n\n\n\n\n\n\n\nOur first step is to look over this function to see what the numerator and denominator degrees are. As we can see here, the numerator degree is less than the degree in the denominator. Because of this, the horizontal asymptote is at y=0.\n\n\n\nThe next step is to look at the numerator and denominator again. This time we are interested in seeing if the numerator degree is one more degree than the denominator. Since it isn't, we know that our rational function does not have any oblique asymptotes.\n\n\n\n\n\n\n\nThen we move on to the vertical asymptotes. Before looking for the vertical asymptotes, we have to reduce the function. We look to see if there are any common factors in the numerator and denominator. \n\n\n\nThere is an x that is common, so we cross it out from our numerator and denominator. This x=0 is part of our removable discontinuity. The removable discontinuity is at (0,-1). Once we get that out of the way, we can move on to find the vertical asymptotes.\n\n\n\nThe denominator contains x2-4. To find the asymptotes, we could break this down into x-2 and x+2. Once we have done this, we set each one equal to zero. We find that our vertical asymptotes are at x=2 and x=-2. \n\n\n\n\n\n\n\nWe then start the search for our intercepts. Since we have a constant in the numerator, we don't have any x-intercepts for our function. For the y-intercepts, we find that there aren't any. When we put a zero into the function, we end up with the point defined as our removable discontinuity.\n\n\n\nOur last step before graphing is to find a few random points. Here is the table of points that I came up with.\n\n\n\n\n\n\n\nOur last step is to plot all our points and asymptotes on the graph. \n\n\n\n\n\n\n\nHere is another example (oblique asymptote included).\n\n\n\nNow we'll work with the following rational function. The first thing we will look at is if the function has a horizontal asymptote. We look at the degree in the numerator and the degree in the denominator.\n\n\n\nThe degree in the numerator is higher than the degree in the denominator. This means that this function has no horizontal asymptote.\n\n\n\n\n\n\n\nThe next asymptote we will look for is an oblique asymptote. Since the numerator degree is one higher than the denominator degree, we know that this function has an oblique asymptote.\n\n\n\nTo find the oblique asymptote, we have to use long division. We divide the denominator into the numerator to find the equation for the oblique asymptote. \n\n\n\n\n\n\n\nIn this case, when we divide x+3 into x2+3x-5, we get x. The equation for our oblique asymptote is at y=x. \n\n\n\nThe next item we need to find is the vertical asymptotes. We look at the denominator and set it equal to 0. In this case, our denominator is x+3. \n\n\n\nOur vertical asymptote is at x=-3.\n\n\n\n\n\n\n\nThen we move on to the x and y-intercepts. \n\n\n\nWe'll start with the y-intercepts because those are easier to find. We replace all of the x's in our equation with zeros and solve the equation. \n\n\n\n\n\n\n\nOur y-intercept is at (0,-5\/3).\n\n\n\nThe x-intercept is found by factoring the numerator and then solving it. The numerator can't be broken down by factoring, so we have to use the quadratic formula to find the factors of the numerator.\n\n\n\n\n\n\n\nThe x-intercepts are at (1.2, 0) and (-4.2, 0).\n\n\n\nThe last step is to find random points that can fill in the graph. The table below shows the points that I randomly picked.\n\n\n\n\n\n\n\nThe last step is to plot the points and asymptotes on the graph. Here is what the final graph looks like. \n\n\n\n\n\n\n\nAlways remember that you will only get better at math with practice. Make sure you understand the basics and conquer them!