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Graphing rational functions and asymptotes

Graphing rational functions isn’t really that hard. There is some calculating, depending on the function. First you should find the points in which your function cannot exist. Those are the zeros of the given function’s denominator. Second thing you should do is find the points where the graph intersects the x- axis. Those are the points you get when you equalize numerator with zero. Also, it is useful to know where the graph intersects the y axis. Those points you get when you equalize x with zero. Also, there are things called asymptotes.

graph-of-function-2-exp-xAsymptotes are lines which that tell you where your function can never be defined. Graph of the function always approaches the asymptote into infinity, but never reach it. They are divided into three groups: horizontal, vertical and slant asymptotes. During your studies you already, maybe unknowingly, already met asymptotes. For example, function f ( x ) = 2^x has horizontal asymptote y = 0, function f ( x ) = \frac{1}{x^2} has vertical asymptote x = 0, and any kind of hyperbola has slope asymptotes, y = \pm \frac{b}{a}.



We say that the line x = a is the vertical asymptote of the function y = f ( x ) if at least one of the limits lim_{ x -> a } f ( x ) or lim_{ y -> a } + f ( x ) is equal to -\infty or +\infty.

There are few rules that we can use in order to easily find asymptotes.

Graphically this means that some line is a vertical asymptote if, when we look at it, from the right or from the left its function values are constantly increasing or constantly decreasing.

Example 1. Find vertical asymptote of function f ( x ) = \frac{1}{(x + 1)^2}

Basically, you’re always looking at the points where this function is not defined in. For this function this is for . Now we have to find out from which side of the asymptote our graph lies. If lim_{y -> a}+ f ( x ) is equal to – \infty or +\infty, then our graph lies on the right side, and if lim_{x -> a}- f(x) is equal to  -\infty or +\infty our graph lies on the left side, and of course, if both are equal, function lies on both sides. What about where goes our graph go into infinity? Well if this limit goes to +\infty our function values also go to +\infty and opposite.

vertical asimptoteSince we got our point in which our function may have a vertical asymptote it’s time to check it out. Let’s try to see what happens with our graph as we move along the x- axis. First if we take very large negative number for x, the square in the denominator will change it into very large positive number, which will mean that the function value of this function will be very close to zero. But, no matter how much we decrease that number, our function will never reach zero, or value that is lesser than zero. As we go on, when we get to zero, our function value is equal to 1 which means that here is where our function is growing. As we pass 0, we come closer to 1 which is our critical point. As you can notice here is where our function grows even larger. Let’s try to make it clearer.

For \frac{1}{2}, f ( x ) = \frac{1}{(\frac{1}{2} - 1)^2} = 4

For a number that is very close to 1, but not exactly 1, this function will go to +\infty.

We know that this it is not defined in 1, but what about very close to 1 in other way? It will also be +\infty. If we continue, function values will continuously drop, and also reach points that are very close to zero, but will never be zero.

This means that around 1, graph does reach infinity which means that x = 1 is vertical asymptote of function f(x) = \frac{1}{(x - 1)^2}.

We say that a line  y = b is a horizontal asymptote of a function y = f(x) if lim_{x -> +\infty} f(x) = b or lim_{x -> -\infty} f(x) = b

Horizontal asymptote is a horizontal line which the graph of the function is always approaching but never touches.

Example 2. Find asymptotes and draw the graph of the function f(x) = \frac{x}{x + 1}

This graph will have vertical asymptote x = - 1.

To determine horizontal asymptotes we have to find lim_{x -> \pm \infty} f(x).

lim_{x -> \pm \infty} \frac{x}{x + 1} \setminus \frac{: x}{: x} = lim_{x -> \pm \infty} \frac{1}{1 + \frac{1}{x}} = 1

This means that the line y = 1 is the horizontal asymptote.

All you need to do now is find where your graph lies. All these graphs are pretty much the same. You only need to know in which parts you should draw. This graph will be made out of two parts. First we’ll take some point that is lesser than -1, for example -2. f (- 2) = 2 which means that our graph will be in the top left corner. Now for the second part we’ll take 0. f (0) = 0 which means that the second part will lie on the lower right part.y=1-and-x=1-functionsasimptotes-of-function-y=1

We say that a line y = kx + l  is a slant asymptote of a function f (x)  when x -> \infty if lim_{x - > \infty} ( f ( x ) - (kx + l) ) = 0.

Coefficients of this line we determine as:

k = lim_{x - > \infty} \frac{f(x)}{x}, l = lim_{x - > \infty} ( f(x) - kx)


Example 3. Find asymptotes and draw a graph of a function f(x) = \frac{x^2}{x + 1}

Right away we should notice vertical asymptote x = - 1.

Now let’s get to work with calculating the coefficients for the slant asymptote.

k = lim_{x -> \infty} \frac{f(x)}{x}, k = lim_{x -> \infty} \frac{\frac{x^2}{x + 1}}{x} = lim_{x -> \infty} \frac{x^2}{x^2 + x} = 1 -> k = 1

l = lim_{x -> \infty} (x - kx),

l = lim_{x -> \infty} (\frac{x^2}{x + 1} - 1 * x) = lim_{x -> \infty} (\frac{x^2 - x^2 - x}{x + 1} = lim_{x -> \infty} (\frac{- x}{x + 1} = - 1 -> l = -1

We got that our slant asymptote is the line y = x - 1.

slant asymptote

The last steps of drawing graphs of functions is always the same: first you draw the asymptotes, then find few points and fit the graph between the asymptotes.