Graphing rational functions, asymptotes
This section shows another kind of function whose graphs we can understand effectively by our methods. There is one new item here, the idea of asymptote of the graph of a function.
A vertical asymptote of the graph of a function $f$ most commonly occurs when $f$ is defined as a ratio $f(x)=g(x)/h(x)$ of functions $g,h$ continuous at a point $x_o$, but with the denominator going to zero at that point while the numerator doesn't. That is, $h(x_o)=0$ but $g(x_o) \ne 0$. Then we say that $f$ blows up at $x_o$, and that the line $x=x_o$ is a vertical asymptote of the graph of $f$.
And as we take $x$ closer and closer to $x_o$, the graph of $f$ zooms off (either up or down or both) closely to the line $x=x_o$.
A very simple example of this is $f(x)=1/(x-1)$, whose denominator is $0$ at $x=1$, so causing a blow-up at that point, so that $x=1$ is a vertical asymptote. And as $x$ approaches $1$ from the right, the values of the function zoom up to $+\infty$. When $x$ approaches $1$ from the left, the values zoom down to $-\infty$.
A horizontal asymptote of the graph of a function $f$ occurs if either limit $$\lim_{x\rightarrow +\infty}f(x)$$ or $$\lim_{x\rightarrow -\infty}f(x)$$ exists. If $R=\lim_{x\rightarrow +\infty}f(x)$, then $y=R$ is a horizontal asymptote of the function, and if $L=\lim_{x\rightarrow -\infty}f(x)$ exists then $y=L$ is a horizontal asymptote.
As $x$ goes off to $+\infty$ the graph of the function gets closer and closer to the horizontal line $y=R$ if that limit exists. As $x$ goes of to $-\infty$ the graph of the function gets closer and closer to the horizontal line $y=L$ if that limit exists.
So in rough terms asymptotes of a function are straight lines which the graph of the function approaches at infinity. In the case of vertical asymptotes, it is the $y$-coordinate that goes off to infinity, and in the case of horizontal asymptotes it is the $x$-coordinate which goes off to infinity.
Example
Find asymptotes, critical points, intervals of increase and decrease, inflection points, and intervals of concavity up and down of $f(x)={x+3\over 2x-6}$: First, let's find the asymptotes. The denominator is $0$ for $x=3$ (and this is not cancelled by the numerator) so the line $x=3$ is a vertical asymptote. And as $x$ goes to $\pm\infty$, the function values go to $1/2$, so the line $y=1/2$ is a horizontal asymptote.
The derivative is $$f'(x)={1\cdot(2x-6)-(x+3)\cdot 2\over (2x-6)^2}={-12\over (2x-6)^2}$$ Since a ratio of polynomials can be zero only if the numerator is zero, this $f'(x)$ can never be zero, so there are no critical points. There is, however, the discontinuity at $x=3$ which we must take into account. Choose auxiliary points $0$ and $4$ to the left and right of the discontinuity. Plugging in to the derivative, we have $f'(0)=-12/(-6)^2 <0$, so the function is decreasing on the interval $(-\infty,3)$. To the right, $f'(4)=-12/(8-6)^2 <0$, so the function is also decreasing on $(3,+\infty)$.
The second derivative is $f''(x)=48/(2x-6)^3$. This is never zero, so there are no inflection points. There is the discontinuity at $x=3$, however. Again choosing auxiliary points $0,4$ to the left and right of the discontinuity, we see $f''(0)=48/(-6)^3 <0$ so the curve is concave downward on the interval $(-\infty,3)$. And $f''(4)=48/(8-6)^3>0$, so the curve is concave upward on $(3,+\infty)$.
Plugging in just two or so values into the function then is enough to enable a person to make a fairly good qualitative sketch of the graph of the function.
Exercises
- Find all asymptotes of $f(x)={ x-1 \over x+2 }$.
- Find all asymptotes of $f(x)={ x+2 \over x-1 }$.
- Find all asymptotes of $f(x)={ x^2-1 \over x^2-4 }$.
- Find all asymptotes of $f(x)={ x^2-1 \over x^2+1 }$.