The rational function: domain, asymptotes and a removable point

There is no division by 0, so $\displaystyle {f(x)}={\frac{1}{x}}$ and $\displaystyle {g(x)}={\frac{1}{x^2}}$ are defined where $\displaystyle x\neq0$. What happens to their values:

• As $\displaystyle x$ gets close to 0 from the right, $\displaystyle \frac{1}{x}$ grows without limit: $\displaystyle {\frac{1}{0.01}}={100}$, $\displaystyle {\frac{1}{0.001}}={1000}$. As $\displaystyle x$ gets close to 0 from the left, $\displaystyle \frac{1}{x}$ is negative and its absolute value grows without limit.
• As $\displaystyle x$ grows without end, $\displaystyle \frac{1}{x}$ gets close to 0: $\displaystyle {\frac{1}{1000}}={0.001}$. So too as $\displaystyle x$ falls without end.
• $\displaystyle \frac{1}{x^2}$ is always positive, and on both sides of 0 it grows without limit.

A line that the graph gets closer and closer to is called an asymptote. Both functions have the vertical asymptote $\displaystyle {x}={0}$ and the horizontal asymptote $\displaystyle {y}={0}$.

• $\displaystyle \frac{1}{x}$ is odd: its graph is symmetric about the origin. $\displaystyle \frac{1}{x^2}$ is even: its graph is symmetric about the $\displaystyle y$-axis.
• $\displaystyle \frac{1}{x}$ falls on each side of 0: its average rate of change from $\displaystyle x_1$ to $\displaystyle x_2$ on the same side is $\displaystyle \frac{-1}{x_1x_2}$, always negative.
• Even so, it does not fall “everywhere”: $\displaystyle {f(-1)}={-1}$ is less than $\displaystyle {f(1)}={1}$. Between the two sides is $\displaystyle {x}={0}$, where it is not defined.

Shifts move the asymptotes too:

• The vertical asymptote is $\displaystyle {x}={p}$, and the horizontal one $\displaystyle {y}={k}$.
• $\displaystyle a$ stretches the graph, and $\displaystyle a<0$ reflects it in the horizontal asymptote.

On the graph below, $\displaystyle {y}={\frac{2}{x-1}+3}$: the asymptotes are $\displaystyle {x}={1}$ and $\displaystyle {y}={3}$.

The graph of $\displaystyle \frac{1}{f(x)}$ is read from the graph of $\displaystyle f$:

• The zeros of $\displaystyle f$ leave the domain, and at each of them there is usually a vertical asymptote.
• The sign of $\displaystyle \frac{1}{f(x)}$ is the sign of $\displaystyle f(x)$.
• Where $\displaystyle f(x)$ is very large, $\displaystyle \frac{1}{f(x)}$ is close to 0.
• Where $\displaystyle f$ rises (and is not 0), $\displaystyle \frac{1}{f}$ falls, and the other way round.

On the graph below, $\displaystyle {f(x)}={x^2-4}$ (dashed) and $\displaystyle \frac{1}{x^2-4}$:

• $\displaystyle {x^2-4}={0}$ where $\displaystyle {x}={\pm2}$: the vertical asymptotes $\displaystyle {x}={-2}$ and $\displaystyle {x}={2}$.
• $\displaystyle x^2-4$ is negative between $\displaystyle -2$ and 2, so $\displaystyle \frac{1}{x^2-4}$ is negative there too.
• The parabola has the minimum $\displaystyle (0,-4)$, and there $\displaystyle \frac{1}{x^2-4}$ has the maximum $\displaystyle \left(0,-\frac{1}{4}\right)$.
• As $\displaystyle x$ grows or falls without end, $\displaystyle x^2-4$ grows, and $\displaystyle \frac{1}{x^2-4}$ gets close to 0: the horizontal asymptote $\displaystyle {y}={0}$.

A rational function is a quotient of two polynomials, like $\displaystyle \frac{2x+1}{x-3}$ or $\displaystyle \frac{x^2-x-2}{x^2-4}$. Its domain: every $\displaystyle x$ where the denominator is not 0.

At a zero $\displaystyle {x}={a}$ of the denominator there are two cases:

• The numerator is not 0 at $\displaystyle a$: near $\displaystyle a$ the denominator is close to 0 and the numerator is not, and the quotient’s absolute value grows without limit. The line $\displaystyle {x}={a}$ is a vertical asymptote.
• The numerator is 0 at $\displaystyle a$ too: the factor $\displaystyle x-a$ cancels. For example $\displaystyle {\frac{x^2-4}{x-2}}={\frac{(x-2)(x+2)}{x-2}}={x+2}$ for every $\displaystyle x\neq2$. The graph is the line $\displaystyle {y}={x+2}$ without the point $\displaystyle (2,4)$ – a hole in the graph. Such a point is called a removable discontinuity.

The reduced function, together with the condition $\displaystyle x\neq2$, is the same function – and the reduced form is easier to work with.

When $\displaystyle x$ is very large, the highest power decides in every polynomial, and the rest is negligible beside it. For example in $\displaystyle \frac{2x^2+1}{x^2-3x}$, when $\displaystyle {x}={1000}$: $\displaystyle {2x^2}={2{,}000{,}000}$, and 1 is negligible beside it; $\displaystyle {x^2}={1{,}000{,}000}$, and $\displaystyle {3x}={3000}$ is very small beside it. So the quotient is close to $\displaystyle {\frac{2x^2}{x^2}}={2}$.

• The numerator’s degree is lower than the denominator’s: the denominator grows faster, and the quotient gets close to 0 – the asymptote $\displaystyle {y}={0}$.
• The same degree: the quotient gets close to the quotient of the leading coefficients – $\displaystyle {y}={2}$ in the example.
• The numerator’s degree is higher: for example in $\displaystyle \frac{x^3}{x^2+1}$ the numerator grows faster, and the values grow without limit – there is no horizontal asymptote.

$\displaystyle \infty$ is not a number: say “as $\displaystyle x$ grows without end”, and do not calculate with $\displaystyle \infty$.

$\displaystyle {f(x)}={\frac{x^2-x-2}{x^2-4}}$.
The example below investigates $\displaystyle f$ without a derivative:

• $\displaystyle {\frac{x^2-x-2}{x^2-4}}={\frac{(x-2)(x+1)}{(x-2)(x+2)}}$, and the domain: $\displaystyle x\neq2$, $\displaystyle x\neq-2$.
• At $\displaystyle {x}={2}$ the numerator is 0 too: once reduced, $\displaystyle {f(x)}={\frac{x+1}{x+2}}$, and $\displaystyle {\frac{2+1}{2+2}}={\frac{3}{4}}$. A removable discontinuity at $\displaystyle \left(2,\frac{3}{4}\right)$.
• At $\displaystyle {x}={-2}$ the numerator is $\displaystyle {(-2-2)(-2+1)}={4\neq0}$: the vertical asymptote $\displaystyle {x}={-2}$.
• The numerator and the denominator are of degree 2, and the coefficients of $\displaystyle x^2$ are 1 and 1: the horizontal asymptote $\displaystyle {y}={1}$.
• The zero: $\displaystyle {x}={-1}$. And on the $\displaystyle y$-axis: $\displaystyle {f(0)}={\frac{-2}{-4}}={\frac{1}{2}}$.
• The sign, by $\displaystyle \frac{x+1}{x+2}$: positive where $\displaystyle x<-2$ and where $\displaystyle x>-1$ (without $\displaystyle {x}={2}$), and negative where $\displaystyle -2<x<-1$.

The board does not show a hole in a graph: the point $\displaystyle \left(2,\frac{3}{4}\right)$ is marked on it only to show where the hole is.

The function is $\displaystyle {f(x)}={\frac{3}{x+2}-1}$.

Write on the board the equations of the two asymptotes of $\displaystyle f$.

Find the zero of $\displaystyle f$.

The board shows the graph of $\displaystyle {f(x)}={\frac{1}{x}}$, and two graphs made from it by shifts, stretches and reflections.
Write a function of the form $\displaystyle {y}={\frac{a}{x-p}+k}$ for each of them.

The function is $\displaystyle {g(x)}={\frac{1}{x^2-2x-3}}$.

Find the zeros of the denominator.

Write on the board the equations of all the asymptotes of $\displaystyle g$.

Solve the inequality $\displaystyle g(x)>0$.
Check the answer by the sign of the parabola $\displaystyle {y}={x^2-2x-3}$.

The function is $\displaystyle {f(x)}={\frac{x^2-9}{x-3}}$.

Reduce the fraction.

What is the domain of $\displaystyle f$?
Describe its graph, and write the coordinates of the removable discontinuity.

The function is $\displaystyle {f(x)}={\frac{2x^2-8}{x^2+x-6}}$.

Find the zeros of the denominator, and write the domain of $\displaystyle f$.

Factor the numerator and the denominator, and reduce.

Write on the board the equations of the vertical asymptote and of the horizontal asymptote.

Write the coordinates of the removable discontinuity, the zero of $\displaystyle f$, and where it is positive and negative.

Find the horizontal asymptote of each function, or explain why it has none.
Explain by the highest powers in the numerator and the denominator.

• $\displaystyle {f(x)}={\frac{3x+1}{x^2+4}}$
• $\displaystyle {g(x)}={\frac{4x^2-x}{2x^2+5}}$
• $\displaystyle {h(x)}={\frac{x^3}{x^2+1}}$