Rational functions near undefined points and at the ends

Near a zero of the denominator that is not a zero of the numerator, the fraction’s denominator is tiny and its numerator is not, so $\displaystyle f(x)$ grows or falls without end. Which of the two happens on each side is decided by the signs:

• the numerator’s sign near the point;
• the denominator’s sign just left and just right of the point.

The example below looks at both sides of $\displaystyle {x}={2}$ for $\displaystyle {f(x)}={\frac{x+1}{x-2}}$:

• near $\displaystyle {x}={2}$ the numerator is close to 3, positive;
• just right of 2 (say $\displaystyle {x}={2.01}$) the denominator is a small positive number: $\displaystyle {f(x)}={\frac{3.01}{0.01}}={301}$, and $\displaystyle f(x)$ grows without end;
• just left of 2 (say $\displaystyle {x}={1.99}$) the denominator is a small negative number: $\displaystyle {f(x)}={\frac{2.99}{-0.01}}={-299}$, and $\displaystyle f(x)$ falls without end.

When the factor of the denominator has an even power, it is positive on both sides, so the graph goes the same way on both sides. For $\displaystyle {g(x)}={\frac{3}{(x-1)^2}}$, $\displaystyle g(x)$ grows without end on both sides of $\displaystyle {x}={1}$.

As $\displaystyle x$ grows or falls without end, only the highest powers of the numerator and of the denominator matter, so their degrees decide the horizontal asymptote:

• the numerator’s degree is lower: $\displaystyle f(x)$ approaches 0, the asymptote is $\displaystyle {y}={0}$;
• the degrees are equal: the asymptote is $\displaystyle {y}={b}$, where $\displaystyle b$ is the ratio of the coefficients of the highest powers;
• the numerator’s degree is higher: $\displaystyle f(x)$ grows or falls without end, and there is no horizontal asymptote.

The sign of $\displaystyle f(x)-b$ says from which side the graph approaches $\displaystyle {y}={b}$: above it where $\displaystyle f(x)-b>0$, below it where $\displaystyle f(x)-b<0$.

For example, $\displaystyle {f(x)}={\frac{2x^2}{x^2+1}}$ has the asymptote $\displaystyle {y}={2}$, and $\displaystyle {f(x)-2}={\frac{2x^2-2x^2-2}{x^2+1}}={\frac{-2}{x^2+1}<0}$: the graph is below the asymptote at both ends.

A horizontal asymptote describes the graph only far to the left and far to the right, so in between the graph may cross it. The example below finds where $\displaystyle {f(x)}={\frac{x^2+x}{x^2+1}}$ meets its asymptote $\displaystyle {y}={1}$:

• $\displaystyle {\frac{x^2+x}{x^2+1}}={1}$ gives $\displaystyle {x^2+x}={x^2+1}$, so $\displaystyle {x}={1}$: the graph crosses the asymptote at $\displaystyle (1,1)$;
• $\displaystyle {f(x)-1}={\frac{x-1}{x^2+1}}$: positive where $\displaystyle x>1$ – far to the right the graph approaches $\displaystyle {y}={1}$ from above;
• negative where $\displaystyle x<1$ – far to the left it approaches $\displaystyle {y}={1}$ from below.

A number that is a zero of both the numerator and the denominator gives no asymptote when the common factor cancels: the graph has a hole there. For example, $\displaystyle {f(x)}={\frac{x^2-1}{x^2-x}}$:

• $\displaystyle {f(x)}={\frac{(x-1)(x+1)}{x(x-1)}}={\frac{x+1}{x}}$, provided $\displaystyle x\neq1$;
• $\displaystyle {x}={0}$ is a zero of the denominator only: a vertical asymptote $\displaystyle {x}={0}$;
• $\displaystyle {x}={1}$ is a zero of both: a hole at $\displaystyle {\left(1,\frac{1+1}{1}\right)}={(1,2)}$;
• near $\displaystyle {x}={1}$ the values of $\displaystyle f$ are close to 2, from both sides.

The function is $\displaystyle {f(x)}={\frac{2x-4}{x+1}}$.

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

Solve the inequality $\displaystyle f(x)>0$.

On which side of $\displaystyle {x}={-1}$ does $\displaystyle f(x)$ grow without end, and on which does it fall without end?
Explain by the signs of the numerator and the denominator.

The function is $\displaystyle {f(x)}={\frac{x^2+2x}{x^2+1}}$.
Its horizontal asymptote is $\displaystyle {y}={1}$.

Solve the equation $\displaystyle {f(x)}={1}$.

Mark on the board the point where the graph crosses its asymptote.

Work out $\displaystyle f(x)-1$.
From which side does the graph approach the asymptote far to the right, and from which far to the left?

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

Write the equations of its asymptotes on the board.

Solve the inequality $\displaystyle f(x)>0$.

What does $\displaystyle f(x)$ do on each side of $\displaystyle {x}={2}$? Explain.

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

Reduce the fraction.

Write the domain of $\displaystyle f$, its asymptotes and the point where its graph has a hole.

Write a rational function with all these properties:

• the vertical asymptote $\displaystyle {x}={1}$, with $\displaystyle f(x)$ growing without end on both its sides;
• the horizontal asymptote $\displaystyle {y}={0}$.

Explain why your function fits.