Rational Functions: Practice

The domain excludes any input that makes the denominator zero.

$x - 5 = 0 \;\Rightarrow\; x = 5$

So $x = 5$ is excluded.

Set the denominator to zero to find the only excluded value.

$x - 2 = 0 \;\Rightarrow\; x = 2$

The domain is every real number except $2$.

Substitute $x = 6$ and simplify.

$f(6) = \dfrac{3}{6} = \dfrac{1}{2}$

A vertical asymptote sits where the denominator is zero.

$x = 0$

As $x$ nears $0$, the value $\tfrac{1}{x}$ grows without bound, so the curve races along the y-axis.

Test each point in $y = \tfrac{1}{x}$.

$\frac{1}{2} = \tfrac{1}{2} \; \checkmark$

The point $\left(2, \tfrac{1}{2}\right)$ works, while $(0, 0)$ is impossible because $x = 0$ is excluded.

Substitute $x = -2$.

$f(-2) = \dfrac{1}{-2} = -\dfrac{1}{2}$

The denominator is zero at one point.

$x - 1 = 0 \;\Rightarrow\; x = 1$

The domain is every real number except $1$.

By definition a rational function is one polynomial divided by another.

$f(x) = \frac{p(x)}{q(x)}$

The denominator $q(x)$ must not be the zero polynomial.

The function is undefined where the denominator is zero.

$x - 6 = 0 \;\Rightarrow\; x = 6$

Dividing $1$ by a fraction inverts that fraction.

$f\!\left(\tfrac{1}{3}\right) = \dfrac{1}{1/3} = 3$

An input is excluded exactly when it makes the denominator zero, because dividing by zero has no value.

$x - 1 = 0 \;\Rightarrow\; x = 1$

At $x = 1$ the denominator is $0$, so $x = 1$ is thrown out.

The denominator is $x$, which is zero at one input.

$x = 0$

The domain is every real number except $0$.