Rational Functions: Practice

Factor and cancel the repeated factor.

$\frac{x^2 - x - 6}{x - 3} = \frac{(x - 3)(x + 2)}{x - 3} = x + 2 \quad (x \neq 3)$

The hole sits at $x = 3$, and the reduced form gives its height $3 + 2 = 5$, so the hole is $(3, 5)$.

Factor top and bottom and cancel the shared factor.

$\frac{(x - 5)(x + 5)}{(x - 5)(x + 1)} = \frac{x + 5}{x + 1} \quad (x \neq 5)$

The hole is at $x = 5$ with height $\frac{5 + 5}{5 + 1} = \frac{10}{6} = \frac{5}{3}$, so it is $\left(5, \tfrac{5}{3}\right)$.

Multiply every term by $x - 4$ to clear denominators.

$x = 4 + 2(x - 4) \;\Rightarrow\; x = 2x - 4 \;\Rightarrow\; x = 4$

But $x = 4$ makes the original denominator zero, so it is extraneous and the equation has no solution.

Divide the numerator by the denominator.

$x^2 + 1 = (x - 1)(x + 1) + 2 \;\Rightarrow\; \frac{x^2 + 1}{x - 1} = x + 1 + \frac{2}{x - 1}$

For large $x$ the remainder term fades, so the graph lines up with $y = x + 1$.

Match $\tfrac{a}{x - h} + k$: here $h = 1$ and $k = 3$, and the negative $a$ only flips the branches.

$x = h = 1, \qquad y = k = 3$

So the asymptotes are $x = 1$ and $y = 3$.

Factor the denominator and cancel.

$\frac{x - 1}{x^3 - x} = \frac{x - 1}{x(x - 1)(x + 1)} = \frac{1}{x(x + 1)} \quad (x \neq 1)$

The factor $x - 1$ cancels (a hole), leaving vertical asymptotes at $x = 0$ and $x = -1$, so two.

Check each for a degree-based $y = 0$ and a denominator zero at $x = 3$.

$\frac{1}{x - 3}: \ n < m \Rightarrow y = 0, \quad x - 3 = 0 \Rightarrow x = 3$

The others give $y = 1$, $y = 2$, or no horizontal asymptote, so only $\tfrac{1}{x - 3}$ fits.

Factor and cancel.

$\frac{x + 1}{x^2 - x - 2} = \frac{x + 1}{(x - 2)(x + 1)} = \frac{1}{x - 2} \quad (x \neq -1)$

The factor $x + 1$ cancels, giving a hole at $x = -1$, while $x - 2$ survives as a vertical asymptote at $x = 2$.

Cross-multiply.

$4(x - 2) = 3(x + 1) \;\Rightarrow\; 4x - 8 = 3x + 3 \;\Rightarrow\; x = 11$

Checking, $\frac{4}{12} = \frac{1}{3}$ and $\frac{3}{9} = \frac{1}{3}$, so $x = 11$ is valid.

Factor and cancel the common $x$.

$\frac{2x^2 + 3x}{x} = \frac{x(2x + 3)}{x} = 2x + 3 \quad (x \neq 0)$

The graph is the line $y = 2x + 3$, but $x = 0$ is excluded, leaving a hole at $(0, 3)$.

Multiply every term by $x - 5$.

$x - 2(x - 5) = 5 \;\Rightarrow\; -x + 10 = 5 \;\Rightarrow\; x = 5$

But $x = 5$ makes the denominator zero, so it is extraneous and there is no solution.

The numerator has degree $2$ and the denominator degree $1$, so $n > m$.

$\frac{x^2}{x} = x$

So for very large $x$ the expression behaves like $x$, which grows without bound, and a top-heavy fraction has no horizontal asymptote.