Rational Functions: Practice

Factor the denominator, a difference of squares, and set each factor to zero.

$x^2 - 9 = (x - 3)(x + 3) = 0 \;\Rightarrow\; x = 3 \text{ or } x = -3$

Both are excluded, so the domain is every real number except $3$ and $-3$.

Compare the degrees: the top has degree $1$ and the bottom degree $2$, so $n < m$.

$y = 0$

The faster-growing denominator drives the ratio toward zero for large $x$.

Factor the denominator and set each factor to zero.

$x^2 + x - 6 = (x - 2)(x + 3) = 0 \;\Rightarrow\; x = 2 \text{ or } x = -3$

The domain is every real number except $2$ and $-3$.

The factor $x - 3$ appears on both the top and the bottom, so it cancels.

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

A cancelled factor leaves a hole at $x = 3$, not an asymptote.

In the form $\tfrac{a}{x - h} + k$, the constant $k$ lifts the horizontal asymptote from $y = 0$.

$y = k = 4$

The vertical asymptote sits where the denominator is zero.

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

The y-intercept is the value at $x = 0$.

$f(0) = \frac{0 + 6}{0 - 2} = \frac{6}{-2} = -3$

So the graph crosses the y-axis at $(0, -3)$.

Set the numerator to zero and check the denominator is not zero there.

$2x - 8 = 0 \;\Rightarrow\; x = 4$

At $x = 4$ the denominator is $5 \neq 0$, so the x-intercept is $(4, 0)$.

Factor and cancel first.

$\frac{x + 2}{x^2 - 4} = \frac{x + 2}{(x - 2)(x + 2)} = \frac{1}{x - 2} \quad (x \neq -2)$

The factor $x + 2$ cancels (a hole at $x = -2$), so the only vertical asymptote is $x = 2$.

Match $\tfrac{a}{x - h} + k$: here $h = 3$ and $k = -1$.

$(h, k) = (3, -1)$

The asymptotes $x = 3$ and $y = -1$ cross at $(3, -1)$.

The top has the higher degree, $3$ against $2$, so $n > m$.

$\text{top-heavy} \;\Rightarrow\; \text{no horizontal asymptote}$

The fraction grows without bound for large $x$.

The degrees match at $2$, so take the ratio of leading coefficients.

$y = \frac{5}{2}$