Rational Functions

Approach: For a rational function the domain excludes exactly the zeros of the denominator, and those same points carry the vertical asymptotes as long as the numerator does not vanish there. The horizontal asymptote is read off by comparing the degrees of numerator and denominator.

Steps:

1. The denominator is $x - 2$, so $f$ is undefined at $x = 2$ and the domain is $\mathbb{R} \setminus \{2\}$.
2. At $x = 2$ the numerator equals $1 \neq 0$, so $f$ is unbounded there and $x = 2$ is a vertical asymptote.
3. The numerator has degree $0$ and the denominator degree $1$, so the numerator degree is smaller and the horizontal asymptote is $y = 0$.
4. End behavior: $f(x) \to -\infty$ as $x \to 2^-$ and $f(x) \to +\infty$ as $x \to 2^+$; as $x \to +\infty$ the graph approaches $0$ from above, and as $x \to -\infty$ it approaches $0$ from below.

Answer:

Domain: $\mathbb{R} \setminus \{2\}$. Vertical asymptote: $x = 2$. Horizontal asymptote: $y = 0$.

Approach: Factor numerator and denominator and cancel the common factor only to read the shape of the graph. Keep the domain of the original expression: a cancelled factor leaves a hole in the graph, not a vertical asymptote.

Steps:

1. Factor the numerator: $x^2 - 1 = (x-1)(x+1)$, so for every $x \neq 1$ we get $f(x) = \dfrac{(x-1)(x+1)}{x-1} = x + 1$.
2. The original expression is undefined at $x = 1$, so the domain is $\mathbb{R} \setminus \{1\}$.
3. The graph is the line $y = x+1$ with one point deleted. The deleted point is read from the limit: $\lim\limits_{x\to1} f(x) = 1 + 1 = 2$, so the hole sits at $(1, 2)$.
4. Because the zero of the denominator at $x = 1$ is cancelled by the numerator, there is no vertical asymptote there — the discontinuity is removable. Since the numerator has degree $2$ and the denominator degree $1$, there is no horizontal asymptote either, and the graph follows the line $y = x+1$ itself apart from the deleted point.

Answer:

$f(x) = x + 1$ for every $x \neq 1$; the domain is $\mathbb{R} \setminus \{1\}$, and the graph is the line $y = x+1$ with a removable hole at $(1, 2)$.

Approach:

For a rational function the domain excludes exactly the zeros of the denominator, and those same points are where the vertical asymptotes stand.

Steps:

1. Factor the denominator: $x^2 - 4 = (x-2)(x+2)$, so the domain excludes $x = \pm 2$.
2. Since the numerator $2x + 3$ does not vanish at $x = \pm 2$ (it equals $7$ and $-1$ respectively), the function really blows up there.
3. Hence the vertical asymptotes are the lines $x = -2$ and $x = 2$.

Answer:

$x^2-4 = (x-2)(x+2)$ ⇒ domain excludes $\pm2$. Vertical asymptotes at $x = -2$ and $x = 2$.

Approach:

Divide the numerator by the denominator; the quotient gives the linear part of the function, and the remainder over the denominator tends to zero at infinity, so the quotient is the asymptote.

Steps:

1. Long divide: $x^2 + 1 = (x-1)(x+1) + 2$.
2. Therefore $f(x) = x + 1 + \dfrac{2}{x-1}$.
3. As $x \to \pm\infty$ the remainder term $\dfrac{2}{x-1}$ tends to $0$, so the graph approaches the line $y = x + 1$.

Answer:

Long divide: $x^2 + 1 = (x-1)(x+1) + 2$. So $f(x) = x + 1 + \dfrac{2}{x-1}$ ⇒ oblique asymptote $y = x + 1$.