Graphs of Rational Functions: Additional Practice (Set 1)

Factor both parts. The numerator is $(x+3)(x-2)$ and the denominator is $x(x+3)$, so the domain excludes $0$ and $-3$.

$f(x) = \frac{(x+3)(x-2)}{x(x+3)} = \frac{x-2}{x} \qquad (x \neq 0,\; x \neq -3)$

The factor $(x+3)$ cancels completely, so $x=-3$ is a hole. Its height is the simplified expression evaluated there.

$\frac{-3-2}{-3} = \frac{-5}{-3} = \frac{5}{3}$

The hole is at $\left(-3, \frac53\right)$, and $x=0$ is a vertical asymptote.

The numerator's degree is one more than the denominator's, so divide. The first step gives $x$, since $x(x+3) = x^2+3x$, and subtracting leaves $-3x-5$. The second step gives $-3$, since $-3(x+3) = -3x-9$, and subtracting leaves the remainder $4$.

$f(x) = x - 3 + \frac{4}{x+3}$

Check by multiplying back: $(x-3)(x+3) + 4 = x^2-9+4 = x^2-5$. The correction term dies away far out, so the oblique asymptote is $y = x-3$.

Factor both parts. The numerator is $x(x-2)$ and the denominator is $(x-2)(x+2)$, so the domain excludes $2$ and $-2$.

$f(x) = \frac{x(x-2)}{(x-2)(x+2)} = \frac{x}{x+2} \qquad (x \neq 2,\; x \neq -2)$

The factor $(x-2)$ cancels completely, so $x=2$ is a hole, at height $\frac{2}{2+2} = \frac12$. The factor $(x+2)$ survives, so $x=-2$ is a vertical asymptote. Note that $x=2$ is not an $x$-intercept: it is not in the domain.

The degrees tie and both leading coefficients are $1$, so the horizontal asymptote is $y=1$. Set the function equal to $1$ and clear the denominator, which is never zero.

$x^2 - 4x = x^2 + 4 \;\Longrightarrow\; -4x = 4 \;\Longrightarrow\; x = -1$

Check it: $f(-1) = \frac{1+4}{1+4} = 1$, so the curve passes through $(-1,1)$, a point on the asymptote. The point $(4,1)$ is a trap: $x=4$ is an $x$-intercept, where $f(4) = 0$, not $1$.

Factor the denominator: $x^3 - x = x(x-1)(x+1)$, so the domain excludes $0$, $1$, and $-1$: three candidates.

$f(x) = \frac{x+1}{x(x-1)(x+1)} = \frac{1}{x(x-1)} \qquad (x \neq 0,\; x \neq 1,\; x \neq -1)$

The factor $(x+1)$ cancels completely, so $x=-1$ is a hole, at height $\frac{1}{(-1)(-2)} = \frac12$. The factors $x$ and $(x-1)$ survive in the denominator, so $x=0$ and $x=1$ are vertical asymptotes.

Nothing cancels, so $x=4$ is a vertical asymptote and the values run off in size on both sides. The sign is what needs care, and the squared factor decides it: $(x-4)^2$ is positive on both sides of $4$, never negative, and near $x=4$ the numerator is close to $5$, also positive.

$f(3.9) = \frac{4.9}{0.01} = 490, \qquad f(4.1) = \frac{5.1}{0.01} = 510$

A positive over a positive is positive, so the graph rises without bound on both sides. An odd power in the denominator would flip the sign as you crossed it, but an even power does not.

A vertical asymptote needs a denominator factor that survives the cancelling. Check the third choice.

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

The only denominator factor cancels completely, so that graph is a line with a hole at $(2,4)$ and no asymptote anywhere. The others all keep a factor downstairs: $\frac{x-2}{(x-2)(x+2)} = \frac{1}{x+2}$ has one at $x=-2$, $\frac{1}{x^2-4}$ has two, and $\frac{x+2}{(x+2)^2} = \frac{1}{x+2}$ has one at $x=-2$.

Factor both parts. The numerator is $x(x^2-2x+1) = x(x-1)^2$ and the denominator is $(x-1)(x+1)$, so the domain excludes $1$ and $-1$.

$f(x) = \frac{x(x-1)^2}{(x-1)(x+1)} = \frac{x(x-1)}{x+1} \qquad (x \neq 1,\; x \neq -1)$

The denominator's copy of $(x-1)$ cancels completely (the numerator had two to spare), so $x=1$ is a hole. Its height is the simplified expression there, $\frac{1(0)}{2} = 0$, so the hole sits at $(1,0)$, right on the $x$-axis. That point is still a hole and not an $x$-intercept, because $x=1$ is not in the domain. The factor $(x+1)$ survives, so $x=-1$ is a vertical asymptote.

The numerator $x^2+4$ is never zero for a real $x$, so there are no $x$-intercepts, and it shares no factor with the denominator $(x-2)(x+2)$, so there are no holes: $x=2$ and $x=-2$ are vertical asymptotes. The degrees tie, so the horizontal asymptote is $y = \frac 11 = 1$, not $-1$ (the value $-1$ is the $y$-intercept, $f(0) = \frac{4}{-4}$).

$\frac{x^2+4}{x^2-4} = 1 \;\Longrightarrow\; x^2+4 = x^2-4 \;\Longrightarrow\; 4 = -4$

That is false for every $x$, so the graph never meets its horizontal asymptote.

The degree comparison ($2$ against $1$) suggests an oblique asymptote, but factor before you trust it.

$f(x) = \frac{(3x-2)(2x+1)}{2x+1} = 3x-2 \qquad \left(x \neq -\tfrac12\right)$

The denominator cancels completely, so there is no vertical asymptote and no oblique asymptote either: the graph is exactly the line $y = 3x-2$, with the single point at $x=-\frac12$ removed. Its height there is $3\left(-\frac12\right) - 2 = -\frac72$, so the hole is at $\left(-\frac12, -\frac72\right)$. A line is its own best approximation, and a curve does not need an asymptote when it already is one.

The degrees are $3$ and $2$, a gap of one, so there is an oblique asymptote and long division will produce it. The first step gives $2x$, since $2x(x^2+1) = 2x^3+2x$, and subtracting leaves $x^2-2x-1$. The second step gives $1$, since $1(x^2+1) = x^2+1$, and subtracting leaves the remainder $-2x-2$.

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

The correction term is a proper fraction, so it dies away far out (its top has degree $1$ and its bottom degree $2$), and the curve settles onto the line $y = 2x+1$.

The denominator factors as $(x-1)(x-2)$, and the numerator $x^2+1$ has no real zeros, so it shares no factor with it. Nothing cancels, and both excluded inputs give vertical asymptotes: $x=1$ and $x=2$.

$\frac{x^2+1}{x^2-3x+2} = \frac{1 + \dfrac{1}{x^2}}{\,1 - \dfrac{3}{x} + \dfrac{2}{x^2}\,}$

Far out, every term with an $x$ underneath it dies away, leaving the ratio of the leading coefficients, $\frac 11 = 1$. The degrees tie, so the horizontal asymptote is $y=1$. (This curve does cross that asymptote: setting $x^2+1 = x^2-3x+2$ gives $x = \frac13$, and $f\left(\frac13\right) = 1$.)