Graphs of Rational Functions: Practice

Factor both parts completely.

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

The factor $(x+3)$ appears once on top and once on the bottom, so it cancels completely and $x=-3$ is a hole. The height of the hole is the value of the simplified expression there.

$\frac{-3-4}{-3-3} = \frac{-7}{-6} = \frac{7}{6}$

So the hole is at $\left(-3, \frac 76\right)$.

The original denominator $(x-3)(x+3)$ is zero at $x=3$ and $x=-3$, so both are excluded from the domain. Which of them gives an asymptote depends on what cancels.

$f(x) = \frac{(x-4)(x+3)}{(x-3)(x+3)} = \frac{x-4}{x-3} \qquad (x \neq 3,\; x \neq -3)$

The factor $(x+3)$ cancelled completely, so $x=-3$ is a hole, not an asymptote. The factor $(x-3)$ survives in the denominator, so the only vertical asymptote is $x=3$.

This is the trap the lesson warns about: the factor $(x-2)$ does appear in the numerator, but appearing is not enough. Count the copies, one on top and two on the bottom, and cancel.

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

One copy of $(x-2)$ is still sitting in the denominator, so the values still blow up near $x=2$. The factor did not cancel completely, so $x=2$ is a vertical asymptote, not a hole.

The degrees tie, so the horizontal asymptote is the ratio of the leading coefficients, $y = 3$. To find a crossing, set the function equal to $3$ and clear the denominator (which is never zero, so no solutions are lost).

$3x^2 + 2 = 3(x^2+1) \;\Longrightarrow\; 3x^2 + 2 = 3x^2 + 3 \;\Longrightarrow\; 2 = 3$

That is false for every $x$, so the equation has no solutions and the graph never meets $y=3$. A curve is allowed to cross a horizontal asymptote, but this particular one does not.

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.

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

Check it: $f(2) = \frac{4+4}{4+4} = 1$, so the curve really does pass through $(2,1)$, a point on the asymptote. The point $(0,0)$ is on the graph, but it is not on the line $y=1$.

Factor and cancel first, because a zero of the numerator only counts if it is still in the domain.

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

The original denominator excludes $x=2$, so $x=2$ can never be an $x$-intercept even though it is a zero of the original numerator: it is a hole, at height $\frac{2+2}{2+1} = \frac 43$. The surviving numerator is zero at $x=-2$, which is in the domain, so the only $x$-intercept is $(-2,0)$.

Factor the top and the bottom.

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

The factor $(x-3)$ cancels completely, so $x=3$ is a hole, at height $\frac{2}{3+3} = \frac 13$. The factor $(x+3)$ survives, so $x=-3$ is a vertical asymptote. Note that $x=3$ is not an $x$-intercept: it is not in the domain at all.

Divide $3x^2-2x+1$ by $x+2$. The first step gives $3x$, since $3x(x+2) = 3x^2+6x$, and subtracting leaves $-8x+1$. The next step gives $-8$, since $-8(x+2) = -8x-16$, and subtracting leaves the remainder $17$.

$f(x) = 3x - 8 + \frac{17}{x+2}$

Check by multiplying back: $(3x-8)(x+2) + 17 = 3x^2-2x-16+17 = 3x^2-2x+1$. The correction term dies away far out, so the oblique asymptote is $y = 3x-8$.

Factor everything. The denominator is $x^3 - x = x(x-1)(x+1)$ and the numerator is $(x-1)(x+1)$, so the domain excludes $0$, $1$, and $-1$.

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

Both $(x-1)$ and $(x+1)$ cancel completely, so $x=1$ and $x=-1$ are holes (at heights $1$ and $-1$). Only the factor $x$ survives in the denominator, so there is exactly one vertical asymptote, $x=0$.

The numerator has degree $2$ and the denominator degree $1$, so expect an oblique asymptote and divide. Dividing $x^2+1$ by $x-1$ gives quotient $x+1$ and remainder $2$, since $(x+1)(x-1) + 2 = x^2 - 1 + 2 = x^2+1$.

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

The correction $\frac{2}{x-1}$ shrinks toward nothing as $x$ runs off in either direction, so the curve settles onto the line $y = x+1$.

The sign can change only at the zero of the numerator, $x=-4$, and at the vertical asymptote, $x=2$. Test one point in each of the three intervals they create.

$f(-5) = \frac{-1}{-7} > 0, \qquad f(0) = \frac{4}{-2} < 0, \qquad f(3) = \frac{7}{1} > 0$

The function is positive on the outer two intervals, so the complete solution set is $x < -4$ or $x > 2$. Naming only one of them leaves out half the answer.

Build the features one at a time. A hole at $x=-1$ needs $(x+1)$ to cancel completely, and a vertical asymptote at $x=3$ needs $(x-3)$ to survive.

$\frac{x+1}{(x+1)(x-3)} = \frac{1}{x-3} \qquad (x \neq -1,\; x \neq 3)$

That leaves degree $1$ underneath and degree $0$ on top, so the horizontal asymptote is $y=0$, as required. The second choice cancels $(x-3)$ instead, swapping the two features; the third leaves a copy of $(x+1)$ downstairs, so $x=-1$ becomes a second asymptote; and the fourth has no common factor at all, so it has no hole and no horizontal asymptote.