Graphs of Rational Functions: Practice

A hole at $x=2$ needs the factor $(x-2)$ to cancel, so $(x-2)$ must divide the numerator. By the factor theorem, that happens exactly when the numerator is zero at $x=2$.

$2^2 + 2k - 6 = 0 \;\Longrightarrow\; 2k - 2 = 0 \;\Longrightarrow\; k = 1$

Check it: with $k=1$ the numerator is $x^2+x-6 = (x+3)(x-2)$, so $f(x) = x+3$ for $x \neq 2$, a line with a hole at $(2,5)$. Any other $k$ leaves a nonzero remainder, so the factor stays in the denominator and $x=2$ is an asymptote.

Look at the denominator first: $x^2+9$ is never zero for a real $x$, since $x^2 = -9$ has no real solution. Nothing is excluded from the domain, so there are no vertical asymptotes and no holes. The zeros $x=3$ and $x=-3$ belong to the numerator, so they are $x$-intercepts.

$\frac{x^2-9}{x^2+9} = \frac{1 - \dfrac{9}{x^2}}{\,1 + \dfrac{9}{x^2}\,}$

Dividing top and bottom by $x^2$ shows what happens far out: the terms $\frac{9}{x^2}$ die away and the ratio of the leading coefficients, $\frac 11 = 1$, is left. The degrees tie, so the horizontal asymptote is $y=1$. It is never crossed, because $x^2-9 = x^2+9$ would force $-9 = 9$. At $x=0$ the graph is at $f(0) = -1$, nowhere near the asymptote.

The sign can change only at the zero of the numerator, $x=-1$, and at the vertical asymptotes, $x=-3$ and $x=2$. Those three values cut the line into four intervals, so test one point in each.

$f(-4) = \frac{-3}{(-6)(-1)} < 0, \quad f(-2) = \frac{-1}{(-4)(1)} > 0, \quad f(0) = \frac{1}{(-2)(3)} < 0, \quad f(3) = \frac{4}{(1)(6)} > 0$

The function is positive on the second and fourth intervals, so the complete solution set is $-3 < x < -1$ or $x > 2$.

Factor both parts and count copies of $(x-1)$: one in the numerator, two in the denominator.

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

Cancelling uses up the numerator's only copy and still leaves one $(x-1)$ in the denominator, so the values blow up near $x=1$ and the graph has a vertical asymptote there. A factor has to cancel completely to give a hole. (The $x$-intercept is at $x=-1$, where the surviving numerator is zero.)

The oblique asymptote is the quotient of the long division, so run the division on the last choice. Dividing $2x^2+x+2$ by $x+1$: the first step gives $2x$ (since $2x(x+1) = 2x^2+2x$, leaving $-x+2$), and the second gives $-1$ (since $-1(x+1) = -x-1$, leaving the remainder $3$).

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

Check by multiplying back: $(2x-1)(x+1) + 3 = 2x^2+x-1+3 = 2x^2+x+2$. The other choices divide out to quotients $2x-3$, $2x+3$, and $x$, so their asymptotes are different lines.

Take the requirements one at a time. Vertical asymptotes at $x=1$ and $x=-4$ need the denominator $(x-1)(x+4)$, which rules out the first choice, whose asymptotes are at $-1$ and $4$. The only $x$-intercept at $3$ needs the numerator to be a power of $(x-3)$, which rules out the last choice, whose intercept is at $-3$.

$y = \frac{\text{leading coefficient of } P}{\text{leading coefficient of } Q} = \frac{2}{1} = 2$

The horizontal asymptote $y=2$ needs equal degrees with the leading coefficients in the ratio $2$ to $1$, which the second choice fails (its ratio is $1$ to $1$, giving $y=1$). Only $\frac{2(x-3)^2}{(x-1)(x+4)}$ satisfies all three.

Factor: the numerator is $3(x+2)$ and the denominator is $(x+2)(x-1)$, so the domain excludes $-2$ and $1$.

$f(x) = \frac{3(x+2)}{(x+2)(x-1)} = \frac{3}{x-1} \qquad (x \neq -2,\; x \neq 1)$

The factor $(x+2)$ cancels completely, so $x=-2$ is a hole, at height $\frac{3}{-2-1} = -1$. The factor $(x-1)$ survives, so $x=1$ is a vertical asymptote. The simplified numerator is the constant $3$, which is never zero, so there are no $x$-intercepts at all, and the degree comparison ($0$ against $1$) gives the horizontal asymptote $y=0$.

The denominator factors as $(x-2)(x+2)$. Neither factor divides the numerator (at $x=2$ the numerator is $16-2=14$, and at $x=-2$ it is $-14$), so nothing cancels and both give vertical asymptotes. The degrees are $3$ and $2$, a gap of one, so there is an oblique asymptote as well. Divide.

$\frac{2x^3-x}{x^2-4} = 2x + \frac{7x}{x^2-4}$

Check by multiplying back: $2x(x^2-4) + 7x = 2x^3-8x+7x = 2x^3-x$. The correction term dies away far out, so the oblique asymptote is $y=2x$. (Its remainder $7x$ is zero at $x=0$, so this curve actually crosses that asymptote, at the origin.)

When the degrees tie, dividing the top and the bottom by $x^4$ kills every term except the leading ones, and the horizontal asymptote is the ratio of the leading coefficients.

$y = \frac{-6}{3} = -2$

Nothing else about $P$ and $Q$ matters for the end behavior: not their constant terms, not their roots, and not any common factor, since cancelling one would lower both degrees and divide both leading coefficients by the same number, leaving the ratio unchanged.

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

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

The factor $(x-3)$ cancels completely, so $x=3$ is a hole, at height $\frac{3+2}{3(6)} = \frac{5}{18}$. The factors $x$ and $(x+3)$ survive, so the vertical asymptotes are $x=0$ and $x=-3$: two of them.

Factor both parts. The numerator is a difference of squares, $(2x-1)(2x+1)$, and the denominator factors as $(2x-1)(x-2)$.

$f(x) = \frac{(2x-1)(2x+1)}{(2x-1)(x-2)} = \frac{2x+1}{x-2} \qquad \left(x \neq \tfrac 12,\; x \neq 2\right)$

The factor $(2x-1)$ cancels completely, and it is zero at $x = \frac 12$, so that is where the hole is. Its height is the simplified expression evaluated there.

$\frac{2\left(\frac12\right)+1}{\frac12 - 2} = \frac{2}{-\frac{3}{2}} = -\frac{4}{3}$

The hole is at $\left(\frac12, -\frac43\right)$, and $x=2$ is a vertical asymptote.

Nothing cancels, so $x=2$ is a vertical asymptote and the values must run off to infinity in size. The only question is the sign, and the sign chart settles it: just to the left of $2$ the numerator $x-3$ is about $-1$ (negative) and the denominator $x-2$ is a tiny negative number, so the quotient is a large positive number.

$f(1.9) = \frac{-1.1}{-0.1} = 11, \qquad f(1.99) = \frac{-1.01}{-0.01} = 101$

The values climb past any height you name, so the graph rises without bound on that side. (On the right of $2$ the denominator is a tiny positive, so the graph falls without bound instead.)