Graphs of Rational Functions: Free Response

The domain comes from the denominator you were handed, before anything cancels. That denominator vanishes at $x=4$, at $x=-3$ and at $x=-1$, and nowhere else, so those three inputs are out and no others are.

Now count copies of each factor, above the bar and below.

$x=4:\ 2 \text{ above},\ 1 \text{ below}; \qquad x=-3:\ 1,\ 2; \qquad x=-1:\ 0,\ 1$

A factor decides its own input by whether it cancels completely. At $x=4$ the two copies above outnumber the single copy below, so cancelling clears $(x-4)$ out of the denominator entirely and the graph keeps an ordinary height there: a hole. At $x=-3$ one copy above cancels one of the two below and a copy survives downstairs, so the values blow up and $x=-3$ is a vertical asymptote, even though $(x+3)$ does appear upstairs. At $x=-1$ there is nothing above to cancel with at all, so that is a vertical asymptote too.

Cancel one copy of $(x-4)$ and one copy of $(x+3)$, remembering that the restrictions stay behind:

$f(x) = \frac{x-4}{(x+3)(x+1)} \qquad (x \neq 4,\; x \neq -3,\; x \neq -1)$

The height of the missing point is the reduced expression evaluated at $x=4$:

$\frac{4-4}{(4+3)(4+1)} = \frac{0}{35} = 0$

so the point the graph is missing is $(4,0)$.

For the $x$-axis, a graph meets it where the reduced numerator is zero at an input the function is allowed to use. Here the reduced numerator $x-4$ is zero only at $x=4$, which is exactly the input that was thrown away, so there is no $x$-intercept anywhere. Notice what that combines with: the original numerator has degree $3$ and the original denominator degree $4$, so the bottom outgrows the top and $y=0$ is the horizontal asymptote. This curve chases the $x$-axis far out in both directions and never once lands on it.

Each edit moves one exponent, so re-count the copies at the input that exponent belongs to, and compare the new pair.

Edit (i) touches $(x+3)$, so look at $x=-3$. The numerator now carries two copies against the denominator's two, so cancelling clears $(x+3)$ out of the denominator entirely and the values no longer blow up there. That input changes from a vertical asymptote to a hole.

Edit (ii) touches $(x-4)$ downstairs, so look at $x=4$. The numerator's two copies now face three, so cancelling leaves one copy behind in the denominator and the values blow up. That input changes from a hole to a vertical asymptote.

Edit (iii) touches $x=4$ as well, but from above: three copies now face one. Cancelling still clears $(x-4)$ out of the denominator, so nothing about the classification moves. The curve itself is different, but the input is a hole before the edit and a hole after it.

Put the three comparisons side by side, in the order the edits were listed:

$2 \ge 2, \qquad 2 < 3, \qquad 3 \ge 1$

Two of the three reversed which side has enough copies, and those are exactly the two that changed a classification. So an edit does not have to make an exponent bigger or smaller to matter, it has to flip that comparison. An edit that leaves the comparison alone may still change the height of a hole, the intercepts and the end behavior, while leaving every input classified exactly as it was.

A square is never negative, so $x^2+3 \ge 3$ for every real $x$ and the denominator never vanishes. No input is excluded, which means no holes and no vertical asymptotes to hunt for.

The numerator has degree $3$ and the denominator degree $2$. The top outgrows the bottom, so the values grow without bound far out and no horizontal line can hold them. The excess is exactly one degree, which is the case that produces a slanted line instead, provided the division leaves something behind.

Divide. The first step takes $x(x^2+3) = x^3+3x$ and leaves $-x^2+3x-9$; the second takes $-1(x^2+3) = -x^2-3$ and leaves $3x-6$, whose degree has dropped below the divisor's, so the division stops.

$f(x) = x-1+\frac{3x-6}{x^2+3}$

Check by multiplying back: $(x-1)(x^2+3)+3x-6 = x^3-x^2+3x-3+3x-6$, which is the numerator we started with. The correction term is a proper fraction whose bottom grows quadratically against a top that grows linearly, so it dies away far out, and what is left is the line $y=x-1$.

Subtracting the quotient from the function leaves nothing but the correction term, so the vertical gap between curve and line is that one fraction:

$f(x)-(x-1) = \frac{3x-6}{x^2+3}$

A fraction is zero exactly where its numerator is zero, and the denominator here is never zero, so the curve meets the line exactly where $3x-6=0$.

$3x-6=0 \;\Longrightarrow\; x=2$

Every real number is in the domain, so this meeting is genuine. The line's height there is $2-1=1$, and evaluating the original rule agrees:

$f(2) = \frac{8-4+12-9}{4+3} = \frac{7}{7} = 1$

So the graph cuts through $y=x-1$ at $(2,1)$, and nowhere else, because $3x-6$ has only that one zero. Past it the curve settles back onto the line, approaching from the other side.

Read the definition again: a line is an asymptote when the distance from the curve to it becomes and stays smaller than any distance you care to name, once you travel far enough out along the curve. That is a claim about the far ends only. It puts no condition at all on what happens near the origin, so a single crossing at $x=2$ leaves it untouched, and after the crossing the gap resumes shrinking.

In general, division writes the function as its quotient plus a correction:

$f(x) = S(x) + \frac{R(x)}{Q(x)}, \qquad \deg R < \deg Q$

The gap between curve and line is $\frac{R(x)}{Q(x)}$, so the curve meets the line exactly at the real zeros of $R$ that lie in the domain. Missing the line entirely therefore means $R$ has no such zero.

A linear $Q$ forces that outcome. The remainder must have degree below $1$, so it is a constant, and there are only two ways that can go. If the constant is zero the division is exact, and then the graph IS the quotient line, with a hole at each cancelled factor: there is no asymptote to discuss, because nothing is left for the curve to approach. Otherwise the constant is nonzero, and a nonzero constant is never zero anywhere, so the curve never touches its slanted line. Once $\deg Q \ge 2$, as here, the remainder is free to have a root, and this one does.

Take the four requirements one at a time and let each hand you a factor and a side of the bar.

A hole at $x=5$ needs $(x-5)$ present on both sides, in equal numbers, so that it cancels completely. A vertical asymptote at $x=-2$ needs $(x+2)$ downstairs with no partner upstairs. An $x$-intercept at $(1,0)$ needs $(x-1)$ upstairs, and it must be the only zero the reduced numerator has. A horizontal asymptote at a nonzero height needs the two degrees to tie, and then the height is the ratio of the leading coefficients, so a constant multiplier out front sets it.

$f(x) = \frac{3(x-5)(x-1)}{(x-5)(x+2)}$

Check the degrees before going further. Expanded, the numerator is $3x^2-18x+15$ and the denominator is $x^2-3x-10$: both degree $2$, with leading coefficients $3$ and $1$, so the ratio is $3$ as required. Nothing about the multiplier disturbs the other three features, since it changes no factor and no zero.

Cancel the shared factor to get the expression that governs every input except the excluded ones:

$f(x) = \frac{3(x-1)}{x+2} \qquad (x \neq 5,\; x \neq -2)$

The hole's height is that expression at $x=5$:

$\frac{3(5-1)}{5+2} = \frac{12}{7}$

so the hole sits at $\left(5,\tfrac{12}{7}\right)$.

Now test the rest. The reduced numerator $3(x-1)$ is zero only at $x=1$, which is in the domain, so $(1,0)$ is the only $x$-intercept. The factor $(x+2)$ never cancelled, so $x=-2$ really is a vertical asymptote rather than a second hole. And the horizontal asymptote comes from the leading coefficients of the original two polynomials, $3$ and $1$, whose ratio is $3$, not from their constant terms $15$ and $-10$. All four requirements hold.

The requirements are not enough, and the quickest way to show it is to build a second function that satisfies every one of them.

$g(x) = \frac{3(x-5)(x-1)^3}{(x-5)(x+2)^3}$

Run the four checks. The shared $(x-5)$ cancels completely, so $x=5$ is a hole. The three copies of $(x+2)$ downstairs have no partner upstairs, so $x=-2$ is a vertical asymptote. The reduced numerator $3(x-1)^3$ is zero only at $x=1$, so $(1,0)$ is the only $x$-intercept. And both polynomials now have degree $4$, with leading coefficients $3$ and $1$, so the horizontal asymptote is still $y=3$.

The two graphs are not the same, and one measurement settles it: the hole's height.

$\frac{3(5-1)^3}{(5+2)^3} = \frac{192}{343}, \qquad \text{against} \qquad \frac{12}{7} = \frac{588}{343}$

The reason is worth naming. The four requirements say which factors must appear and on which side, and they fix the ratio of the leading coefficients. They say nothing about how many copies of each factor appear, and raising both multiplicities together preserves every feature on the list while changing the function between the interesting points.

Divide $3x^2+5x-8$ by $x-2$. The first step takes $3x(x-2)=3x^2-6x$ and leaves $11x-8$; the second takes $11(x-2)=11x-22$ and leaves $14$, a constant, so the division stops.

$f(x) = 3x+11+\frac{14}{x-2}$

Check by multiplying back: $(3x+11)(x-2)+14 = 3x^2-6x+11x-22+14 = 3x^2+5x-8$, the numerator we began with. So the line the division names is $y=3x+11$. The student's shortcut kept the first term of the quotient and dropped the second.

Before anything is drawn, test the excluded input. The numerator at $x=2$ is $12+10-8=14$, which is not zero, so no factor of $(x-2)$ hides upstairs, nothing cancels, and $x=2$ is a vertical asymptote rather than a hole. The remainder $14$ says the same thing: an exact division is what a cancelling factor would have produced.

Subtract each candidate line from the function. Using the decomposition from part A,

$f(x)-(3x+11) = \frac{14}{x-2}, \qquad f(x)-3x = 11+\frac{14}{x-2}$

Take the first. Name any distance $d$, however small. Once $|x-2| > \frac{14}{d}$ the gap $\left|\frac{14}{x-2}\right|$ is below $d$, and it stays below it for every $x$ further out, in both directions. That is exactly what the definition asks, so $y=3x+11$ is an asymptote.

Now the second. Once $|x-2| > 14$ the term $\frac{14}{x-2}$ is smaller than $1$ in size, so the gap lies between $10$ and $12$:

$\left|11+\frac{14}{x-2}\right| > 10 \qquad \text{for every } |x-2| > 14$

At $x=1002$ it is $11.014$, and at $x=-1000$ it is about $10.986$. Going further out only presses the gap closer to $11$, so out there it never shrinks toward zero, which is exactly what the definition demands and does not get.

Nearer the origin the story is different, and it is worth being exact about that rather than overclaiming. The gap does reach zero once, where $11+\frac{14}{x-2}=0$, that is at $x=\frac{8}{11}$, so the curve really does cross $y=3x$, at $\left(\frac{8}{11},\frac{24}{11}\right)$. That crossing settles nothing either way, because a crossing in the middle neither makes a line an asymptote nor stops one being an asymptote. What disqualifies $y=3x$ is its behavior far out, where the gap holds near $11$ instead of shrinking.

The ratio of leading coefficients earns its keep when the degrees tie. Dividing the top and the bottom by $x^n$ puts every non-leading term over a power of $x$, and those terms die away far out, leaving the two leading coefficients alone:

$\frac{ax^n+\cdots}{bx^n+\cdots} = \frac{a+\cdots}{b+\cdots}$

The dots do not mean the same thing on both sides: on the left they are the lower-degree terms of each polynomial, and on the right they are what those terms became, each now carrying a power of $x$ underneath it. Every one of them dies away far out, so the fraction is as close to $\frac{a}{b}$ as you please once $x$ is large enough, and the two leading coefficients are the whole story.

When the top outgrows the bottom, the same division leaves a positive power of $x$ upstairs, so the values grow without bound and there is no height to settle on. The ratio $\frac{a}{b}$ is then only the leading coefficient of the quotient, not the whole story: division writes

$f(x) = \left(\tfrac{a}{b}x + c\right) + \frac{R(x)}{Q(x)}$

and the constant $c$ is invisible to any comparison of leading terms, because it comes from the second step of the division, where the first step's leftover is measured against the divisor.

So the student's line would have been right only if that constant term $c$ had been zero, which happens when subtracting $3x(x-2)$ from the numerator leaves something of degree lower than the divisor. Here it leaves $11x-8$, whose degree is not lower, so a second step was owed and it contributed the $11$. One condition rides along with that, and it is the same one the previous question needed: the division must still leave a nonzero remainder. If it came out exact there would be no oblique asymptote for any line to be, since the graph would then be the quotient line itself, with a hole at each cancelled factor.

A point of the graph lies on the line exactly when $f(x)=\frac{a}{b}$ at an input $x$ that is in the domain. At such an input $Q(x) \neq 0$, so multiplying through by $Q(x)$ neither gains nor loses solutions:

$f(x)=\frac{a}{b} \iff P(x)-\frac{a}{b}Q(x)=0 \qquad (Q(x) \neq 0)$

Write $D(x)=P(x)-\frac{a}{b}Q(x)$ and look at its degree. The only terms of degree $n$ available are $ax^n$ from $P$ and $\frac{a}{b}\cdot bx^n$ from the subtraction, and those are the same quantity:

$ax^n-\frac{a}{b}\,(bx^n) = ax^n-ax^n = 0$

So every term of $D$ that survives has degree $n-1$ or lower, giving $\deg D \le n-1$ whenever $D$ is not the zero polynomial. A nonzero polynomial of degree $d$ has at most $d$ roots, so the graph meets its horizontal asymptote at most $n-1$ times. It may be fewer: a root of $D$ that is a zero of $Q$ is not in the domain and contributes no point at all.

The remaining case is $D$ zero at every input. Then $P=\frac{a}{b}Q$ identically, so $f(x)=\frac{a}{b}$ at every input in the domain. The graph is a horizontal line with a hole at each zero of $Q$, and it does not approach that height, it holds it, so the line is not an asymptote in the sense the definition intends.

Both polynomials have degree $2$, with leading coefficients $3$ and $1$, so the horizontal asymptote is $y=3$. Nothing cancels: the numerator is $-20$ at $x=0$ and $40$ at $x=6$, so both zeros of the denominator are vertical asymptotes and the domain excludes exactly those two inputs.

Form the difference from part A, with $\frac{a}{b}=3$:

$D(x) = (3x^2-8x-20)-3(x^2-6x) = 10x-20$

The degree-$2$ terms have destroyed each other, as promised, leaving degree $1$.

$10x-20=0 \;\Longrightarrow\; x=2$

The input $x=2$ is neither $0$ nor $6$, so it is in the domain, and the height there checks out:

$f(2) = \frac{12-16-20}{4-12} = \frac{-24}{-8} = 3$

So the graph passes through $(2,3)$, a point sitting exactly on its own horizontal asymptote, and there is no second such point. With $n=2$ part A allows at most $n-1=1$, so this example shows the bound cannot be lowered for degree $2$.

If $x=c$ is a vertical asymptote of $f=\frac{P}{Q}$, then $c$ is a zero of $Q$ whose factor survives the cancelling, and a zero of the original denominator is excluded whatever else happens:

$Q(c)=0 \;\Longrightarrow\; c \notin \text{domain}(f)$

The graph is the set of points $(x,f(x))$ built from inputs $x$ in the domain, so no point of it has first coordinate $c$ at all. There is simply nothing above $x=c$ to cross the line with. Every point of the graph sits at some positive horizontal distance $|x-c|$ from it, and although that distance can be made as small as you please, it is never zero.

Part A's method cannot be transplanted, and the reason is a difference in kind between the two lines. A horizontal line prescribes an OUTPUT, so "the graph meets it" becomes the equation $f(x)=\frac{a}{b}$, and an equation can be cleared of denominators and its roots counted. A vertical line prescribes an INPUT, so there is no equation in $x$ to solve: the question is not what value $f$ takes at $c$, but whether $f$ has a value at $c$ at all, and by construction it does not. That is also why the two claims feel so different. A horizontal asymptote is a statement about the far ends of the curve and leaves the middle free; a vertical asymptote is a statement about which inputs exist, and that admits no exceptions anywhere.