Rational Functions: Free Response

Set the denominator equal to zero and factor.

$x^2 - x - 6 = (x-3)(x+2) = 0 \;\Longrightarrow\; x = 3 \text{ or } x = -2.$

Those are the only two inputs a rational function ever excludes, so the domain is every real number except $3$ and $-2$. To rule out a hole at either one, check the numerator there instead of assuming: $2(3)+5 = 11 \neq 0$ and $2(-2)+5 = 1 \neq 0$. Neither factor of the denominator survives in the numerator, so nothing cancels, and both $x=3$ and $x=-2$ are genuine vertical asymptotes.

The numerator has degree $1$ and the denominator has degree $2$, so $n<m$ and the horizontal asymptote is

$y = 0.$

For the x-intercept, set the numerator to zero: $2x+5=0$ gives $x = -\dfrac{5}{2}$. Check the denominator there: $\left(-\dfrac{5}{2}\right)^2 - \left(-\dfrac{5}{2}\right) - 6 = \dfrac{11}{4} \neq 0$, so this input is in the domain and the x-intercept is $\left(-\dfrac{5}{2}, 0\right)$.

The horizontal asymptote from part B is $y=0$, and the x-intercept from part B is $\left(-\dfrac{5}{2},0\right)$, a point whose height is also $0$. So the graph passes through a point sitting exactly on its own horizontal asymptote.

$y = 0 \quad \text{(horizontal asymptote)}, \qquad \left(-\tfrac{5}{2}, 0\right) \quad \text{(x-intercept)}$

This is allowed because a horizontal asymptote is a claim about what happens as $x$ runs far out to the left or right; it makes no promise about any single finite $x$, so the curve is free to touch or cross that line somewhere in the middle of its domain before settling toward it at the far edges.

A vertical asymptote works differently: it sits at an input where the function is not even defined, since the denominator vanishes there. There is no output to plot at that $x$ at all, so the curve cannot pass through a vertical asymptote the way it just did through this horizontal one.

Match the rule to $f(x)=\dfrac{a}{x-h}+k$: here $h=5$ and $k=1$ (with $a=4$). The denominator's zero moves the vertical asymptote from $x=0$ to

$x = 5,$

and the added constant moves the horizontal asymptote from $y=0$ to

$y = 1.$

The two asymptotes cross at the center, $(5,1)$.

Evaluate at $x=0$:

$g(0) = \frac{4}{0-5}+1 = -\frac{4}{5}+1 = \frac{1}{5}.$

So the graph crosses the y-axis at $\left(0,\tfrac{1}{5}\right)$.

Setting $g(x)$ equal to its own horizontal asymptote, $1$, and solving:

$\frac{4}{x-5}+1 = 1 \;\Longrightarrow\; \frac{4}{x-5} = 0.$

A fraction is $0$ only when its numerator is $0$, and the numerator here is the constant $4$, which is never $0$ regardless of $x$. So no real $x$ solves this equation, and the branches of $g$ never climb or fall back to height $1$.

That conclusion belongs to this particular shape, a nonzero constant over $x-h$, plus $k$: the numerator can never vanish, so the horizontal asymptote can never be reached. It is not a rule for every rational function. A rational function in general CAN cross its horizontal asymptote at some ordinary $x$; only a vertical asymptote, where the function is undefined, is never crossed.

Multiply both sides by $(x-1)(x+2)$ to clear the denominators:

$4(x+2) = 3(x-1).$

Expand and solve:

$4x+8 = 3x-3 \;\Longrightarrow\; x = -11.$

Check against the original equation's excluded values, $x=1$ and $x=-2$: since $-11$ is neither, it belongs to the domain, and substituting it back confirms both sides agree. So $x=-11$ is a genuine solution.

Multiply every term by $x-4$:

$2x - 3(x-4) = 8.$

Expand and solve:

$2x - 3x + 12 = 8 \;\Longrightarrow\; -x = -4 \;\Longrightarrow\; x = 4.$

Check it against the original equation's excluded value: $x=4$ is exactly the input that makes $x-4=0$, so it was never in the domain to begin with. The candidate the algebra produced is extraneous, and the equation has no solution.

Clearing a denominator means multiplying both sides by an expression built from $x$. That expression can equal $0$ for some input, and

$0 \times k = 0 \text{ for every number } k,$

so multiplying both sides of any equation by $0$ turns them into the same true statement, whether or not the two original sides were equal. That is why the step is not reversible: a candidate can satisfy the cleared equation for exactly this reason, without ever having solved the original one.

Part B is exactly this: the candidate $x=4$ solved the version with denominators cleared, but $x=4$ was never in the domain of the original equation at all. Part A shows the other outcome is just as possible, since $x=-11$ passed the same test cleanly.

Because either outcome can happen, and because nothing about a candidate's size, sign, or how it was reached predicts which one occurred, the only dependable check is substituting each candidate directly into the ORIGINAL equation and confirming both sides agree there.

Factor the numerator and the denominator completely.

$x^2+2x-8 = (x+4)(x-2), \qquad x^2-16 = (x+4)(x-4).$

The factor $x+4$ appears in BOTH, so it cancels; the reduced form is $\dfrac{x-2}{x-4}$ for $x\neq -4$. Because $x+4$ cancelled, $x=-4$ is a hole, not a vertical asymptote. The other factor, $x-4$, appears only in the denominator, so it survives, and $x=4$ is a genuine vertical asymptote.

The student never checked the numerator at either excluded value: both denominator zeros were treated the same way, when only the one whose factor does NOT also appear in the numerator earns the asymptote.

The reduced function, valid everywhere except $x=-4$ (where the factor $x+4$ cancelled), is

$\frac{x-2}{x-4}.$

At $x=4$: the factor $x-4$ never cancelled, so this is a genuine vertical asymptote; the reduced form is undefined there just as the original was.

At $x=-4$: the factor $x+4$ DID cancel, so this excluded value is a hole. Its height is the reduced form evaluated there:

$\frac{-4-2}{-4-4} = \frac{-6}{-8} = \frac{3}{4}.$

So the hole sits at $\left(-4, \tfrac{3}{4}\right)$, while $x=4$ remains a genuine vertical asymptote.

The student's rule was: every zero of the denominator is a vertical asymptote. Part A shows that rule is false in general, since $x=-4$ was a counterexample right inside the same problem.

The corrected rule has to check BOTH polynomials, not just one:

$x=a \text{ is a vertical asymptote} \iff (x-a) \text{ does not cancel with the numerator}.$

So the fix is procedural: factor the numerator and the denominator completely, cancel every shared factor first, and only THEN read the vertical asymptotes off whatever is left in the denominator. Skipping the cancellation step is exactly what produced the student's error.

Write $p(x) = a x^n + (\text{lower-degree terms})$ and $q(x) = b x^m + (\text{lower-degree terms})$, and divide every term of both by $x^m$.

$\frac{p(x)}{x^m} = a x^{n-m} + (\text{terms with exponent} < n-m), \qquad \frac{q(x)}{x^m} = b x^{0} + (\text{terms with exponent} < 0).$

The denominator's leading term $bx^m$ becomes exactly $bx^{m-m}=b$, a constant. The numerator's leading term $ax^n$ becomes $ax^{n-m}$, and every other term of $p(x)$, having a smaller original degree than $n$, ends up with an exponent smaller than $n-m$ after the same division.

From part A,

$f(x) = \frac{ax^{n-m} + (\text{terms with exponent} < n-m)}{b + (\text{terms with exponent} < 0)}.$

Every term in the denominator besides $b$ has a negative exponent, so as $x$ grows without bound each one shrinks toward $0$ exactly the way $\dfrac{1}{x}$ did earlier in this lesson. So the denominator always settles at $b$, in all three cases; only the numerator's behavior can differ.

If $n<m$: the exponent $n-m$ is negative, so the numerator's leading term $ax^{n-m}$ shrinks to $0$ along with every smaller term. The whole numerator settles at $0$, so

$f(x) \to \frac{0}{b} = 0.$

If $n=m$: the exponent $n-m$ is $0$, so $ax^{n-m}=a$ stays fixed while every other numerator term (with negative exponent, since it was smaller than the leading term) shrinks to $0$. The numerator settles at $a$, so

$f(x) \to \frac{a}{b}.$

If $n>m$: the exponent $n-m$ is positive, so the leading term $ax^{n-m}$ grows without bound as $x$ grows. Dividing an ever-growing numerator by the fixed denominator $b$ still grows without bound, so $f(x)$ never settles at any fixed height, and there is no horizontal asymptote.

For $h$, the numerator $5x^3-2$ has degree $n=3$ and leading coefficient $a=5$; the denominator $2x^3+7x$ has degree $m=3$ and leading coefficient $b=2$. Since $n=m$, part B's middle case applies:

$y = \frac{a}{b} = \frac{5}{2}.$

Dividing every term by $x^3$ turns the numerator into $5 - 2x^{-3}$ and the denominator into $2+7x^{-2}$; the terms $-2x^{-3}$ and $7x^{-2}$ both carry negative exponents, so they are exactly the ones part B says shrink to $0$, leaving the constants $5$ and $2$ behind.