Rational Expressions and Functions: Chapter Test

An expression of this kind is undefined exactly where its denominator is zero, so factor the denominator and set each factor to zero. Two numbers with product $-63$ and sum $2$ are $9$ and $-7$.

$x^2+2x-63 = (x+9)(x-7)$

That product is zero when $x = -9$ or when $x = 7$. The numerator's zero, $x = 2$, costs nothing: there the expression is $\frac{0}{-55} = 0$, a perfectly good number.

Factor the numerator as a difference of squares, then turn the opposite factor around so that the common factor is visible.

$16-x^2 = (4-x)(4+x) = -(x-4)(x+4)$

The printed denominator is zero only at $x = 4$, so that is the one excluded value. On the rest of the domain $x-4$ is nonzero and may be divided out:

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

A check settles the sign. At $x = 0$ the printed expression is $\frac{16}{-4} = -4$, and $-(0+4) = -4$, while $0+4 = 4$ is wrong.

Count the copies of $x+4$ on each side of the bar: one above and one below, so it cancels completely.

$f(x) = \frac{x-9}{x+1} \qquad (x \neq -4, \; x \neq -1)$

Nothing is left of $x+4$ in the denominator, so the values do not blow up near $-4$; the reduced expression sails through at height $\frac{-13}{-3} = \frac{13}{3}$. The original is still undefined at $-4$, so exactly one point is missing: a hole.

Factor every polynomial before multiplying anything out. The first numerator needs two numbers with product $8$ and sum $9$, and the first denominator has a common factor of $x$:

$\frac{(x+8)(x+1)}{x(x-3)} \cdot \frac{x-3}{x+1}$

Now multiply across and divide out the factors common to the top and the bottom:

$\frac{(x+8)(x+1)(x-3)}{x(x-3)(x+1)} = \frac{x+8}{x}$

That is as far as it goes. The $x$ left underneath is a factor of the whole denominator, while the $x$ in $x+8$ is one term of a sum, so nothing further cancels: at $x = 2$ the answer is $5$, not $8$.

Factor both denominators completely, coefficients included.

$6x = 2 \cdot 3 \cdot x, \qquad 4x^2-4x = 4x(x-1) = 2^2 \cdot x \cdot (x-1)$

Take each distinct factor to the highest power it reaches in a single denominator: $2^2$, one $3$, one $x$, one $x-1$. The coefficients contribute $\operatorname{lcm}(6,4) = 12$, so the least common denominator is $12x(x-1)$. Both denominators divide into it, $12x(x-1) = 6x \cdot 2(x-1)$ and $12x(x-1) = 3 \cdot 4x(x-1)$.

Factor the quadratic denominator first: $x^2+3x-10 = (x-2)(x+5)$, so $x = 2$ and $x = -5$ are excluded and the LCD is $(x-2)(x+5)$. Multiply every term by it.

$x(x+5) - 2(x-2) = 14 \;\Longrightarrow\; x^2 + 3x - 10 = 0 \;\Longrightarrow\; (x-2)(x+5) = 0$

The candidates are $x = 2$ and $x = -5$. Held against the excluded list one number at a time, both are on it, so both are extraneous however correctly they were derived, and nothing survives.

This is not the same as an identity. There the variable cancels and a true statement is left, so every value in the domain is a solution; here the variable never cancels, two ordinary candidates arrive, and the domain rejects them both.

Compare the degrees: the numerator has degree $2$ and the denominator degree $3$, so the bottom outgrows the top. Divide both halves by $x^3$, the highest power below the bar.

$f(x) = \frac{\dfrac{6}{x} - \dfrac{1}{x^2} + \dfrac{4}{x^3}}{\,3 + \dfrac{5}{x^3}\,}$

Every numerator term now has a power of $x$ underneath it and dies away, while the denominator settles on $3$. The values shrink toward $0$, so the horizontal asymptote is the $x$-axis, $y = 0$.

Factor both halves of the left side.

$\frac{2x^2-8}{x^2+x-2} = \frac{2(x-2)(x+2)}{(x+2)(x-1)}$

The printed denominator is zero at $x = -2$ and $x = 1$, so those two inputs are out of the domain before anything is canceled. Everywhere else $x+2$ is nonzero and divides out, leaving $\frac{2(x-2)}{x-1}$. At $x = -2$ the right side returns $\frac{-8}{-3} = \frac{8}{3}$ while the left side returns $\frac{0}{0}$, which is no number, so the statement fails there; at $x = 1$ neither side exists.

The denominators already match, so put the numerators over the single denominator and bracket the one being subtracted.

$\frac{(7x-1) - (3x+15)}{x^2-4x} = \frac{7x - 1 - 3x - 15}{x^2-4x} = \frac{4x-16}{x(x-4)}$

Now factor the numerator: $4x-16 = 4(x-4)$, which shares the factor $x-4$ with the denominator.

$\frac{4(x-4)}{x(x-4)} = \frac{4}{x} \qquad (x \neq 0, \; x \neq 4)$

The restriction $x \neq 4$ has left the page and has to be carried by hand.

Factor all four polynomials and read the restrictions off the two printed denominators, before any canceling.

$\frac{(x-7)(x+2)}{(x+4)(x-1)} \cdot \frac{(x+4)(x+1)}{(x-7)(x-2)}$

The first denominator gives $x \neq -4$ and $x \neq 1$; the second gives $x \neq 7$ and $x \neq 2$. Canceling $x+4$ and $x-7$ produces the stated answer, whose own denominator advertises only $1$ and $2$. The other two restrictions survive nowhere but in writing.

Factor both halves and count copies of each denominator factor.

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

The original denominator is zero at $x = 0$ and $x = -3$, so both are excluded. The factor $x+3$ appears twice above and once below, so it cancels completely and $x = -3$ is a hole. The factor $x$ appears twice below and not at all above, so a copy survives and $x = 0$ is a vertical asymptote. Reduced, $f(x) = \frac{x+3}{x^2}$, whose height at $-3$ is $0$.

The only denominator is $x+13$, so $x = -13$ is excluded. Multiply every term by it.

$2x + 26 = 2(x+13) \;\Longrightarrow\; 2x + 26 = 2x + 26$

The variable cancels and what is left is true for every value, so every candidate survives the clearing. An identity holds only where both sides exist, and at $x = -13$ the left side is not a number, so the answer is the domain itself: every real number except $-13$.

Combine the top and the bottom into single fractions, then divide.

$\frac{\dfrac{x^2-16}{x^2}}{\dfrac{x+4}{x}} = \frac{(x-4)(x+4)}{x^2} \cdot \frac{x}{x+4} = \frac{x-4}{x}$

Now hunt every denominator that appeared. The inner ones, $x^2$ and $x$, force $x \neq 0$. The bottom of the long bar cannot be zero either, and $\frac{x+4}{x} = 0$ exactly at $x = -4$, so $x \neq -4$ as well. That second restriction is invisible in the answer, which returns $\frac{-4-4}{-4} = 2$ there.

Check for a cancellation before dividing, because that is what decides whether there is anything to approach. Two numbers with product $2 \cdot 12 = 24$ and sum $11$ are $3$ and $8$, so

$2x^2+11x+12 = (2x+3)(x+4), \qquad f(x) = 2x+3 \quad (x \neq -4).$

The division is exact, so the remainder is zero and there is no correction term to die away. The graph is not near the line $y = 2x+3$, it is that line, with a hole at $(-4,-5)$. An oblique asymptote needs a nonzero remainder.

Call the number $n$. Adding it to both parts gives an equation with $n$ in a denominator.

$\frac{5+n}{12+n} = \frac{3}{4} \;\Longrightarrow\; 4(5+n) = 3(12+n) \;\Longrightarrow\; 20+4n = 36+3n \;\Longrightarrow\; n = 16$

The only excluded value is $n = -12$, and $16$ is not it, so the candidate is genuine, and it is positive as the question requires. Checking, $\frac{5+16}{12+16} = \frac{21}{28} = \frac{3}{4}$.

Factor the first denominator: $x^2-9x+18 = (x-3)(x-6)$, which already contains the second, so it is the least common denominator. The second fraction is missing a factor of $x-6$.

$\frac{x}{(x-3)(x-6)} - \frac{2(x-6)}{(x-3)(x-6)} = \frac{x - (2x-12)}{(x-3)(x-6)} = \frac{12-x}{(x-3)(x-6)}$

The minus reaches both terms of $2x-12$, so the $-12$ becomes $+12$. Nothing cancels, since $12-x$ vanishes only at $x = 12$.

Factor both halves completely, taking the $3$ out of the numerator first.

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

The excluded inputs are $5$ and $-3$. The factor $x+3$ appears once above and once below, so it cancels completely and $x = -3$ is the hole; $x-5$ survives, so $x = 5$ is a vertical asymptote. The hole's height is the reduced expression evaluated there:

$\frac{3(-3-3)}{-3-5} = \frac{-18}{-8} = \frac{9}{4}$

Factor all four polynomials and read the three sets of conditions off the printed quotient.

$\frac{(x+9)(x-3)}{x(x-2)} \div \frac{x(x+9)}{(x-2)(x+2)}$

The dividend's denominator gives $x \neq 0$ and $x \neq 2$. The divisor's denominator gives $x \neq 2$ and $x \neq -2$. The divisor's numerator gives $x \neq 0$ and $x \neq -9$, because a divisor may not be zero. Four values in all: $-9$, $-2$, $0$ and $2$. The zero $x = 3$ belongs to the dividend's numerator, which is free to vanish.

The two denominators are opposites, $7-x = -(x-7)$, so subtracting the second fraction adds it back with the sign turned.

$\frac{x}{x-7} + \frac{5}{x-7} = \frac{x+5}{x-7} = 3$

The only excluded value is $x = 7$. Multiplying by $x-7$ gives $x+5 = 3x-21$, so $26 = 2x$ and $x = 13$, which is not excluded. Checking in the original, $\frac{13}{6} - \frac{5}{-6} = \frac{13}{6} + \frac{5}{6} = 3$.

Factor and read the restrictions off both printed denominators first: $(x-10)(x+10)$ gives $x \neq 10$ and $x \neq -10$, and $x-10$ gives $x \neq 10$ again.

$\frac{x(x-10)}{(x-10)(x+10)} \cdot \frac{x+10}{x-10} = \frac{x}{x-10}$

In the reduced rule $x-10$ still sits below the bar, so $x = 10$ is a vertical asymptote, while $x+10$ has canceled completely, making $x = -10$ a hole at height $\frac{-10}{-20} = \frac{1}{2}$. The reduced degrees tie at $1$ with leading coefficients $1$ and $1$, so the horizontal asymptote is $y = 1$.