Rational Functions — Free Algebra 2 Review Games.

The denominator x-3 = 0 when x = 3, so x = 3 is excluded.

Factor: \((x+2)(x-2)/(x-2) = x+2\) (\(x \neq 2\)).

A vertical asymptote occurs where the denominator is zero (and numerator isn't).

Set denominator to 0: x + 5 = 0, x = -5.

Degree of numerator < degree of denominator, so y = 0.

Same degree: HA = leading coefficients ratio = 2/1 = 2.

Factor: \((x+3)(x-3)/((x-3)(x+2)) = (x+3)/(x+2)\).

2 = 6x, x = 2/6 = 1/3.

Set denominator to 0: x = 1 and x = -2.

When numerator degree exceeds denominator by 1, there's a slant asymptote.

Multiply by (x-2): x = 3 + 2(x-2), x = 3+2x-4, -x = -1, x = 1. Wait: check. Actually x = 3+2x-4 => x=2x-1 => -x=-1 => x=1. Hmm let me recheck with x=1: 1/(-1)=-1, 3/(-1)+2=-3+2=-1. Yes, x=1. But answer says 7/2. Let me redo. x/(x-2) = 3/(x-2) + 2. Subtract 3/(x-2): (x-3)/(x-2) = 2. x-3=2(x-2)=2x-4. -x=-1. x=1. The answer should be 1. Let me fix.

LCD \(= (x+1)(x-1)\). \((x-1+x+1)/(x^2-1) = 2x/(x^2-1)\).

Long division: \(x^2+2x+1\) divided by \(x+3\) = \(x - 1\) remainder \(4\). Slant asymptote: \(y = x - 1\).

Cross multiply: \((x+1)^2 = (x+3)(x-2)\), \(x^2+2x+1 = x^2+x-6\), \(x = -7\). Hmm: \(2x+1=x-6\), \(x=-7\). Check: \(-6/-9=2/3\), \(-4/-6=2/3\). Yes \(x=-7\). Let me just fix the answer to match.

A hole occurs when a factor cancels from both numerator and denominator.

The expression is undefined when the denominator equals zero. Setting (x-1)(x+4) = 0 gives x = 1 or x = -4. Note that x = -3 makes the numerator zero, which gives a value of 0 rather than making the expression undefined.

Factor the numerator as a difference of squares: \(x^2 - 25 = (x-5)(x+5)\). Then cancel the common factor \((x+5)\) to get \(x - 5\), with the restriction \(x \neq -5\). A common error is writing \(x + 5\) as the answer by canceling incorrectly.

Factor the denominator: \(x^2 - 16 = (x-4)(x+4)\). Setting each factor to zero gives \(x = 4\) and \(x = -4\). Both values make the denominator zero and must be excluded. The mistake of writing \(x = 16\) comes from not factoring the difference of squares.

Factor the numerator: 6x^3 - 3x^2 = 3x^2(2x - 1). Dividing by 3x^2 gives 2x - 1, with the restriction x not equal to 0. A common error is writing 2x^2 - 1 by subtracting exponents incorrectly instead of canceling the entire 3x^2 factor.

When the degree of the numerator equals the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients. Here that ratio is 4/2 = 2, so y = 2. The answer y = 0 would apply only if the numerator had a smaller degree than the denominator.

To find the y-intercept, substitute x = 0: y = (0 + 2)/(0 - 4) = 2/(-4) = -1/2. A common mistake is reading off 2 or -2 from the constants in the numerator or denominator without actually substituting x = 0.

Factor the numerator: \(x^2 - x = x(x - 1)\). Dividing by \(x\) gives \(x - 1\), with the restriction \(x \neq 0\). A common error is treating \(x^2/x\) as \(x^2 - 1\) instead of correctly computing \(x^2/x = x\).

Factor both polynomials: numerator \(= (x+2)(x+3)\) and denominator \(= (x+1)(x+2)\). Cancel the common factor \((x+2)\) to get \((x+3)/(x+1)\), with \(x \neq -2\). A frequent error is canceling only one term rather than an entire binomial factor.

Factor the denominator: \(x^2 - 4 = (x-2)(x+2)\). Neither factor shares a root with the numerator \((x+1)\), so both \(x = 2\) and \(x = -2\) are vertical asymptotes. The numerator zero \(x = -1\) produces an x-intercept, not an asymptote.

Cross-multiply to get 3(x - 1) = 1(x + 2), which gives 3x - 3 = x + 2. Subtracting x from both sides: 2x - 3 = 2, so 2x = 5 and x = 5/2. Check: neither x = 2 nor x = -1 was introduced as an excluded value issue, so x = 5/2 is valid.

Since both the numerator and denominator have degree 2, the horizontal asymptote equals the ratio of the leading coefficients: \(3/1 = 3\), so \(y = 3\). The answer \(y = 0\) only applies when the numerator degree is strictly less than the denominator degree.

Factor \(x^2 - 9 = (x-3)(x+3)\). The product becomes \((x-3)(x+3)(x+2) / ((x+2)(x-3))\). Cancel \((x-3)\) and \((x+2)\) to get \(x + 3\). Students who only cancel one factor may arrive at \(x - 3\) or leave an unsimplified fraction.

Factor both: numerator \(= (x-2)(x+2)\), denominator \(= (x-2)(x+3)\). The common factor \((x-2)\) cancels, creating a hole at \(x = 2\). The remaining denominator factor \((x+3)\) gives a vertical asymptote at \(x = -3\). Students who confuse holes and asymptotes may select \(x = -3\) instead.

The LCD is \(x^2 - 4 = (x+2)(x-2)\). Multiplying through: \((x-2) + (x+2) = 8\), which simplifies to \(2x = 8\), giving \(x = 4\). Check: \(x = 4\) makes no denominator zero. Note that \(x = 2\) would make denominators zero and is excluded from the domain.

Dividing by a fraction means multiplying by its reciprocal: \((x^2-1)/(x+2)\) times \((x^2-4)/(x-1)\). Factor: \((x-1)(x+1)/(x+2)\) times \((x-2)(x+2)/(x-1)\). Cancel \((x-1)\) and \((x+2)\) to get \((x+1)(x-2) = x^2 - x - 2\). The choice \(x^2 + x - 2\) results from a sign error in expanding.

Set the numerator equal to zero: \(x^2 - 5x + 6 = (x-2)(x-3) = 0\), giving \(x = 2\) and \(x = 3\). Neither value equals \(-2\) (the excluded value), so both are valid x-intercepts. The value \(x = -2\) is excluded from the domain entirely and is not an intercept.

Multiply both sides by (x - 1): x + 2 = 2(x - 1) = 2x - 2. Subtract x from both sides: 2 = x - 2, so x = 4. Check: (4+2)/(4-1) = 6/3 = 2. The value x = 1 would make the denominator zero and is excluded.

As x approaches infinity, if the denominator grows faster than the numerator, the ratio shrinks toward zero. This occurs when degree of p(x) is less than degree of q(x). Equal degrees produce a nonzero horizontal asymptote (ratio of leading coefficients), and higher numerator degree means no horizontal asymptote.

The LCD is \(x^2 - 1 = (x-1)(x+1)\). Multiplying through: \(2(x+1) - (x-1) = 1\), which gives \(2x+2 - x+1 = 1\), so \(x+3 = 1\) and \(x = -2\). Check: \(x = -2\) makes none of the denominators zero (\(-3\), \(-1\), and \(3\) respectively), so \(x = -2\) is valid.

Perform polynomial long division of 2x^2 - 3x + 1 by x - 2. First term: 2x^2 / x = 2x; 2x(x-2) = 2x^2 - 4x; subtract to get x + 1. Next term: x/x = 1; 1(x-2) = x - 2; subtract to get remainder 3. The quotient is 2x + 1, so the slant asymptote is y = 2x + 1.

Multiply both sides by (x - 2): x + (x - 2) = 2, which gives 2x - 2 = 2, so 2x = 4 and x = 2. However, x = 2 makes the original denominators zero, so it is an extraneous solution. The equation has no valid solution. Selecting x = 2 is the most common error — students solve algebraically without checking for extraneous roots.

Multiply the numerator and denominator of the main fraction by the LCD of all inner fractions, which is \(xy\). Numerator: \((1/x - 1/y)(xy) = y - x\). Denominator: \((1/x + 1/y)(xy) = y + x\). Result: \((y - x)/(y + x)\). Note that \((x - y)/(x + y)\) differs only in sign and equals \(-(y-x)/(y+x)\), not the same value.

Factor: numerator \(= (x+2)(x-1)\), denominator \(= (x-1)(x+1)\). The common factor \((x-1)\) cancels, producing a hole at \(x = 1\), not a vertical asymptote. The remaining factor \((x+1)\) in the denominator creates a vertical asymptote at \(x = -1\). Since the degrees are equal (both 2), the horizontal asymptote is the ratio of leading coefficients: \(1/1 = 1\), giving \(y = 1\).

The LCD is \(x^2 - 4 = (x-2)(x+2)\). Multiply through: \((x+3)(x+2) - (x-1)(x-2) = 12\). Expand: \((x^2+5x+6) - (x^2-3x+2) = 12\), which gives \(8x + 4 = 12\), so \(8x = 8\) and \(x = 1\). Check: \(x = 1\) makes none of the denominators zero, so it is a valid solution.

The expression changes sign at x = -1 (excluded, denominator = 0) and x = 3 (numerator = 0). Test three intervals: for x < -1, both factors are negative so the ratio is positive; for -1 < x < 3, the numerator is negative and denominator is positive so the ratio is negative; for x > 3, both factors are positive so the ratio is positive. The solution is x < -1 or x > 3, with endpoints excluded since the inequality is strict.

The denominator x - 5 equals zero when x = 5, making the expression undefined. Therefore x = 5 must be excluded. Choice B is a sign error — it solves x + 5 = 0 rather than x - 5 = 0.

Factor the numerator using difference of squares: \(x^2 - 4 = (x + 2)(x - 2)\). Cancel the common factor \((x + 2)\), leaving \(x - 2\). Choice B is wrong because the numerator contains \((x - 2)\), not a second \((x + 2)\). Choice C results from subtracting 4 from \(x\) without factoring.

Vertical asymptotes occur where the denominator equals zero. Setting x + 7 = 0 gives x = -7. Choice B is a sign error — it would be correct if the denominator were x - 7. Choices C and D describe horizontal asymptotes, not vertical ones.

\((x^2 + 1)/(x - 1)\) is fully simplified because \(x^2 + 1\) does not factor over the real numbers and shares no factor with \(x - 1\). Choice A reduces to \(x + 1\) since \(x^2 - 1 = (x - 1)(x + 1)\). Choice B reduces to \(1/(x - 2)\) after canceling \((x + 2)\). Choice D reduces to \(x - 1\) after factoring out \(x\).

Factor the numerator: \(x^2 - x = x(x - 1)\). Then cancel the common factor \(x\), leaving \(x - 1\). Choice C results from dividing \(x^2\) by \(x\) but ignoring the \(-x\) term. Choice D reverses the sign, which would require factoring out \(-x\) rather than \(x\).

The denominator equals zero at x = 1 and x = -2. Since neither value cancels with the numerator factor (x + 3), both produce vertical asymptotes. The zero of the numerator at x = -3 gives an x-intercept, not an asymptote. Choice B incorrectly counts only one asymptote.

When the numerator and denominator have the same degree, the horizontal asymptote equals the ratio of leading coefficients: 4/2 = 2. Choice B uses the numerator's leading coefficient alone. Choice C inverts the ratio. Since the degrees are equal (both degree 1), a horizontal asymptote does exist.

Factor the numerator as a difference of squares: \(x^2 - 9 = (x - 3)(x + 3)\). Dividing by \((x - 3)\) cancels that factor, leaving \(x + 3\). Choice B would mean the wrong factor canceled. Choice D is just the factored form of the numerator before any cancellation occurs.

The LCD is \((x + 1)(x - 1) = x^2 - 1\). Rewrite each fraction over the LCD: \(\frac{2(x - 1)}{x^2 - 1} + \frac{3(x + 1)}{x^2 - 1}\). Add numerators: \((2x - 2) + (3x + 3) = 5x + 1\). Result: \(\frac{5x + 1}{x^2 - 1}\). Choice A has a sign error in the constant term. Choice C adds only the numerator coefficients without distributing through the binomials.

The LCD is \((x - 2)(x + 2) = x^2 - 4\). Expand the numerators: \((x + 1)(x + 2) = x^2 + 3x + 2\) and \((x - 1)(x - 2) = x^2 - 3x + 2\). Subtract: \((x^2 + 3x + 2) - (x^2 - 3x + 2) = 6x\). Result: \(\frac{6x}{x^2 - 4}\). Choice B gives \(2x\), which comes from incorrectly subtracting only constant terms. Choice D drops the variable from the numerator.

Factor the numerator: \(x^2 - 4 = (x - 2)(x + 2)\). Simplify: \((x - 2)(x + 2) / [(x - 2)(x + 3)] = (x + 2)/(x + 3)\), with a hole at \(x = 2\). Since \((x - 2)\) cancels completely, \(x = 2\) is a hole, not an asymptote. Only \(x = -3\) (where the remaining denominator equals zero) is a vertical asymptote. Choice B incorrectly treats the hole at \(x = 2\) as a vertical asymptote.

Cross-multiply: 3(x + 1) = 2(x - 4). Distribute: 3x + 3 = 2x - 8. Subtract 2x: x + 3 = -8, so x = -11. Verify: 3/(-15) = -1/5 and 2/(-10) = -1/5. Check. Choice B has the wrong sign. Choice C is the common value both sides equal at the solution, not the value of x itself.

Substitute \(x = 0\): \(y = (0 - 4)/(0 - 0 - 2) = (-4)/(-2) = 2\). The y-intercept is \((0, 2)\). The denominator at \(x = 0\) equals \(-2\), not zero, so the function is defined there and a y-intercept exists. Choice B gets the sign wrong. Choice C inverts the fraction.

Factor: numerator \(= 2(x - 2)(x + 2)\), denominator \(= (x - 2)^2\). After canceling one \((x - 2)\): \(y = 2(x + 2)/(x - 2)\). A factor of \((x - 2)\) remains in the denominator, giving a vertical asymptote at \(x = 2\) — not a hole. A hole requires the factor to cancel completely from both numerator and denominator. Choice B incorrectly assumes complete cancellation.

Factor all four polynomials: \((x - 3)(x + 2) / [(x + 2)(x - 1)] \times [(x + 1)(x - 1)] / [(x - 3)(x + 3)]\). Cancel common factors: \((x - 3)\), \((x + 2)\), and \((x - 1)\) each appear in both a numerator and denominator. The result is \((x + 1)/(x + 3)\). Choice B results from mixing up the signs when canceling binomial factors.

Factor the denominator: \(x^2 - 5x + 6 = (x - 2)(x - 3)\). The numerator \((x - 1)\) shares no common factor with either denominator factor, so both roots of the denominator are vertical asymptotes: \(x = 2\) and \(x = 3\). Choice B confuses the zero of the numerator (\(x = 1\), an x-intercept) with a vertical asymptote.

Multiply both sides by the LCD x(x + 1): 2(x + 1) + x = 3. Distribute: 2x + 2 + x = 3, so 3x = 1 and x = 1/3. Since x = 1/3 does not equal 0 or -1 (the excluded values), it is valid. Choice B comes from incorrectly treating 3x = 1 as x = 3.

Multiply both sides by \((x - 3)(x + 3)\): \(x(x + 3) + 2(x - 3) = 18\). Expand: \(x^2 + 5x - 6 = 18\), so \(x^2 + 5x - 24 = 0\). Factor: \((x + 8)(x - 3) = 0\), giving \(x = -8\) or \(x = 3\). Since \(x = 3\) makes the original denominators zero, it is extraneous. The only valid solution is \(x = -8\). Choice C fails to reject the extraneous solution.

A hole at \(x = 2\) requires \((x - 2)\) to be a factor of the numerator, meaning \(x = 2\) must be a root. Substitute \(x = 2\) into the numerator: \(4 + 2k + 4 = 0\), so \(2k = -8\) and \(k = -4\). With \(k = -4\) the numerator factors as \((x - 2)^2\) and the common factor cancels to produce a hole. Choice B gives \(k = 4\), making the numerator \(x^2 + 4x + 4 = (x + 2)^2\), whose root is \(x = -2\), not \(x = 2\).

Divide \(x^2 + x - 2\) by \(x + 3\) using polynomial long division. First step: \(x \times (x + 3) = x^2 + 3x\); remainder \(= -2x - 2\). Second step: \(-2 \times (x + 3) = -2x - 6\); remainder \(= 4\). So \(y = x - 2 + 4/(x + 3)\). As \(x\) approaches infinity, \(4/(x + 3)\) approaches \(0\), and the slant asymptote is \(y = x - 2\). Choice B has the wrong constant, and Choice D simply copies the denominator.

Critical values are x = 1 (zero of numerator) and x = -2 (undefined). Test intervals: for x < -2 both factors are negative, giving a positive quotient (fails). For -2 < x < 1, numerator is negative and denominator is positive, giving a negative quotient (satisfies). For x > 1 both are positive (fails). Include x = 1 since the expression equals 0 there, but exclude x = -2 since the expression is undefined. Solution: -2 < x <= 1. Choice B incorrectly includes x = -2.

Simplify the numerator: \(1/(x-1) - 1/x = [x - (x-1)] / [x(x-1)] = 1 / [x(x-1)] = 1/(x^2 - x)\). Now the complex fraction becomes \([1/(x^2 - x)]\) divided by \([1/(x^2 - x)] = 1\). Any nonzero quantity divided by itself equals 1. Choices B and C arise from errors when combining fractions in the numerator before dividing.

For a hole at x = -3, the factor (x + 3) must appear in both numerator and denominator so it cancels completely. For a vertical asymptote at x = 2, (x - 2) must remain in the denominator after simplification. Only choice A satisfies both: (x + 3) cancels to leave 1/(x - 2) with a hole at x = -3 and a vertical asymptote at x = 2. Choice C has vertical asymptotes at both x = 2 and x = -3, with no hole. Choice D has an x-intercept at x = -3, not a hole.

Cross-multiply: \((x + 1)(x - 5) = (x - 2)^2\). Expand both sides: \(x^2 - 4x - 5 = x^2 - 4x + 4\). Subtract \(x^2 - 4x\) from both sides: \(-5 = 4\). This is a contradiction, so no value of \(x\) can satisfy the equation. The equation has no solution. This happens because the equation implicitly requires two equal differences with different constant terms.

Factor the numerator: \(x^2 - 2x - 8 = (x - 4)(x + 2)\). Factor the denominator: \(x^2 - x - 12 = (x - 4)(x + 3)\). The factor \((x - 4)\) cancels completely, creating a hole at \(x = 4\). The remaining denominator factor \((x + 3)\) equals zero at \(x = -3\), giving a vertical asymptote there. Choice B treats \(x = 4\) as a vertical asymptote, failing to recognize that the \((x - 4)\) factor cancels.

The domain excludes values that make the denominator zero. Setting (x - 3)(x + 2) = 0 gives x = 3 or x = -2. The numerator zero at x = -5 does not affect the domain — it creates an x-intercept, not an excluded value. Choice D incorrectly includes x = -5 as a domain restriction.

When the numerator and denominator have the same degree (both degree 1), the horizontal asymptote is the ratio of leading coefficients: 5/3. Choice B (y = 0) is incorrect — that rule applies only when the degree of the numerator is less than the degree of the denominator.

Factor the numerator: \(x^2 - 9 = (x - 3)(x + 3)\). The common factor \((x + 3)\) cancels with the denominator, leaving \(x - 3\). Choice A (\(x + 3\)) is the factor that was canceled, not the remaining quotient. Choice D is the original numerator left unfactored.

A horizontal asymptote of \(y = 0\) occurs when the degree of the numerator is strictly less than the degree of the denominator. Choice B has a degree-1 numerator and degree-2 denominator, so \(y = 0\) is the horizontal asymptote. Choice A has equal degrees (\(y = 3\)). Choice C has numerator degree greater than denominator degree (no horizontal asymptote). Choice D simplifies to \(x + 2\), a linear function with no horizontal asymptote.

Set \(x^2 - 16 = 0\). Factor: \((x - 4)(x + 4) = 0\), giving \(x = 4\) or \(x = -4\). Both values make the denominator zero and must be excluded. Choice C misses \(x = -4\). Choice B is the value that makes the numerator zero — that is an x-intercept, not a domain restriction.

Divide coefficients and subtract exponents: (6/2) * x^(3-2) = 3x. Choice B (3x^2) results from incorrectly keeping the exponent as 3 - 1 = 2 instead of 3 - 2 = 1. Choice D (3/x) results from subtracting exponents in the wrong order.

Vertical asymptotes occur where the denominator equals zero with no corresponding factor cancellation. For \(y = 5/(x^2 + 9)\), the equation \(x^2 + 9 = 0\) has no real solutions, so there are no vertical asymptotes. Choice A has a vertical asymptote at \(x = 3\). Choice B factors as \((x-1)/((x-1)(x+1))\), giving a hole at \(x = 1\) and a vertical asymptote at \(x = -1\). Choice D has vertical asymptotes at \(x = 2\) and \(x = 3\).

The LCD is (x + 2)(x - 1). Rewrite each fraction: 3(x - 1)/LCD + 5(x + 2)/LCD = (3x - 3 + 5x + 10)/LCD = (8x + 7)/LCD. Choice A has the wrong constant: -3 + 10 = +7, not -7. Choice C incorrectly adds only the numerators without adjusting for the LCD.

Factor: numerator \(= (x - 2)(x - 3)\), denominator \(= (x - 2)(x + 2)\). The factor \((x - 2)\) cancels, creating a hole at \(x = 2\) — not an x-intercept. The simplified numerator \((x - 3) = 0\) gives the only x-intercept: \(x = 3\). Choice A incorrectly counts \(x = 2\) as an x-intercept when it is actually a hole in the graph.

Factor each denominator: \(x^2 - 1 = (x + 1)(x - 1)\) and \(x^2 + 2x + 1 = (x + 1)^2\). The LCD must include each factor to its highest power: \((x + 1)^2\) from the second denominator and \((x - 1)\) from the first. LCD \(= (x + 1)^2(x - 1)\). Choice B omits the \((x - 1)\) factor. Choice C only uses each factor to the first power, missing the squared \((x + 1)\).

Multiply by the reciprocal: \([(x^2 - 4)/(x + 3)] \times [(x^2 + 6x + 9)/(x - 2)]\). Factor: \(x^2 - 4 = (x + 2)(x - 2)\) and \(x^2 + 6x + 9 = (x + 3)^2\). The expression becomes \((x + 2)(x - 2)(x + 3)^2 / [(x + 3)(x - 2)]\). Cancel \((x - 2)\) and one \((x + 3)\) to get \((x + 2)(x + 3)\). Choice A results from not using the reciprocal and instead canceling incorrectly.

Factor: \(x^2 - 1 = (x - 1)(x + 1)\) and \(2x^2 + x - 1 = (2x - 1)(x + 1)\). The common factor \((x + 1)\) cancels, creating a hole at \(x = -1\). After simplification, the function is \((x - 1)/(2x - 1)\). Since both polynomials have degree 2, the horizontal asymptote is the ratio of leading coefficients: \(1/2\). Choice B misses the hole. Choice C inverts the leading coefficient ratio.

Note \(x^2 - 9 = (x - 3)(x + 3)\), so the LCD is \((x - 3)(x + 3)\). Rewrite: \(\frac{x(x + 3)}{\text{LCD}} - \frac{9}{\text{LCD}} = \frac{x^2 + 3x - 9}{\text{LCD}}\). Choice B has a sign error: expanding \(x(x + 3)\) gives \(+3x\), not \(-3x\). Choice C incorrectly suggests full cancellation that does not occur.

Multiply both sides by \(x^2 - 4 = (x - 2)(x + 2)\): \((x + 2) + (x - 2) = 4\), which gives \(2x = 4\) and \(x = 2\). However, \(x = 2\) makes the original denominators zero, so it is an extraneous solution. The equation has no solution. Choice A incorrectly accepts the extraneous solution without checking it in the original equation.

Multiply by the reciprocal and factor all expressions: \((x - 3)(x + 2) / ((x - 3)(x + 3)) \times (x + 2)(x + 3) / ((x + 2)(x - 1))\). Cancel \((x - 3)\), \((x + 3)\), and one \((x + 2)\) to get \((x + 2)/(x - 1)\). Choice A results from canceling \((x + 2)\) instead of \((x + 3)\) in one of the steps, producing an incorrect result.

Factor: \(3x^2 - 12 = 3(x - 2)(x + 2)\) and \(x^2 - 2x - 8 = (x - 4)(x + 2)\). The common factor \((x + 2)\) cancels, creating a hole at \(x = -2\). After simplification, the only vertical asymptote is \(x = 4\). The horizontal asymptote equals the ratio of leading coefficients: \(3/1 = 3\). Choice C incorrectly treats \(x = -2\) as a vertical asymptote rather than a hole.

Multiply both sides by (x - 1): x + 4 = 2(x - 1) = 2x - 2. Rearrange: 4 + 2 = 2x - x, so 6 = x. Check: (6 + 4)/(6 - 1) = 10/5 = 2. Choice C gives (2 + 4)/(2 - 1) = 6/1 = 6, not 2.

Factor \(x^2 - 4 = (x - 2)(x + 2)\), so the LCD is \((x - 2)(x + 2)\). Rewrite: \(\frac{2}{\text{LCD}} + \frac{3(x - 2)}{\text{LCD}} = \frac{2 + 3(x - 2)}{\text{LCD}} = \frac{2 + 3x - 6}{\text{LCD}} = \frac{3x - 4}{\text{LCD}}\). Choice A has a sign error: \(2 + 3(-2) = -4\), not \(+8\). Choice B incorrectly adds only numerators without converting to a common denominator.

Subtract 1 from both sides: (x + 3)/(x - 2) - 1 > 0. Combine over a common denominator: [(x + 3) - (x - 2)]/(x - 2) > 0, which simplifies to 5/(x - 2) > 0. Since 5 is always positive, this inequality holds when x - 2 > 0, meaning x > 2. There is no sign change at x = -3 because (x + 3) is not in the denominator. Choice C is a common error from treating this as a product of two factors.

Perform polynomial long division of \(2x^3 - x^2 - 1\) by \(x^2 + 1\). First step: \(2x^3 / x^2 = 2x\); subtract \(2x(x^2 + 1) = 2x^3 + 2x\) to get remainder \(-x^2 - 2x - 1\). Second step: \(-x^2 / x^2 = -1\); subtract \(-1(x^2 + 1) = -x^2 - 1\) to get remainder \(-2x\). The quotient is \(2x - 1\), so the slant asymptote is \(y = 2x - 1\). Choice B (\(y = 2x + 1\)) results from a sign error in the second step of the division.

Factor: \(2x^2 + x - 6 = (2x - 3)(x + 2)\), giving zeros at \(x = 3/2\) and \(x = -2\). Factor: \(x^2 - x - 2 = (x - 2)(x + 1)\), giving vertical asymptotes at \(x = 2\) and \(x = -1\) (no cancellation occurs). Since both polynomials have degree 2, the horizontal asymptote is the ratio of leading coefficients: \(2/1 = 2\). Choice B lists \(x = 3\) as a zero instead of \(x = 3/2\) — a common error when forgetting to divide by the leading coefficient 2.

Note \(x^2 - x - 6 = (x + 2)(x - 3)\), so multiply both sides by the LCD \((x + 2)(x - 3)\): \(3(x - 3) - 2(x + 2) = 5x - 8\). Expand: \(3x - 9 - 2x - 4 = 5x - 8\), giving \(x - 13 = 5x - 8\), then \(-4x = 5\), so \(x = -5/4\). Since \(-5/4\) is neither \(-2\) nor \(3\), it is a valid solution. Choice A (\(x = 5/4\)) results from a sign error when solving \(-4x = 5\).

Simplify the numerator: x/(x + 1) - (x + 1)/(x + 1) = (x - x - 1)/(x + 1) = -1/(x + 1). Simplify the denominator: (x + 1)/(x + 1) - 1/(x + 1) = x/(x + 1). Divide: [-1/(x + 1)] / [x/(x + 1)] = -1/(x + 1) times (x + 1)/x = -1/x. Choice A (1/x) omits the negative sign produced when simplifying the numerator of the complex fraction.

Factor: \(x^3 - 8 = (x - 2)(x^2 + 2x + 4)\) and \(x^2 - 4 = (x - 2)(x + 2)\). Cancel \((x - 2)\) to get \(\frac{x^2 + 2x + 4}{x + 2}\), with a hole at \(x = 2\). The hole's y-value is \(\frac{4 + 4 + 4}{2 + 2} = \frac{12}{4} = 3\), so the hole is at \((2, 3)\). Divide \(x^2 + 2x + 4\) by \((x + 2)\): quotient \(x\), remainder \(4\), giving slant asymptote \(y = x\). The remaining denominator \((x + 2)\) yields vertical asymptote \(x = -2\). Choice A incorrectly computes the slant asymptote as \(y = x + 2\).

A hole at x = 2 requires the numerator to equal zero when x = 2 (so the factor (x - 2) cancels). Substitute x = 2: k(4) - 7(2) + 2 = 4k - 14 + 2 = 4k - 12 = 0, giving k = 3. Verify: 3x^2 - 7x + 2 = (3x - 1)(x - 2), confirming (x - 2) cancels. Choice B (k = 7/2) results from incorrectly ignoring the constant term +2 and solving 4k = 14.

A vertical asymptote occurs where the denominator equals zero and the factor does not cancel. Setting x - 5 = 0 gives x = 5. The line x = -5 is incorrect because that would require the denominator to be x + 5. The answers y = 5 and y = 0 confuse vertical asymptotes (vertical lines) with horizontal asymptotes (horizontal lines).

Factor the numerator as a difference of squares: x² - 9 = (x - 3)(x + 3). Then (x - 3)(x + 3)/(x + 3) = x - 3, for x ≠ -3. A common error is choosing x + 3, which would result from canceling incorrectly. The choice (x - 3)/(x + 3) reflects a failure to factor the numerator at all.

Factor the denominator: x² - x - 6 = (x - 3)(x + 2). The function is undefined where the denominator equals zero: x = 3 and x = -2. Note that x = -2 also makes the numerator zero, so there is actually a hole at x = -2, but x = -2 is still excluded from the domain. The choice x = -3 and x = 2 comes from incorrect factoring.

When the degree of the numerator (0, since the numerator is the constant 3) is less than the degree of the denominator (2), the horizontal asymptote is y = 0. The value y = 3 is the leading coefficient of the numerator, not the asymptote. Since the degree of the denominator is greater, the function shrinks toward zero as x grows large.

Factor the numerator: 2x² + 6x = 2x(x + 3). Then (2x(x + 3))/(2x) = x + 3, for x ≠ 0. The choice x² + 3 results from canceling only one factor of 2x incorrectly. The choice 2x + 3 results from dividing only the first term by 2x and leaving 6x unchanged.

When the degrees of the numerator and denominator are equal, the horizontal asymptote is the ratio of the leading coefficients: 5/1 = 5, so y = 5. The answer y = 0 applies only when the numerator's degree is strictly less than the denominator's degree. The value y = -2 is the constant term in the numerator, not the asymptote.

A hole (removable discontinuity) occurs when the same linear factor appears in both the numerator and denominator and cancels, making the function undefined at that single point while behaving like a simpler function everywhere else. A vertical asymptote, by contrast, occurs when the denominator factor does not cancel, causing the function to grow without bound.

To find the y-intercept, substitute x = 0: y = (0 + 4)/(0 - 2) = 4/(-2) = -2. The y-intercept is (0, -2). The choice (0, 4) ignores the denominator. The choice (0, -4) might arise from evaluating -(x + 4) instead of dividing by (x - 2).

Factor each piece: x² - 4 = (x - 2)(x + 2), x² + 3x = x(x + 3), x² + 2x - 3 = (x + 3)(x - 1). Multiply and cancel: [(x - 2)(x + 2)(x + 3)(x - 1)] / [x(x + 3)(x + 2)] = (x - 2)(x - 1)/x. The factor (x + 2) and (x + 3) both cancel. Choosing (x + 2)(x - 1)/x results from forgetting to cancel (x + 2).

Cross-multiply: 2(x + 2) = 3(x - 1). Expand: 2x + 4 = 3x - 3. Solve: 4 + 3 = 3x - 2x, so x = 7. Check that x = 7 is not an excluded value (x ≠ 1 and x ≠ -2), so x = 7 is valid. The answer x = -7 is a sign error, and x = 1 would make the original equation undefined.

The LCD is (x + 1)(x - 1) = x² - 1. Rewrite each fraction: 3(x - 1)/[(x + 1)(x - 1)] + 2(x + 1)/[(x + 1)(x - 1)] = [3x - 3 + 2x + 2]/(x² - 1) = (5x - 1)/(x² - 1). The choice (5x + 1)/(x² - 1) comes from a sign error when distributing 3(x - 1): using +3 instead of -3.

Factor the numerator: x² + 5x + 6 = (x + 2)(x + 3). Factor the denominator: x² + 7x + 12 = (x + 3)(x + 4). Cancel the common factor (x + 3): result is (x + 2)/(x + 4), for x ≠ -3. Choosing (x + 3)/(x + 4) incorrectly cancels (x + 2) instead of (x + 3).

Factor: 2x - 6 = 2(x - 3) and x² - 9 = (x - 3)(x + 3). After canceling (x - 3), the simplified form is 2/(x + 3). Since (x - 3) canceled, x = 3 is a hole (removable discontinuity), not a vertical asymptote. The remaining denominator gives a vertical asymptote at x = -3.

Multiply both sides by (x + 3): x + (x + 3) = 6. Simplify: 2x + 3 = 6, so 2x = 3, giving x = 3/2. Since x = 3/2 ≠ -3, it is a valid solution. The choice x = -3 would make the original denominators zero, so it is an excluded value, not a solution.

Factor x² - 4 = (x - 2)(x + 2). The LCD is (x - 2)(x + 2). Rewrite the second fraction: 1/(x + 2) = (x - 2)/[(x - 2)(x + 2)]. Subtract: [(x + 1) - (x - 2)]/(x² - 4) = [x + 1 - x + 2]/(x² - 4) = 3/(x² - 4). The distractor (2x + 3)/(x² - 4) results from adding instead of subtracting the adjusted numerators.

Factor: 3x + 9 = 3(x + 3) and x² - x - 12 = (x - 4)(x + 3). Cancel (x + 3): the simplified form is 3/(x - 4). Since (x + 3) canceled, x = -3 is a hole. Since (x - 4) remains in the denominator, x = 4 is a vertical asymptote. Choosing vertical asymptotes at both values ignores that (x + 3) cancels.

Dividing by a fraction means multiplying by its reciprocal: [(x² + 2x - 8)/(x² - 4)] · [(x - 2)/(x + 4)]. Factor: x² + 2x - 8 = (x + 4)(x - 2) and x² - 4 = (x + 2)(x - 2). Multiply and cancel: [(x + 4)(x - 2)(x - 2)] / [(x + 2)(x - 2)(x + 4)] = (x - 2)/(x + 2). Leaving (x + 4) uncanceled gives the distractor (x - 2)(x + 4)/(x + 2).

Perform polynomial long division: x² + 3x - 2 divided by x + 1. x² + 3x - 2 = (x + 1)(x + 2) + (-4), so y = x + 2 - 4/(x + 1). As x → ∞, the term -4/(x + 1) → 0, leaving the oblique asymptote y = x + 2. The choice y = x + 3 is a common error from not completing the division correctly.

Factor x² - 4 = (x - 2)(x + 2). The LCD is (x - 2)(x + 2). Multiply through: 2 + (x + 2) = 3(x - 2). Simplify: x + 4 = 3x - 6, so 10 = 2x, giving x = 5. Check: x = 5 satisfies x ≠ 2 and x ≠ -2, so it is valid. The choice x = 2 would make the original equation undefined.

Critical values are x = 1 (numerator = 0) and x = -3 (denominator = 0, excluded). Test each interval: for x < -3, (neg)/(neg) = positive (not ≤ 0); for -3 < x < 1, (neg)/(pos) = negative (satisfies ≤ 0); for x > 1, (pos)/(pos) = positive (not ≤ 0). At x = 1 the expression equals 0 (included), but x = -3 is excluded. Solution: -3 < x ≤ 1. The choice -3 ≤ x ≤ 1 incorrectly includes x = -3.

Multiply both the numerator and denominator of the complex fraction by xy (the LCD): numerator becomes y - x, denominator becomes y + x. Result: (y - x)/(y + x). The choice (x - y)/(x + y) is the negative of the correct answer, arising from reversing the subtraction order. Note that (y - x)/(y + x) and -(x - y)/(x + y) are equivalent; the presented form matches the factored result directly.

Factor: numerator = (x - 1)(x - 2) and denominator = (x - 1)(x - 3). The factor (x - 1) appears in both, so it cancels, creating a hole at x = 1. The remaining denominator factor (x - 3) gives a vertical asymptote at x = 3. Choosing vertical asymptotes at both x = 1 and x = 3 ignores that (x - 1) cancels.

Factor the numerator: x² + 2x - 3 = (x + 3)(x - 1). Cancel (x - 1): the simplified form is y = x + 3, a line, for x ≠ 1. At x = 1, the y-value on the line would be 1 + 3 = 4, so the hole is at (1, 4). Because the degree of the numerator exceeds the denominator's degree, there is no horizontal asymptote — the graph is a slanted line with one missing point.

For a hole at x = 2, the factor (x - 2) must divide the numerator exactly. Substitute x = 2 into the numerator and set it equal to zero: 4 - 2k + 4 = 0, so 8 - 2k = 0, giving k = 4. Verify: x² - 4x + 4 = (x - 2)², so y = (x - 2)²/(x - 2) = x - 2 (linear) with a hole at x = 2. The value k = 2 gives x² - 2x + 4, which does not factor with (x - 2).

Factor x² - 4 = (x - 2)(x + 2). Multiply through by (x - 2)(x + 2): (x + 1)(x + 2) - (x - 1)(x - 2) = 8. Expand: (x² + 3x + 2) - (x² - 3x + 2) = 8. Simplify: 6x = 8, so x = 4/3. Check: 4/3 ≠ 2 and 4/3 ≠ -2, so the solution is valid. The distractor x = 8/3 likely comes from dividing 8 by 3 without the factor of 6 in the numerator.

The function is undefined wherever the denominator equals zero. Setting x + 5 = 0 gives x = -5. Therefore the domain excludes x = -5. Choice B (x ≠ 5) is wrong because x + 5 = 0 gives -5, not 5.

To find the y-intercept, substitute x = 0: y = (0 - 3)/(0 + 2) = -3/2. Choice A (3/2) is the result of forgetting the negative sign in the numerator.

When the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. In choice C, the numerator has degree 0 and the denominator has degree 2, so y → 0. Choices A and D have equal degrees (HA at ratio of leading coefficients), and choice B has numerator degree greater than denominator degree (oblique asymptote, not horizontal).

Vertical asymptotes occur where the denominator equals zero and the numerator does not. Setting x - 4 = 0 gives x = 4. Choice A is wrong because vertical asymptotes are vertical lines (x = constant), not horizontal lines (y = constant). Choice C uses the wrong sign.

Divide coefficients and subtract exponents: (10/2) * x^(3-2) = 5x¹ = 5x. Choice B (5x²) incorrectly adds the exponents instead of subtracting. Choice D (5/x) incorrectly subtracts 3 - 2 as 2 - 3.

The x-intercept occurs where the numerator equals zero (and denominator is non-zero). Set 2x - 4 = 0, which gives x = 2. At x = 2, the denominator is 2 + 3 = 5 ≠ 0, so (2, 0) is valid. Choice A confuses the zero of the denominator (which gives a vertical asymptote) with the x-intercept.

When the numerator and denominator have equal degrees, the horizontal asymptote is the ratio of the leading coefficients: 5/5 = 1, so y = 1. Choice B (y = 5) incorrectly uses the leading coefficient itself rather than the ratio. Choice A applies only when the numerator degree is less than the denominator degree.

Factor the numerator: x² - 1 = (x + 1)(x - 1). The expression becomes [(x + 1)(x - 1)/(x + 3)] * [(x + 3)/(x - 1)]. Cancel (x + 3) and (x - 1) from numerator and denominator to get x + 1, with restrictions x ≠ 1 and x ≠ -3. Choice C is partially simplified but still has a common factor.

Factor the denominator: x² - 3x - 10 = (x - 5)(x + 2). The denominator equals zero at x = 5 and x = -2, so both values are excluded from the domain even though x = -2 also makes the numerator zero (creating a hole, not a vertical asymptote). The domain excludes all values where the denominator is zero. Choice A misses x = -2.

Factor: x² - 4 = (x + 2)(x - 2), so y = (x + 2)(x - 2)/(x - 2) = x + 2 with a hole at x = 2. To find the y-coordinate of the hole, substitute x = 2 into the simplified form: y = 2 + 2 = 4. The hole is at (2, 4). Choice A uses y = 2 instead of evaluating the simplified expression.

The LCD is (x - 3)(x + 3) = x² - 9. Rewrite each fraction: 4(x + 3)/[(x - 3)(x + 3)] + 2(x - 3)/[(x - 3)(x + 3)] = [4x + 12 + 2x - 6]/(x² - 9) = (6x + 6)/(x² - 9). Choice A incorrectly adds only the numerators without converting to the LCD first.

Factor the denominator: x² + 2x - 3 = (x + 3)(x - 1). Set each factor to zero: x = -3 and x = 1. The numerator x + 1 equals zero at x = -1, which does not match either denominator zero, so there are no holes and both are vertical asymptotes. Choice D incorrectly uses the wrong signs.

Cross multiply: 4(x + 3) = 2(x - 1). Expand: 4x + 12 = 2x - 2. Subtract 2x from both sides: 2x + 12 = -2. Subtract 12: 2x = -14, so x = -7. Verify: neither x - 1 nor x + 3 is zero at x = -7. Choice B uses a sign error when solving.

Factor numerator: 3x² - 12 = 3(x² - 4) = 3(x + 2)(x - 2). Factor denominator: x² + x - 6 = (x + 3)(x - 2). Cancel the common factor (x - 2): 3(x + 2)/(x + 3), with restriction x ≠ 2. Choice C incorrectly cancels (x + 2) instead of (x - 2). Choice D drops the factor of 3.

Subtract 2/x from both sides: 1/3 = 5/x - 2/x = 3/x. Cross multiply: x = 9. Verify: x = 9 is non-zero. Check: 2/9 + 1/3 = 2/9 + 3/9 = 5/9 = 5/9. Choice C results from incorrectly solving 1/3 = 1/x.

Factor: numerator = 2(x + 3), denominator = (x + 3)(x - 3). The common factor (x + 3) creates a hole at x = -3. The remaining denominator factor (x - 3) creates a vertical asymptote at x = 3. Choice C is wrong because x = -3 is a hole (common factor), not a vertical asymptote.

When the numerator and denominator have equal degrees (both degree 2), the horizontal asymptote equals the ratio of their leading coefficients: 1/2. Choice B (y = 1) uses only the numerator's leading coefficient. Choice D (y = 2) inverts the ratio.

The LCD is (x + 2)(x - 1). Rewrite: x(x - 1)/[(x + 2)(x - 1)] - 3(x + 2)/[(x + 2)(x - 1)] = [x² - x - 3x - 6]/[(x + 2)(x - 1)] = (x² - 4x - 6)/[(x + 2)(x - 1)]. Choice B has a sign error: -3(x + 2) = -3x - 6, not -3x + 6.

Note x² - 4 = (x + 2)(x - 2), so multiply both sides by (x + 2)(x - 2): 3(x - 2) - 1(x + 2) = 8. Expand: 3x - 6 - x - 2 = 8, giving 2x - 8 = 8, so 2x = 16 and x = 8. Check: x = 8 makes no denominator zero, so it is valid. Choice B (x = 4) results from an arithmetic error when simplifying 2x - 8 = 8.

Critical values are x = -2 (numerator zero) and x = 3 (denominator zero, excluded). Test three intervals: x < -2 (e.g., x = -4): (-2)/(-7) > 0, satisfies; -2 < x < 3 (e.g., x = 0): (2)/(-3) < 0, does not satisfy; x > 3 (e.g., x = 4): (6)/(1) > 0, satisfies. Include x = -2 (makes fraction 0, satisfying ≥ 0) but exclude x = 3. Solution: x ≤ -2 or x > 3. Choice D incorrectly includes x = 3 where the function is undefined.

Perform polynomial long division of x² - 2x + 5 by x - 1. x² ÷ x = x; x(x - 1) = x² - x; subtract to get -x + 5. Then -x ÷ x = -1; -1(x - 1) = -x + 1; subtract to get remainder 4. The quotient is x - 1, so the oblique asymptote is y = x - 1. Choice B (y = x - 2) uses the middle coefficient incorrectly rather than performing the division.

Multiply numerator and denominator of the complex fraction by 2x (the LCD of all inner fractions): numerator becomes 2 + x, denominator becomes 2 - x. The result is (x + 2)/(2 - x). Choice B (x + 2)/(x - 2) is incorrect because the denominator 2 - x equals -(x - 2), making these expressions negatives of each other.

Multiply both sides by (x - 3): (2x + k) + (x - 3) = 5. Simplify: 3x + k - 3 = 5, so 3x = 8 - k, giving x = (8 - k)/3. This solution is extraneous (no valid solution) when x = 3, because x = 3 makes the original denominator zero. Set (8 - k)/3 = 3: 8 - k = 9, so k = -1. Choice B (k = 1) gives x = 7/3, which is a valid solution.

The denominator x² - 9 = (x - 3)(x + 3) has a zero at x = 3. For x = 3 to be a hole, the numerator must also have (x - 3) as a factor. Factor: x² + bx - 12 = (x - 3)(x + c) means -3c = -12 so c = 4. Then (x - 3)(x + 4) = x² + x - 12, giving b = 1. Choice C confuses the second factor value (4) with b.

Note x² - 4 = (x - 2)(x + 2). Multiply both sides by (x - 2)(x + 2): (x + 2) + (x - 2) = 4. Simplify: 2x = 4, so x = 2. However, x = 2 makes the original denominators (x - 2) and (x² - 4) equal to zero, so x = 2 is an extraneous solution. The equation has no valid solution. Choice A lists the extraneous solution without checking the original equation.

The denominator x + 3 equals zero when x = -3, making the function undefined there. All other real numbers are valid inputs, so the domain is all real numbers except x = -3. Choice C is wrong because setting x + 3 = 0 gives x = -3, not x = 3.

A vertical asymptote occurs where the denominator equals zero. Setting x - 7 = 0 gives x = 7. The asymptote is a vertical line, so it is written as x = 7, not y = 7. Choice B uses the wrong sign.

Factor the numerator using difference of squares: x² - 16 = (x - 4)(x + 4). Cancel the common factor (x - 4) to get x + 4. Because x = 4 was excluded from the original domain, the restriction x ≠ 4 must be kept. Choice D omits this required restriction.

Factor the denominator: x² - 3x = x(x - 3). This equals zero when x = 0 or x = 3, so both must be excluded. The numerator value x = -5 does not restrict the domain. Choice D incorrectly uses the numerator zero instead of the second denominator zero.

Compare the degrees: the numerator has degree 0 (a constant) and the denominator has degree 2. When the numerator's degree is strictly less than the denominator's degree, the horizontal asymptote is y = 0. Choice A incorrectly uses the numerator's constant value as the asymptote.

Factor the numerator: 4x² - 8x = 4x(x - 2). Divide by 2x: 4x(x - 2)/(2x) = 2(x - 2) = 2x - 4. The restriction x ≠ 0 must be stated because x = 0 was excluded from the original expression. Choice A incorrectly divides only part of the numerator, leaving a factor of 2 undivided.

A hole (removable discontinuity) occurs when the same factor appears in both the numerator and denominator and cancels. The function is undefined at that x-value but the limit exists, leaving just a missing point. This differs from a vertical asymptote (choice A), where the factor does not cancel and the function grows without bound.

Factor the numerator: x² + x - 6 = (x + 3)(x - 2). The factor (x - 2) cancels, leaving f(x) = x + 3 with x ≠ 2. To find the y-coordinate of the hole, substitute x = 2 into the simplified form: 2 + 3 = 5. The hole is at (2, 5). Choice D incorrectly places the hole on the x-axis.

When the numerator and denominator have equal degrees (both degree 1 here), the horizontal asymptote equals the ratio of the leading coefficients: 5/2. Choice C inverts the ratio. Choice B incorrectly uses the constant terms from the expression instead of the leading coefficients.

Factor each part: x² - 9 = (x - 3)(x + 3) and 2x + 8 = 2(x + 4). Multiply: [(x - 3)(x + 3)/(x + 4)] · [2(x + 4)/(x + 3)]. The factors (x + 4) and (x + 3) cancel, leaving 2(x - 3) with restrictions x ≠ -4 and x ≠ -3. Choice A has the wrong sign inside the factor.

The LCD is (x - 1)(x + 2) = x² + x - 2. Rewrite: 2(x + 2)/[(x-1)(x+2)] + 3(x - 1)/[(x-1)(x+2)] = (2x + 4 + 3x - 3)/(x² + x - 2) = (5x + 1)/(x² + x - 2). Choice B has a sign error: the constant in the numerator is 4 - 3 = +1, not -1.

Cross-multiply: 2(x + 2) = x(x - 1). Expand: 2x + 4 = x² - x. Rearrange: x² - 3x - 4 = 0, which factors as (x - 4)(x + 1) = 0. So x = 4 or x = -1. Check both: x = 4 gives denominators 3 and 6 (neither zero); x = -1 gives denominators -2 and 1 (neither zero). Both solutions are valid. Choice A finds only one of the two roots.

Factor the denominator: x² + 2x - 8 = (x + 4)(x - 2). The expression is undefined when the denominator equals zero: x = -4 or x = 2. Although x = 2 creates a hole rather than a vertical asymptote (it also zeroes the numerator), the function is still undefined there and must be excluded. Choice D uses the wrong sign for the second factor.

Dividing by a fraction means multiplying by its reciprocal: (x² - 1)/(x + 2) · (x² + 5x + 6)/(x - 1). Factor: x² - 1 = (x - 1)(x + 1) and x² + 5x + 6 = (x + 2)(x + 3). Multiply and cancel (x - 1) and (x + 2): the result is (x + 1)(x + 3) with restrictions x ≠ -2 and x ≠ 1. Choice A uses (x - 1) instead of (x + 1).

Factor the denominator: x² - x - 6 = (x - 3)(x + 2). Because (x + 2) also appears in the numerator, it cancels, leaving 1/(x - 3) with a hole at x = -2. The only vertical asymptote is x = 3. Choice A incorrectly classifies x = -2 as a vertical asymptote; since the factor cancels, x = -2 is a removable discontinuity (hole), not an asymptote.

X-intercepts occur where the numerator equals zero and the denominator is nonzero. Factor the numerator: x² - 5x + 6 = (x - 2)(x - 3), giving x = 2 or x = 3. Check denominators: 2 + 2 = 4 and 3 + 2 = 5, both nonzero. Both x-intercepts are valid. Choice A gives the vertical asymptote location rather than the x-intercepts.

Factor the numerator: x² - 2x - 8 = (x - 4)(x + 2). The common factor (x - 4) cancels, leaving f(x) = x + 2 with x ≠ 4. This is a linear function (not a parabola), with slope 1 and y-intercept 2, but with a hole at x = 4. The y-coordinate of the hole is 4 + 2 = 6, so the hole is at (4, 6). Choice D confuses the hole with a vertical asymptote.

Since the numerator's degree exceeds the denominator's by exactly 1, there is an oblique asymptote found by polynomial long division. Divide x² + 3x - 4 by (x + 1): x² + 3x - 4 = (x + 1)(x + 2) + (-6). The quotient is x + 2 with remainder -6, so the oblique asymptote is y = x + 2. Choice C mistakenly uses the coefficient of x in the numerator rather than performing the division.

Note that x² - x - 2 = (x - 2)(x + 1), which is the LCD. Multiply every term by (x - 2)(x + 1): x(x + 1) - 3(x - 2) = 5. Expand: x² + x - 3x + 6 = 5, giving x² - 2x + 1 = 0, or (x - 1)² = 0, so x = 1. Since x = 1 does not make any original denominator zero, it is valid. Choices A and B are the excluded values of the domain.

Simplify the numerator: x/y - y/x = (x² - y²)/(xy) = (x - y)(x + y)/(xy). Simplify the denominator: 1/y - 1/x = (x - y)/(xy). Divide the numerator by the denominator: [(x - y)(x + y)/(xy)] divided by [(x - y)/(xy)] equals (x - y)(x + y)/(xy) times xy/(x - y). The (x - y) and xy factors cancel, leaving x + y. Choice C results from canceling only part of the expression.

An oblique asymptote exists when the numerator's degree exceeds the denominator's by exactly 1 with no cancellation. For the asymptote to disappear, (x + 1) must be a factor of the numerator, meaning x = -1 is a root. Substitute: (-1)² + c(-1) + 2 = 0 gives 3 - c = 0, so c = 3. With c = 3, the numerator becomes x² + 3x + 2 = (x + 1)(x + 2), which simplifies to the line y = x + 2, giving no oblique asymptote. Choice A (c = 2) yields numerator x² + 2x + 2, which does not factor with (x + 1).

Critical values are x = 1 (numerator zero) and x = -3 (denominator zero, excluded). Test three intervals: for x less than -3, both factors are negative so the quotient is positive (fails ≤ 0); for -3 less than x less than 1, the numerator is negative and denominator is positive so the quotient is negative (satisfies ≤ 0); for x greater than 1, both factors are positive (fails ≤ 0). Include x = 1 since it yields 0, but exclude x = -3 since the function is undefined there. Solution: (-3, 1]. Choice A incorrectly includes x = -3.

Factor: x² - 4 = (x - 2)(x + 2) and x² + x - 6 = (x + 3)(x - 2). Cancel the common factor (x - 2): the simplified form is (x + 2)/(x + 3), x ≠ 2. The vertical asymptote is x = -3. For the hole at x = 2, substitute into the simplified form: (2 + 2)/(2 + 3) = 4/5. The hole is at (2, 4/5). Choice D incorrectly places the hole at y = 0 on the x-axis.

Factor: 2x² - 7x + 3 = (2x - 1)(x - 3) and x² - 4x + 3 = (x - 3)(x - 1). Cancel (x - 3): f(x) = (2x - 1)/(x - 1), x ≠ 3. Vertical asymptote: x = 1. Horizontal asymptote: ratio of leading coefficients = 2/1, so y = 2. Hole at x = 3: (2(3) - 1)/(3 - 1) = 5/2, giving the hole at (3, 5/2). Choice C is wrong because the horizontal asymptote is y = 2, not y = 1.

The LCD is (x + 1)(x - 2). Multiply every term through: 2(x - 2) - 1(x + 1) = 3. Expand: 2x - 4 - x - 1 = 3, giving x - 5 = 3, so x = 8. Check: x = 8 does not make any denominator zero (8 + 1 = 9 and 8 - 2 = 6). Choice A results from the error of not adding 5 to both sides after reaching x - 5 = 3.

Factor the numerator as a difference of squares: x² - 9 = (x + 3)(x - 3). Cancel the common factor (x - 3) to get x + 3, where x ≠ 3. Choice B, x - 3, is the cancelled factor itself — a common error of keeping the wrong factor after cancellation.

A rational function is undefined where its denominator equals zero. Setting x + 7 = 0 gives x = -7. Choice B (x = 7) is a common sign error — substituting x = 7 gives 4/14, which is defined.

When the degree of the numerator equals the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients. Here both degrees are 1, so y = 5/2. Choice D (y = 0) applies only when the numerator's degree is strictly less than the denominator's degree.

Divide coefficients: 12/4 = 3. Subtract exponents for like bases: x³/x² = x¹ = x, and y²/y⁵ = y^(2-5) = y^(-3) = 1/y³. The result is 3x/y³. Choice B flips the y exponent to the numerator, which is a subtraction-direction error.

Multiply numerators and denominators: [(x + 2)(x - 4)] / [(x - 4)(x + 5)]. Cancel the common factor (x - 4) to get (x + 2)/(x + 5), where x ≠ 4. Choice C (1) would only be correct if (x + 2) = (x + 5), which is never true.

Set the denominator equal to zero: x - 6 = 0, so x = 6. Choice B (x = -6) is a sign error. Choice D (y = 0) is the horizontal asymptote, not the vertical asymptote, since the numerator degree is less than the denominator degree.

Factor x from the numerator: x(x + 5)/x = x + 5, where x ≠ 0. Choice B (x² + 5) is incorrect — it results from incorrectly dividing only the first term. Choice C is nonsensical because like terms were not combined.

The LCD is (x + 2)(x - 3). Rewrite each fraction: 3(x - 3)/[(x + 2)(x - 3)] - (x + 2)/[(x + 2)(x - 3)]. Combine: [3x - 9 - x - 2]/[(x + 2)(x - 3)] = (2x - 11)/[(x + 2)(x - 3)]. Choice C forgets to distribute the negative sign to the +2, incorrectly yielding -7 instead of -11.

Both the numerator and denominator have degree 2, so the horizontal asymptote is the ratio of leading coefficients: 1/3. Choice B inverts the ratio. Choice C applies when the numerator degree is lower. Choice D applies when the numerator degree is higher (which would give an oblique asymptote instead).

Factor the numerator: x² - 2x - 15 = (x - 5)(x + 3). Factor the denominator: x² - 9 = (x - 3)(x + 3). Cancel (x + 3) to get (x - 5)/(x - 3), where x ≠ -3. Choice C incorrectly places (x + 3) in the denominator after cancellation rather than (x - 3).

Factor the numerator: x² - 4x + 3 = (x - 1)(x - 3). Cancel (x - 1) to get f(x) = x - 3 with a hole at x = 1. The y-coordinate of the hole is found by substituting x = 1 into the simplified form: 1 - 3 = -2. Choice B is wrong because when a factor cancels, it creates a hole (removable discontinuity), not a vertical asymptote.

Cross-multiply: 3(x + 1) = 1(x - 2), giving 3x + 3 = x - 2. Subtract x and subtract 3 from both sides: 2x = -5, so x = -5/2. Since -5/2 ≠ 2 and -5/2 ≠ -1, the solution is valid. Choice B uses the wrong signs when solving 2x = -5.

A horizontal asymptote of y = 0 occurs when the degree of the numerator is strictly less than the degree of the denominator. In choice B, the numerator has degree 1 and the denominator has degree 2, so y = 0. Choice A has equal degrees (y = 2), choice C has a higher numerator degree (oblique asymptote), and choice D has equal degrees (y = 2).

X-intercepts occur where the numerator equals zero (and the denominator is nonzero). Factor x² - 3x - 10 = (x - 5)(x + 2) = 0, giving x = 5 or x = -2. Neither value makes the denominator zero (x ≠ -1), so both are valid intercepts. Choice D confuses the signs of the factors.

Factor each part: x² - 16 = (x + 4)(x - 4) and x² - x - 12 = (x - 4)(x + 3). The expression becomes [(x + 4)(x - 4)(x + 3)] / [(x - 4)(x + 3)(x + 4)] = 1, where x ≠ 4, -3, -4. All three binomial factors cancel completely.

Multiply every term by the LCD, which is 3x: 3 + x = 4. Solving gives x = 1. Verify: 1/1 + 1/3 = 4/3, and 4/(3·1) = 4/3. Since x = 1 ≠ 0, the solution is valid. Choice C (x = 3) results from forgetting to distribute the LCD to all terms.

Factor the numerator: 2x² + x - 3 = (2x + 3)(x - 1). Factor the denominator: x² - 1 = (x + 1)(x - 1). Cancel (x - 1) to get (2x + 3)/(x + 1), where x ≠ 1. Choice B incorrectly cancels (x + 1) instead of (x - 1).

Factor the denominator: x² - 2x - 8 = (x - 4)(x + 2). Setting each factor to zero gives x = 4 and x = -2. These are the excluded values. Choice B results from factoring the denominator with wrong signs: (x + 4)(x - 2) ≠ x² - 2x - 8.

Perform polynomial long division of 2x² - 3x + 1 by (x - 2). First step: 2x² ÷ x = 2x; multiply 2x(x - 2) = 2x² - 4x; subtract to get x + 1. Second step: x ÷ x = 1; multiply 1(x - 2) = x - 2; subtract to get remainder 3. The quotient is 2x + 1, so the oblique asymptote is y = 2x + 1. Choice B comes from a sign error in the subtraction step.

Find critical values: x = -2 (numerator zero) and x = 4 (denominator zero, excluded). Test each interval: for x < -2, both factors are negative so the ratio is positive (satisfies >0). For -2 < x < 4, numerator is positive and denominator is negative, so the ratio is negative. For x > 4, both are positive so the ratio is positive. The solution is x < -2 or x > 4. Choice B gives where the expression is negative.

For a hole at x = -3, the factor (x + 3) must divide the numerator evenly. Substitute x = -3 into the numerator and set it equal to zero: (-3)² + k(-3) - 6 = 9 - 3k - 6 = 3 - 3k = 0, giving k = 1. Verify: x² + x - 6 = (x + 3)(x - 2), so f(x) = x - 2 with a hole at x = -3. Choice B gives k = -1, for which the numerator at x = -3 equals 3 + 3 - 6 = 0... wait let me recheck. x=-3: 9 + 3 - 6 = 6 ≠ 0. So k = -1 does not work.

Multiply both numerator and denominator by the LCD of all inner fractions, which is 2x. Numerator becomes 2 + x. Denominator becomes 2 - x. The result is (x + 2)/(2 - x). Choice D, (x + 2)/(x - 2), has the denominator signs flipped — note that 2 - x and x - 2 differ by a factor of -1.

Factor: numerator = 3(x + 2)(x - 2), denominator = (x + 3)(x - 2). Cancel (x - 2) to get 3(x + 2)/(x + 3) with a hole at x = 2. The remaining denominator gives a vertical asymptote at x = -3. Since both original degrees are equal, the horizontal asymptote is the ratio of leading coefficients: 3/1 = 3. Choice B incorrectly treats x = 2 as a vertical asymptote rather than a hole.

Multiply through by the LCD (x + 2)(x - 1): 3(x - 1) + 2(x + 2) = 8. Expand: 3x - 3 + 2x + 4 = 8, so 5x + 1 = 8, giving 5x = 7 and x = 7/5. Since 7/5 ≠ -2 and 7/5 ≠ 1, the solution is valid. Choice B (x = 1) is an extraneous value that would make the original denominators undefined.

Factor denominator: x² - 1 = (x + 1)(x - 1). Critical values: x = -1 and x = 1 (excluded from domain) and x = 3 (where expression equals zero, included). Sign analysis: on x < -1 the expression is negative (satisfies ≤ 0); on -1 < x < 1 it is positive; on 1 < x < 3 it is negative; on x > 3 it is positive. At x = 3, the expression equals zero (included). Solution: x < -1 or 1 < x ≤ 3. Choice B incorrectly includes x = -1 and x = 1, which are excluded because the denominator is zero there.

A vertical asymptote occurs where the denominator equals zero. Setting x - 5 = 0 gives x = 5. The asymptote is a vertical line, written as x = 5, not y = 5. Choice A is incorrect because the zero of (x - 5) is positive 5, not -5.

Factor the numerator using difference of squares: x² - 9 = (x - 3)(x + 3). Then (x - 3)(x + 3)/(x + 3) = x - 3, since the (x + 3) factors cancel (valid because x ≠ -3). Choice A is wrong — x + 3 would require the numerator to be (x + 3)², which it is not.

The domain excludes values that make the denominator zero. Factor: x² - 4 = (x - 2)(x + 2), which equals zero at x = 2 and x = -2. Both must be excluded. Choice D is wrong because substituting x = 4 gives 4² - 4 = 12, not zero.

Since both fractions share the denominator (x + 2), subtract numerators directly: (3x - 6)/(x + 2). Factor the numerator: 3(x - 2)/(x + 2). This is fully simplified because x - 2 and x + 2 share no common factors. Choice B is incorrect because it drops the -6 from the numerator entirely.

The LCD is (x - 1)(x + 2) = x² + x - 2. Rewrite each fraction: 2(x + 2)/[(x - 1)(x + 2)] + 3(x - 1)/[(x - 1)(x + 2)] = (2x + 4 + 3x - 3)/(x² + x - 2) = (5x + 1)/(x² + x - 2). Choice C reflects a sign error — using (x + 1) instead of (x - 1) in the second numerator yields 5x + 7, which is wrong.

Dividing by a fraction means multiplying by its reciprocal: (x² - 4)/(x + 1) times (x² - 1)/(x - 2). Factor: (x - 2)(x + 2)/(x + 1) times (x + 1)(x - 1)/(x - 2). The (x - 2) and (x + 1) factors cancel, leaving (x + 2)(x - 1). Choice B results from dividing straight across without taking the reciprocal.

When numerator and denominator have equal degrees, the horizontal asymptote is the ratio of leading coefficients: 4/2 = 2, so y = 2. Choice A is wrong because both polynomials are degree 3 — degrees are equal, not unequal. Choice D is a common misconception: equal degrees do produce a horizontal asymptote; only when the numerator's degree exceeds the denominator's is there no horizontal asymptote.

The LCD is (x - 3)(x + 3) = x² - 9. Multiply through: 2(x + 3) - (x - 3) = 12. Simplify: 2x + 6 - x + 3 = 12, so x + 9 = 12, giving x = 3. However, substituting x = 3 into the original equation causes division by zero in every term. Therefore x = 3 is an extraneous solution, and the equation has no valid solution. Choice A is the extraneous result that students often report without checking.

Factor both expressions: numerator = (x - 3)(x + 2), denominator = (x - 2)(x + 2). The shared factor (x + 2) cancels, producing a hole at x = -2. The factor (x - 2) remains only in the denominator, producing a vertical asymptote at x = 2. Choice A is incorrect because x = -2 is a removable discontinuity (hole), not an asymptote, since the (x + 2) factor appears in both numerator and denominator.

For choice C: factor as (x - 3)(x + 2)/[(x - 3)(x + 3)]. The shared factor (x - 3) cancels, creating a hole at x = 3. The remaining (x + 3) in the denominator creates a vertical asymptote at x = -3 — so C has both. Choice A simplifies to x + 2 with only a hole at x = 2. Choice B simplifies to 1 with holes at x = 2 and x = -2 but no asymptote. Choice D has no real zeros in its denominator since x² + 1 > 0 for all real x, giving no discontinuities.