Separate holes, poles and end behavior

Try

Compare f(x)=(x²−9)/(x−3) and g(x)=1/(x−3). Both original denominators exclude x=3. Predict whether that shared exclusion forces the same type of graph behavior. Factor the numerator of f before deciding.

For x≠3, f(x)=x+3. Its nearby values approach 6, leaving a missing point on a line. By contrast, g(x) becomes arbitrarily large in magnitude near 3. An undefined input can correspond to a hole or a pole; the denominator alone does not decide which.

Predict

Consider h(x)=(x²−1)/[(x−1)(x+2)]. Predict what happens at each denominator zero. The factor x−1 cancels completely, while x+2 remains below the fraction bar. Predict the sign near −2 from each side after cancellation, keeping the original exclusions recorded.

Experiment

For allowed inputs, h(x)=(x+1)/(x+2). Compute selected nearby values.

Input x	Simplified quotient	Interpretation
0	1/2	Ordinary graph point
1	2/3 in the extension only	Original has a hole
−21/10	11	Left of the pole
−19/10	−9	Right of the pole

Sketch the vertical line x=−2, the horizontal line y=1 and the hole at (1,2/3). The numerator near −2 is negative. On the left the denominator is negative, producing positive values; on the right it is positive, producing negative values. Sign analysis should agree with the sample rows.

Understand

First list every original denominator zero. Then factor and cancel common factors on the remaining domain. A completely canceled denominator zero leaves a removable hole when the reduced expression has a finite value there. If a denominator factor remains after cancellation and the reduced numerator is nonzero at that input, the graph has a vertical asymptote there.

Multiplicity matters. In (x−1)/(x−1)², one factor cancels but one remains, so x=1 is still a pole, not a hole. A single canceled factor is not enough to classify the original point; inspect the fully reduced expression.

End behavior follows from leading powers or division. For numerator and denominator of equal degree, their leading-coefficient ratio gives a horizontal asymptote. If the numerator degree is smaller, the ratio approaches zero. If the numerator degree is larger, perform polynomial division instead of claiming no useful asymptotic description exists.

For (x²+2)/(x−1), division gives x+1+3/(x−1). The remainder term approaches zero at both infinite ends, so y=x+1 is a slant asymptote. The original domain still excludes 1. For higher degree differences the polynomial quotient can have degree above one, producing a polynomial asymptotic model rather than a horizontal or slant line.

An asymptote is not universally a forbidden crossing. For x/(x²+1), the horizontal asymptote is y=0 and the graph crosses it at x=0. To decide whether a particular graph meets its asymptote, set their formulas equal and solve on the allowed domain.

One sign chart connects zeros, multiplicities, inequalities and limits

For R(x)=((x−1)²(x+2))/((x−1)(x+1)²), first exclude 1 and −1. On the remaining domain, R(x)=(x−1)(x+2)/(x+1)². At 1 the denominator factor cancels completely but a numerator factor remains: there is a hole at (1,0), not a zero of the original function. Both one-sided limits there equal zero. At −1 the reduced denominator has even multiplicity two and the numerator tends to −2; both one-sided limits are −∞. In symbols, lim as x→−1⁻ R(x)=lim as x→−1⁺ R(x)=−∞, while lim as x→1⁻ R(x)=lim as x→1⁺ R(x)=0. The two infinite-end limits are both 1, by equal degree and equal leading coefficients.

For the inequality R(x)≥0, the reduced denominator is positive wherever defined, so use the numerator’s signs. The product (x−1)(x+2) is positive outside [−2,1], negative inside, and zero at its endpoints. Include −2 and exclude 1. The solution is (−∞,−2] together with (1,∞). Multiplying by the squared denominator is safe only after recording its zero exclusion; blindly multiplying by a general denominator of unknown sign is not safe.

The same workflow covers the degree comparisons. For 2/(x²+1), both end limits are 0. For (x²+1)/(x−1), polynomial division gives x+1+2/(x−1), so subtracting the line x+1 leaves a remainder tending to 0; the function itself tends to +∞ at +∞ and −∞ at −∞. For (x³+1)/(x−1), the quotient x²+x+1 similarly controls both positive infinite ends. These are different statements from the limit of the difference between the function and its polynomial asymptotic model. The registered changed task requires every exclusion, sign interval and one-sided/end limit, not just a memorized degree rule.

Apply

Use a checklist of mathematical questions, not visual guesses: original domain, reduced formula, zeros, holes, remaining poles, one-sided signs and end behavior. The written division problem asks you to prove whether a crossing exists. The counterexample problem asks you to distinguish bounded missing-point behavior from unbounded nearby outputs.

Continue

The next function family models repeated multiplication rather than ratios of polynomials. Its base conditions will govern both valid exponent laws and whether equations can be solved by matching exponents.