AP Precalc 1.10-1.11A Holes & BinomialsAP Precalc 1.10-1.11A Holes & Binomials

A P P R E C A L C U L U S · UNIT1

Holes, Equivalent Forms, and the Binomial

Theorem

Topics 1.10 and 1.11A Study Guide

In one sentence: a hole is a missing point where a factor cancelled, factored and standard forms show different features of the same function, and the binomial theorem expands (a + b)ⁿ using Pascal's triangle.

Why it works, in plain language. A hole is what is left when a factor is divided by itself. (x − 3)/(x − 3) is 1 everywhere except at x = 3, where it is 0/0 and undefined. So the graph looks like the simplified function with one pinprick removed. To find the pinprick's height, use the simplified function. For the binomial theorem, (a + b)ⁿ means n copies of (a + b) multiplied together. Every term picks a or b from each copy. The number of ways to pick b exactly k times is the coefficient, which is exactly the Pascal's triangle entry. The powers of a and b always add to n.

Step by step: find a hole. (1) Factor top and bottom. (2) Find common factors. (3) Set the common factor equal to 0 to get the x-value. (4) Cancel it. (5) Plug that x-value into what is left to get the y- value. (6) Write the point (x, y).

Step by step: one term of a binomial. (1) Identify a, b (with signs and coefficients), and n. (2) Pick the row of Pascal's triangle for n. (3) For the k-th term use the k-th number in the row, a raised to n − (k − 1), and b raised to k − 1. (4) Multiply it all out, watching signs.

1.10 Holes

A hole at x = a happens when (x − a) is a common factor of numerator and denominator. To find the hole's coordinates: cancel the factor, then plug x = a into the simplified function. The hole is (a, simplified value).

Near a hole the function approaches a finite number from both sides. Describe it with limx→a⁻ f(x) = limx→a⁺ f(x) = L, while f(a) is undefined.

Examples.

f(x) = (x² + 3x)/(x² − 9) = x(x + 3)/((x + 3)(x − 3)). Cancel (x + 3): x/(x − 3). At x = −3: −3/−6 = 1/2. Hole (−3, 1/2).

d(t) = (t² + t − 20)/(t + 5) = (t + 5)(t − 4)/(t + 5) = t − 4. At t = −5: −9. Hole (−5, −9). h(x) = (x + 5)²/((x + 3)(x + 5)) = (x + 5)/(x + 3). At x = −5: 0. Hole (−5, 0).

AP Precalculus 1 of 4

Hole versus vertical asymptote

Hole Vertical asymptote

Cause Factor appears in both top and bottom Factor remains in the denominator only Behavior Outputs approach a finite number Outputs go to ±∞

Graph Open circle Dashed vertical line

Table example. f(x) = (x² − 16)/(x − 4). At x = 3.9, 3.99, 3.999: 7.9, 7.99, 7.999. At 4.001, 4.01, 4.1: 8.001, 8.01, 8.1. f(4) is undefined. So a hole at x = 4 with limx→4⁻ = limx→4⁺ = 8. f(x) = (x² − 2x − 8)/(4x + 8) = (x − 4)(x + 2)/(4(x + 2)). Hole at x = −2 where both limits are −1.5. Test-prep. f(x) = (x² − 9)/(x² + 8x + 15) = (x − 3)(x + 3)/((x + 5)(x + 3)): a hole at x = −3 (and a VA at −5).

A graph that is a straight line with an open circle at x = 4 and an x-intercept at −2 could be (x + 2)(x − 4)/(x − 4): zero from the numerator, hole from the common factor.

If f(0) = 2 and limx→4⁺ f(x) = 6, then limx→4⁺ f(x) = 3·f(0).

Calculator use (free-response)

"Find all x with g(x) = 4": graph g and the line y = 4, then read the intersection. For g(x) = 2.4x³ − 6.25x² + 3x + 1 this happens at x ≈ 2.296.

End behavior from the leading term: g(x) = 2.4x³ − … has limx→∞ g = ∞; g(x) = −3.2x² + 5x + 7.5 has limx→−∞ g = −∞; g(x) = (4x² − 9)/(2x² + x − 3) has limx→∞ g = 2.

1.11A Equivalent representations

Factored form and standard (general) form are the same function, and each makes different features easy to read.

Feature Read it from

Zeros (x-intercepts) Factored form: set each factor of the numerator to 0 (after cancelling)

Holes, vertical asymptotes Factored form: cancelled vs. remaining denominator factors Degree, end behavior, horizontal

asymptote

Standard form: leading terms

y-intercept Plug in 0 (the constant term in standard form)

Where f(x) ≥ 0 Sign table from factored form

Polynomial. f(x) = 2(x − 4)² = 2x² − 16x + 32. Degree 2. lim = ∞ on both ends. Zero x = 4 (even multiplicity, touches). f(x) ≥ 0 on (−∞, ∞). y-intercept 32.

g(x) = −3x³ + 27x = −3x(x + 3)(x − 3). Degree 3. limx→∞ g = −∞, limx→−∞ g = ∞. Zeros −3, 0, 3. g ≥ 0 on (−∞, −3] ∪ [0, 3]. y-intercept 0.

AP Precalculus 2 of 4

Rational. h(x) = 2(x + 3)(x − 3)/((x + 2)(x + 3)) = (2x² − 18)/(x² + 5x + 6). Domain (−∞, −3) ∪ (−3, −2) ∪ (−2, ∞). Zero x = 3. Hole x = −3. VA x = −2. HA y = 2. y-intercept −18/6 = −3. h(t) = (t³ − 4t²)/(t² + 2t − 15) = t²(t − 4)/((t + 5)(t − 3)). Zeros t = 0 and 4. VA t = −5 and 3. HA none (degree 3 over 2). Domain (−∞, −5) ∪ (−5, 3) ∪ (3, ∞). y-intercept 0.

The binomial theorem

Pascal's triangle gives the coefficients. Row n lists the coefficients of (a + b)ⁿ: n = 31331

n = 414641

n = 515 10 10 5 1

n = 616 15 20 15 6 1

n = 717 21 35 35 21 7 1

n = 818 28 56 70 56 28 8 1

In (a + b)ⁿ the power of a goes down from n to 0 and the power of b goes up from 0 to n. The exponents in each term add to n.

k-th term = (coefficient at position k) · an−(k−1) · bk−1. There are n + 1 terms. Put the whole a and b (with their coefficients and signs) in parentheses before raising to the powers. Full expansions.

(x + 2)⁴ = x⁴ + 8x³ + 24x² + 32x + 16

(x − 3)⁵ = x⁵ − 15x⁴ + 90x³ − 270x² + 405x − 243

(2x + 1)³ = 8x³ + 12x² + 6x + 1

(3x − 2)⁴ = 81x⁴ − 216x³ + 216x² − 96x + 16

One term. (x + 5)⁵, 3rd term: 10·x³·5² = 250x³. (x − 2)⁷, 4th term: 35·x⁴·(−2)³ = −280x⁴. (3x − 1)⁸, 5th term: 70·(3x)⁴·(−1)⁴ = 5670x⁴. (2x + 3)⁴, last term: 3⁴ = 81.

(x − 2)⁶, 4th term: 20·x³·(−2)³ = −160x³, so the coefficient is −160. (2a − b)⁴, 2nd term: 4·(2a)³·(−b) = −32a³b.

Traps. (1) Signs alternate when b is negative. (2) Raise the coefficient too: (3x)⁴ = 81x⁴, not 3x⁴. (3) The 4th term uses the third power of b. (4) If limx→∞ f(x) = ∞, f could be x(x + 2)³(x − 1) (degree 5, positive leading coefficient), but not a function with HA y = 0 or y = 1.

Quick check

1. Hole of (x² − 25)/(x − 5)? (5, 10)

2. How many terms in (x + 1)⁹? 10

3. 2nd term of (x + 4)⁵? 5·x⁴·4 = 20x⁴

4. Which form shows zeros fastest, standard or factored? Factored

AP Precalculus 3 of 4

Practice set