AP Precalc 1.11B-1.12A Slant & ShiftsAP Precalc 1.11B-1.12A Slant & Shifts

A P P R E C A L C U L U S · UNIT1

Slant Asymptotes and Translations

Topics 1.11B and 1.12A Study Guide

In one sentence: polynomial division rewrites a rational function as a quotient plus a remainder, the quotient is the slant asymptote, and translations slide a graph by adding to the input or to the output. Why it works, in plain language. Dividing polynomials is just long division with x's. The quotient tells you what the function looks like for huge |x|, because the remainder fraction (remainder over divisor) shrinks to 0. So f(x) = quotient + remainder/divisor behaves like the quotient far from the origin. When the quotient is a line, that line is a slant asymptote. Translations work on the rule: g(x) = f(x − 3) asks "what did f do 3 units ago?", so the picture slides right. That is why the horizontal change feels backwards. Adding outside, f(x) + k, directly adds to every output, so it moves up by k. Step by step: synthetic division. (1) Divisor x − c: write c. (2) Write the dividend's coefficients with 0 for gaps. (3) Bring down the first. (4) Multiply by c, add to the next. Repeat. (5) The last value is the remainder; the others are the quotient's coefficients.

Step by step: transform a graph or table. (1) Write g in terms of f. (2) For any x you want, rewrite the input to f. (3) Look up that f value. (4) Apply the outside changes (add, multiply). (5) For domain, move x-values using the inside change. For range, apply the outside change to y-values. 1.11B Polynomial division

Long division works for any divisor. Synthetic division is a shortcut when the divisor is x − c (use c). Synthetic division recipe

1. Write the coefficients of the dividend in order. Use 0 for a missing power.

2. Bring down the first coefficient. Multiply by c, add to the next coefficient, repeat. 3. The last number is the remainder. The others are the coefficients of the quotient, one degree lower. Example. (x³ − 4x² + 6x − 4)/(x − 2): c = 2. Rows: 1, −2, 2, remainder 0. Quotient x² − 2x + 2. Long division example. (3x³ − 4x² − 3)/(x² + 5x + 1) = 3x − 19 + (92x + 16)/(x² + 5x + 1). Write 0x as a placeholder.

Quotient and remainder form: f(x) = quotient + remainder/divisor. For (x² + 5x + 2)/(x + 5): c = −5 gives 1, 0, remainder 2, so f(x) = x + 2/(x + 5).

Trap. If the divisor is 2x + 1 or any divisor with a leading coefficient other than 1, divide the numerator and denominator by that coefficient first (or use long division). For (4x² + 12x − 6)/(2x + 1): divide by 2 to get (2x² + 6x − 3)/(x + 1/2), then c = −1/2 gives 2, 5, −5.5. Quotient 2x + 5. AP Precalculus 1 of 4

Slant (oblique) asymptotes

Degree of top vs. bottom Asymptote

Top lower horizontal, y = 0

Equal horizontal, y = ratio of leading coefficients

Top is exactly one degree higher slant asymptote y = the quotient (ignore the remainder) Top is two or more degrees higher neither a horizontal nor a slant asymptote

Examples.

(x³ − 2x² − 4x + 1)/(x² − 2x + 1): divide to get x with remainder −5x + 1. Slant y = x. (x² − 9x + 4)/(x + 6): c = −6 gives 1, −15, 94. Slant y = x − 15.

(9x⁴ − 5x² + 3x − 6)/(3x³ − 4x²): slant y = 3x + 4.

(4x⁵ − 3x³ + 4x + 1)/(5x³ − 2x² + 1): degree 5 over 3, neither. (2x⁴ + x² + 1)/(3x⁴ − 2x² + 5x): equal degrees, horizontal y = 2/3.

Full analysis. f(x) = (3x³ − 12x)/(x² − 2x − 8) = 3x(x + 2)(x − 2)/((x − 4)(x + 2)). Domain (−∞, −2) ∪ (−2, 4) ∪ (4, ∞). Zeros x = 0 and 2. Hole x = −2. VA x = 4. HA none. Slant y = 3x + 6 (from the long division). y-intercept 0. End behavior: limx→−∞ f = −∞, limx→∞ f = ∞.

Test-prep. f(x) = −2x + 1 + (3x + 4)/(x² − 4x − 12). The slant asymptote is y = −2x + 1, which falls from left to right: limx→−∞ f = ∞ and limx→∞ f = −∞.

f(x) = (6x² + ax + 2)/(x + 3) has slant asymptote y = 6x + 3. Synthetic division with c = −3: 6, a − 18. So a − 18 = 3 and a = 21.

1.12A Translations (additive transformations)

Graph g from f using g(x) = f(x − h) + k.

Change Name Effect

f(x) + k vertical translation (adds to the output) up k if k > 0, down if k < 0

f(x + h) horizontal translation (adds to the input) left h if h > 0, right if h < 0 (opposite sign!) −f(x) reflection over the x-axis flips every y-value

Every point (x, y) on f moves to (x − h, y + k) on g(x) = f(x + h) + k. So g(x) = f(x − 2) + 4 moves points 2 right and 4 up.

Domain moves with horizontal shifts. Range moves with vertical shifts (and flips with reflections). Order: reflect first, then shift vertically. g(x) = −f(x) + 5 is a vertical reflection followed by a vertical translation of 5.

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Numeric transformations (use the table)

g(x) = f(x) + 2 and f(5) = −2: g(5) = −2 + 2 = 0.

g(x) = f(x + 2) − 3, find g(1): g(1) = f(3) − 3. With f(3) = 8, g(1) = 5.

g(x) = −f(x − 2), find g(4): g(4) = −f(2). With f(2) = 21, g(4) = −21.

Domain and range

f has domain [−5, 3], range [−4, 8]. g(x) = f(x + 5): domain shifts left 5, [−10, −2], range unchanged [−4, 8].

f has domain (0, 5), range [−10, 4]. g(x) = f(x − 2) + 4: domain (2, 7), range [−6, 8]. f has domain [−2, 4], range (−1, 8). g(x) = −f(x + 3) + 5: domain [−5, 1]. Range: flip (−1, 8) to (−8, 1), then up 5: (−3, 6).

Algebraic transformations

f(x) = 4x + 3, g(x) = f(x) + 5: 4x + 8.

f(x) = 2x − 5, g(x) = f(x + 3) + 4: 2(x + 3) − 5 + 4 = 2x + 5.

f(x) = x³ + 2x², g(x) = −f(x) + 5: −x³ − 2x² + 5.

f(x) = 2x² − 3x + 1, g(x) = f(x − 2) + 5: 2x² − 11x + 20.

Traps. (1) Inside = horizontal and backwards. (2) f(x) + 4 adds to the output (an additive transformation of f's values), but f(x + 4) adds to each input. For f(x) = −x² + 3x + 2, the image after a vertical translation of 4 is p(x) = −x² + 3x + 6. (3) y = f(x + 2) − 1 means left 2 and down 1. (4) A vertical reflection followed by a vertical translation is a valid description of g(x) = −f(x) + 5, but a horizontal translation is not.

Quick check

1. (x² + 3x + 1)/(x − 1): slant asymptote? y = x + 4

2. Which way does g(x) = f(x − 7) shift f? Right 7

3. f has range [2, 9]. Range of g(x) = −f(x) + 1? [−8, −1]

4. Top degree is 4 and bottom degree is 2. Slant asymptote? No

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