University:
York University
Course:
MATH 1013 | Applied Calculus
Academic year:

2023
Views:

119

Pages:

2
Author:

Niklas

IA. Rational Functions LONG DIVISION, OBLIQUE ASYMPTOTES & REMOVABLE DISCONTINUITIES (Ch. 2-6) I. POLYNOMIAL LONG DIVISION: ex. (3x³ - 4x² + 2x + 4) ÷ (x² - 2x + 2) Step 1: Set up in long division notation (if any terms are skipped, a placeholder 0 must be added): x² - 2x + 2 | 3x³ - 4x² + 2x + 4 Step 2: Divide 3x³ (first term of dividend) by x² (leading term of divisor) and place it in the quotient: x² - 2x + 2 | 3x³ - 4x² + 2x + 4 3x Step 3: Multiply that result (3x) by the entire divisor and subtract from the dividend, bring down the next term: x² - 2x + 2 | 3x³ - 4x² + 2x + 4 3x³ - 6x² + 6x ---------------- 2x² - 4x + 4 Step 4: Repeat with new polynomial until there is a remainder: x² - 2x + 2 | 3x³ - 4x² + 2x + 4 3x³ - 6x² + 6x ---------------- 2x² - 4x + 4 2x² - 4x + 4 ---------------- 0 -> R0 Step 5: Record result: 3x + 2 + 0 / (x² - 2x + 2) = 3x + 2 II. OBLIQUE ASYMPTOTES: when a rational function does not have a horizontal asymptote (i.e. the degree of the numerator is greater than that of the denominator), it instead has an Oblique asymptote. to find an oblique asymptote, you must divide the numerator by the denominator. the remainder can be ignored because as the domain increases, the remainder tends to zero. ex. graph \(f(x)=\frac{x^{3}}{x^{2}-4}\) domain: \((-\infty,-2)\cup(-2,2)\cup(2,\infty)\) \[\frac{x^{3}}{x^{2}-4}=\frac{x^{3}}{(x-2)(x+2)}\Rightarrow x\ne-2.\] \[x\ne2\] VD: \(x=-2, x=2\) yint O let \(x=0\) \[\frac{0}{0-4}=0\] xint let \(x^{3}=\underline{0}\) \[x=\sqrt[1]{0}\] \[x=0\] ha: none 0a: \(y=x\) \[x^{2}-4\sqrt{x^{3}+0x^{2}+0x+0}\] \[4x\rightarrow R4x\] \[x+\frac{4x}{x^{2}-4}\] end benavior: as \(x\rightarrow-2^{-} y\rightarrow-\infty\) as \(x\rightarrow2 y\rightarrow-\infty\) as \(x\rightarrow\infty, y\rightarrow\infty\) \(x\rightarrow-\infty y\rightarrow-\infty x\rightarrow-2^{+}, y\rightarrow\infty\) as \(X\rightarrow2^{+} 4^{\rightarrow\infty}\) III. REMOVABLE DISCONTINUITIES: ↳ if factors in the numerator and denominator can be eliminated, there is a removable discontinuity at the root(s) of the factors. graphically, besides the r.d., the original function is identical to the simplified function examine: \(f(x)=x-1\) VS. \[f(x)=\frac{x(x-1)}{x}\] x.h 2-x- →X-2 removable disanhinuity