Rational Functions (I)

I. Rational Functions (Ch. 2-6) I. DEFINITION: rational function - a quotient of polynomial functions: [f(x)=\frac{N(x)}{D(x)}] II. DOMAIN OF RATIONAL FUNCTIONS a. (f(x)=\frac{x}{(x+3)(x-1)}) b. [f(x)=\frac{x-2}{x^{2}+9}] \[(x+3)(x-1)\ne0\] \[x^{2}+9\ne0\] \[x+3\ne0\] \[x-1\ne0\] \[x^{2}\ne-9\] \[x\ne-3\] \[x\notin\mathbb{R}\] \[D\cdot(-60,-3)v(-3,1)v(1,\infty)\] \[D=(-\infty,\infty)\] III. SIMPLIFYING RATIONAL EXPRESSIONS a b. \(f(x)=\frac{x-2}{x^{2}-3x+2}\) \[f(x)=\frac{(2x-1)(x+4)}{(x+4)(x-3)}\] \[f(x)=\frac{2x-1}{x-3},x\ne-4\] \[f(x)=\frac{x-2}{(x-2)(x-1)}\] \[f(x)=\frac{1}{x-1},x+2\] IV. SOLVING RATIONAL EQUATIONS AND INEQUALITIES a. Solve \(\frac{(x+2)(2x-5)}{(x-1)(x+3)}=0\) LCD: (x-1)(x+3) \[\{-2,\frac{5}{2}\}|\] \[(x-1)(x+3)\ne0\] \[(x+2)(2x-5)=0\] \[x-1\ne0\] \[x+2=0\] \[x\ne1\] \[2x-5=0\] \[x+3\ne0\] \[x=\frac{5}{2}\] \[x\ne-3\] \[x=-2\] b. \(\frac{4}{x^{2}+4x-5}=\frac{1}{x^{2}-2x+1}\) LCD \((x+5)(x-1)^{2}\) \[\frac{4}{(x\cdot5)(x-1)}-\frac{1}{(x-1)^{2}}=0\] {3} \[(x+5)(x-1)^{2}\ne0\] \[4(x-1)-(x+5)=0\] \(x\cdot5\ne0\) +0 \[4x-4-x-5=0\] 2 \[3x-9=0\] \[x+1\] \[x=3\] \[x\ne-5\] C. (\frac{x+6}{x^{2}-5x-24}\ge0) LCD ((x-8)(x+3)) \[\frac{x+6}{(x-8)(x+3)}\ge0\] (x-8)(x+3)+0 X-80 x+3+0 (6) (6.3)(8,0) X78 X7-3 IV. GRAPHING RATIONAL FUNCTIONS: \[y=\frac{1}{x}\] X y -5 -2 O undef. 2 5 domain: (-00, 0)(0,00) x-int: none y-int: none asymptotes describe restrictions and end behavior HA \(y=0\) VA \(x=0\) For rational functions in general: \(y=\frac{x}{x^{2}-4}\) let N(x) = x and \(D(x)=x^{2}-4\) domain let D(x)≠0 X x²-4 X (x-2)(x+2) ⇒(x-2)(x^2)=0{x X-270. X≠2 (-00,-2)0(-2,2)0(2,00) yint let \(x=0\) X 00 x²-4 (0)2-4 x-int let \(N(x)=0\) \[\frac{x}{x^{1}\cdot4}\rightarrow x=0\] |- 1- b) to find horizontal asymptotes: If the degree of the denominator is greater than that of the numerator, then there exists a horizontal asymptote at (y=0.) (i.e. it is "bottom heavy") If the degree of the numerator is greater that of the denominator, then there is no horizontal asymptote, instead, there exists an oblique asymptote, or slant asymptote. This is found by dividing the numerator by the denominator and ignoring any remainder. (ie. it is top heavy") If the degree of the numerator is equal to that of the denominator, then there exists a horizontal asymptote at (y=\frac{a}{b}) where a is the leading coefficient of the numerator, and b is the leading coefficient of the denominator.   b) to find vertical asymptotes: after simplifying the function let (D(x)=0) (P(x)=\frac{2x+1}{x+1}) domain: ((-\infty,-1)\cup(-1,\infty)) va: (x=-1) y-int: ((\frac{1}{7},0)) x-int: none ha: (y=2) oa: none end behavior: as (x\rightarrow\infty,) (y\rightarrow2) as (x\rightarrow-\infty) (y\rightarrow2) as (x\rightarrow-1^{-}) (y\rightarrow\infty) as (x\rightarrow-1^{+}) (y\rightarrow-\infty) (P(x)=\frac{x-4}{-4x-16}) domain: ((-\infty,-4)\cup(-4,\infty)) va: (x=-4) y-int: ((\frac{1}{4},0)) x-int: ({4}) ha: none oa: none end behavior: as (x\rightarrow\infty,) (y\rightarrow-\frac{1}{4}) as (x\rightarrow-\infty) (y\rightarrow-\frac{1}{4}) as (x\rightarrow-4^{-}) (y\rightarrow\infty) as (x\rightarrow-4^{+}) (y\rightarrow-\infty)