25 Section 5.4 Rational Functions
a. [latex]k(x)=\frac{70x^5-90x^2+12}{120x^4+23x^3+5}[/latex]
b. [latex]g(x)=\frac{2x^3+4}{-9x^3+10}[/latex]
c. [latex]j(x)=\frac{(2x-5)(x-4)(x+7)}{(3x+1)(x+4)(x-1)}[/latex]
Show Solution
Answers
a. First, notice the degree of the denominator is 4, so divide all terms by [latex]x^4[/latex], then take the limit as [latex]x \to \infty[/latex].
[latex]\displaystyle\lim_{x \to \infty} \frac{\frac{70x^5}{x^4}-\frac{90x^2}{x^4}+\frac{12}{x^4}}{\frac{120x^4}{x^4}+\frac{23x^3}{x^4}+\frac{5}{x^4}}[/latex]
[latex]\displaystyle\lim_{x \to \infty} \frac{70x-\frac{90}{x^2}+\frac{12}{x^4}}{120+\frac{23}{x}+\frac{5}{x^4}}[/latex]
Then as [latex]x \to \infty[/latex], the fractions [latex]-\frac{90}{x^2}[/latex], [latex]\frac{12}{x^4}[/latex], [latex]\frac{23}{x}[/latex], and [latex]\frac{5}{x^4}[/latex] all approach 0.
This leaves: [latex]\displaystyle\lim_{x \to \infty} \frac{70x}{120}[/latex], which is [latex]\infty[/latex]. Thus this function does not have a horizontal asymptote.
Thinking about the domain of this function, the polynomial in the denominator will never equal 0. This makes the domain of the function all real numbers, [latex](-\infty, \infty)[/latex].
b. The numerator and denominator polynomials have the same degree; divide all terms by [latex]x^3[/latex] and take the limit as [latex]x \to \infty[/latex].
[latex]\displaystyle\lim_{x \to \infty} \frac{\frac{2x^3}{x^3}+\frac{4}{x^3}}{\frac{-9x^3}{x^3}+\frac{10}{x^3}}[/latex]
[latex]\displaystyle\lim_{x \to \infty} \frac{2+\frac{4}{x^3}}{-9+\frac{10}{x^3}}[/latex]
Then as [latex]x \to \infty[/latex], the two fractions [latex]\frac{4}{x^3}[/latex] and [latex]\frac{10}{x^3}[/latex] both approach 0, leaving the horizontal asymptote to be [latex]y=-\frac{2}{9}[/latex].
For the domain, look for [latex]x[/latex] values that make the denominator 0, ie solve [latex]-9x^3+10=0[/latex]. This gives [latex]\sqrt[3]{\frac{10}{9}}[/latex].
Thus, the domain is [latex](-\infty, \sqrt[3]{\frac{10}{9}}) \cup (\sqrt[3]{\frac{10}{9}}, \infty)[/latex].
c. First, do NOT use the distributive property to simplify the numerator and denominator. Instead, look at the terms that would create the leading term of each polynomial. In the numerator, you’d have [latex]2x^3[/latex], and in the denominator, [latex]3x^3[/latex]. Since they have the same degree, their coefficients make up the horizontal asymptote, [latex]y=\frac{2}{3}[/latex].
Since the denominator is factored, you can find the domain by setting each factor equal to 0, then excluding those values. The denominator is 0 when [latex]x=-\frac{1}{3}[/latex], [latex]x=-4[/latex], and [latex]x=1[/latex]. This makes the domain [latex](-\infty,-4)\cup (-4,-\frac{1}{3})\cup (-\frac{1}{3},1)\cup (1,\infty)[/latex].