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Chapter 3: Problem 39

Find the horizontal asymptote, if there is one, of the graph of each rational function. $$ g(x)=\frac{12 x^{2}}{3 x^{2}+1} $$

Short Answer

Expert verified
The horizontal asymptote of the given function is \(y = 4\).

Step by step solution

01

Identify the degrees of polynomials

First, you need to identify the degrees of the polynomial in the numerator and the denominator. Here, both the numerator, \(12x^2\), and the denominator, \(3x^2+1\), are of degree 2. Hence, the degrees of both polynomials are equal.
02

Find the ratio of the leading coefficients

Next, you should find the ratio of the leading coefficients of the highest degree terms in the numerator and the denominator. In the given function, the leading coefficient in the numerator is 12 and in the denominator is 3. So, the ratio of the leading coefficients is \(\frac{12}{3} = 4\).
03

Write down the horizontal asymptote

A horizontal line y = c is a horizontal asymptote of the function f, if either \(\lim_{{x}\to {+\infty}} f(x) = c\) or \(\lim_{{x}\to {-\infty}} f(x) = c\). In this case, as the degrees of the numerator and denominator are the same, the ratio of the leading coefficients will give the equation of the horizontal asymptote. So, the horizontal asymptote is \(y = 4\).

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