Question

find all horizontal asymptotes of the following function.
$f(x) = \frac{2(x + 6)(x + 10)}{3(5x - 9)(x + 10)}$
multiply out:
$f(x) = \frac{2x^2 + 32x + 120}{15x^2 + 123x - 270}$
degree of numerator same as denominator:
look at coefficients of highest exponent.
$\frac{2x^2 + 32x + 120}{15x^2 + 123x - 270}$
horizontal asymptote: $y = \frac{2}{15}$

Explanation:

Step1: Simplify the function

First, we can cancel out the common factor \((x + 10)\) in the numerator and the denominator (note that \(x
eq - 10\) to avoid division by zero). So the function \(f(x)=\frac{2(x + 6)}{3(5x-9)}\) (after canceling \((x + 10)\)). When we expand the numerator and the denominator, we get \(f(x)=\frac{2x^{2}+32x + 120}{15x^{2}+123x-270}\) as given.

Step2: Determine the degrees of numerator and denominator

The degree of a polynomial is the highest power of \(x\) in it. For the numerator \(2x^{2}+32x + 120\), the degree is \(2\) (since the highest power of \(x\) is \(x^{2}\)). For the denominator \(15x^{2}+123x - 270\), the degree is also \(2\). When the degrees of the numerator and the denominator of a rational function \(y = \frac{f(x)}{g(x)}\) are equal, the horizontal asymptote is given by the ratio of the leading coefficients (the coefficients of the highest - degree terms).

Step3: Find the leading coefficients and the horizontal asymptote

The leading coefficient of the numerator \(2x^{2}+32x + 120\) is \(2\) (the coefficient of \(x^{2}\)). The leading coefficient of the denominator \(15x^{2}+123x - 270\) is \(15\) (the coefficient of \(x^{2}\)). So the horizontal asymptote \(y\) is the ratio of these leading coefficients, i.e., \(y=\frac{2}{15}\).

Answer:

The horizontal asymptote of the function \(f(x)\) is \(y = \frac{2}{15}\)