To determine the horizontal asymptote of the rational function [tex]\( f(x) = \frac{3(x-2)(x+5)}{2(x+4)^2} \)[/tex], follow these steps:
1. Identify the degrees of the numerator and the denominator:
- The numerator of the function is [tex]\( 3(x-2)(x+5) \)[/tex]. When expanded, the highest power term (degree) is [tex]\( 3x^2 \)[/tex], so the degree of the numerator is 2.
- The denominator of the function is [tex]\( 2(x+4)^2 \)[/tex]. When expanded, the highest power term (degree) is [tex]\( 2x^2 \)[/tex], so the degree of the denominator is also 2.
2. Compare the degrees of the numerator and the denominator:
- If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is [tex]\( y = 0 \)[/tex].
- If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is given by the ratio of the leading coefficients.
- If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote (there could be an oblique asymptote, but this is not relevant for this function).
3. Determine the leading coefficients and find the ratio:
- The leading coefficient of the numerator is 3 (from the term [tex]\( 3x^2 \)[/tex]).
- The leading coefficient of the denominator is 2 (from the term [tex]\( 2x^2 \)[/tex]).
- The horizontal asymptote is therefore [tex]\( y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}} = \frac{3}{2} \)[/tex].
Thus, the equation of the horizontal asymptote for the function [tex]\( f(x) = \frac{3(x-2)(x+5)}{2(x+4)^2} \)[/tex] is:
[tex]\[ y = \frac{3}{2} \][/tex]
So, the correct answer is:
[tex]\[ y = \frac{3}{2} \][/tex]
