Certainly! Let's analyze the function [tex]\( h(x) = \frac{4x^2 - 100}{8x - 20} \)[/tex] step by step to identify its asymptotic behavior.
1. Simplifying the function: First, observe that the numerator and denominator can be simplified by factoring out a common factor.
- The numerator [tex]\( 4x^2 - 100 \)[/tex] can be factored as [tex]\( 4(x^2 - 25) \)[/tex]. Further, [tex]\( x^2 - 25 \)[/tex] is a difference of squares and can be factored as [tex]\( (x - 5)(x + 5) \)[/tex].
- The denominator [tex]\( 8x - 20 \)[/tex] can be simplified by factoring out the constant 4: [tex]\( 4(2x - 5) \)[/tex].
So, the function becomes:
[tex]\[
h(x) = \frac{4(x - 5)(x + 5)}{4(2x - 5)}
\][/tex]
The 4 in the numerator and denominator cancel out:
[tex]\[
h(x) = \frac{(x - 5)(x + 5)}{2x - 5}
\][/tex]
2. Finding the horizontal asymptote: To determine the horizontal asymptote, we need to compare the degrees of the polynomial in the numerator and the denominator.
- The degree of the numerator [tex]\( (x - 5)(x + 5) \)[/tex] is 2 (as it is equivalent to [tex]\( x^2 - 25 \)[/tex]).
- The degree of the denominator [tex]\( 2x - 5 \)[/tex] is 1.
Since the degree of the numerator (2) is higher than the degree of the denominator (1), this usually suggests no horizontal asymptote. However, when the degree of the numerator is higher by exactly one, there is an oblique (slant) asymptote.
3. Confirming this behavior: To identify the horizontal asymptote more precisely in such cases, recall that if the degrees were equal, the horizontal asymptote would be determined by the ratio of the leading coefficients of [tex]\( x^2 \)[/tex] in the numerator and the leading coefficient of [tex]\( x \)[/tex] in the denominator.
The leading term in the numerator of [tex]\( 4x^2 - 100 \)[/tex] is [tex]\( 4x^2 \)[/tex], and the leading term in the denominator [tex]\( 8x - 20 \)[/tex] is [tex]\( 8x \)[/tex].
The ratio of the leading coefficients:
[tex]\[
\frac{4x^2}{8x} = \frac{4}{8} = \frac{1}{2}
\][/tex]
Thus, the horizontal asymptote is given by [tex]\( y = \frac{1}{2} \)[/tex].
Therefore, the correct answer is:
C. There is a horizontal asymptote at [tex]\( y = \frac{1}{2} \)[/tex].
