Question

determining horizontal asymptotes
find the horizontal asymptote of the function $h(x) = \frac{-x^2 + 5x - 2}{x^2 + x}$.
$y = \square$

Explanation:

Step1: Identify degrees of numerator and denominator

The function is \( h(x)=\frac{-x^{2}+5x - 2}{x^{2}+x} \). The degree of the numerator (highest power of \( x \)) is \( 2 \) (from \( -x^{2} \)) and the degree of the denominator is also \( 2 \) (from \( x^{2} \)).

Step2: Use rule for horizontal asymptote (equal degrees)

When the degrees of the numerator (\( n \)) and denominator (\( m \)) are equal (\( n = m \)), the horizontal asymptote is the ratio of the leading coefficients. The leading coefficient of the numerator is \( - 1 \) (coefficient of \( x^{2} \)) and the leading coefficient of the denominator is \( 1 \) (coefficient of \( x^{2} \)). So the horizontal asymptote is \( y=\frac{-1}{1}=-1 \).

Answer:

\( -1 \)