QuestionFind all horizontal and vertical asymptotes (if any). (If an answer does not exist, enter DNE.) $$s(x)=\frac{4 x^{2}+1}{2 x^{2}+9 x-5}$$ vertical asymptote $x=$ $\square$ (smaller value) vertical asymptote $x=$ $\square$ (larger value) horizontal asymptote $\square$

Studdy Solution
To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Both the numerator $4x^2 + 1$ and the denominator $2x^2 + 9x - 5$ are degree 2 polynomials.
When the degrees are equal, the horizontal asymptote is given by the ratio of the leading coefficients:
$$\text{Horizontal Asymptote: } y = \frac{4}{2} = 2$$
The vertical asymptotes are at $x = -5$ and $x = \frac{1}{2}$. The horizontal asymptote is $y = 2$.
Vertical asymptote $x = \boxed{-5}$ (smaller value)
Vertical asymptote $x = \boxed{\frac{1}{2}}$ (larger value)
Horizontal asymptote $y = \boxed{2}$