Question

determine the vertical and horizontal asymptotes of $y = \frac{2x + 5}{x^{2}-6x + 9}$
horizontal asymptote:
select
vertical asymptote:
select

Explanation:

Step1: Find the horizontal asymptote

For a rational function \(y=\frac{f(x)}{g(x)}\) where \(f(x)=2x + 5\) (degree \(n = 1\)) and \(g(x)=x^{2}-6x + 9=(x - 3)^{2}\) (degree \(m=2\)).
Since \(nightarrow\pm\infty}\frac{2x + 5}{x^{2}-6x + 9}=0\). So the horizontal asymptote is \(y = 0\).

Step2: Find the vertical asymptote

Set the denominator equal to zero: \(x^{2}-6x + 9=(x - 3)^{2}=0\). Solving \((x - 3)^{2}=0\) gives \(x=3\).
Check that the numerator \(2x+5\) is not zero at \(x = 3\) (\(2\times3+5=6 + 5=11
eq0\)). So the vertical asymptote is \(x = 3\).

Answer:

Horizontal asymptote: \(y = 0\)
Vertical asymptote: \(x = 3\)