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

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

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

Answer

Answer:

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

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 (n<m), by the rule of horizontal asymptotes for rational functions (\lim_{x\rightarrow\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\neq0)). So the vertical asymptote is (x = 3).