The vertical asymptotes of the function are 3 and 6
Given the function
[tex]\frac{x+6}{x^2-9x+18}[/tex]
The vertical asymptotes of the function occur at the point where the denominator is zero.
Equate the denominator to zero.
x^2 - 9x + 18 = 0
Factorize the quadratic expression:
x^2 - 3x - 6x + 18 =0
x(x-3)-6(x-3) = 0
(x-3)(x-6) = 0
x - 3 = 0 and x - 6 = 0
x = 3 and 6
Hence the vertical asymptotes of the function are 3 and 6
Learn more on vertical asymptotes here: https://brainly.com/question/21678319
