find the vertical asymptotes of the following rational function(remember VA is X) x+6/(x^2+11x+30 ) my answer is x+6/(x+6)(x+5) cancel alike term i get 1/(x+5 )and V.A:x=-5 is that right thanks

y=x+6/(x^2+11x+30)

Yes, you are correct! To find the vertical asymptotes of the rational function, you need to examine the denominator of the function and determine the values of x that make the denominator equal to zero.

Given the rational function: f(x) = (x+6)/(x^2+11x+30)

To find the vertical asymptotes:
1. Begin by factoring the denominator (x^2+11x+30) to determine when it is equal to zero.
x^2 + 11x + 30 = 0

2. Factorize the quadratic equation:
(x+5)(x+6) = 0

3. Set each factor equal to zero:
x+5 = 0 or x+6 = 0

4. Solve for x:
x = -5 or x = -6

Therefore, the vertical asymptotes of the rational function occur at x = -5 and x = -6.

However, in your simplification, you canceled out the common factor of (x+6) in the numerator and denominator, which is incorrect. Dividing by (x+6) would lead to the function having a hole at x = -6, but it would not affect the vertical asymptote. Thus, the correct simplified form of the function is f(x) = 1/(x+5).

In conclusion, the vertical asymptote of the given rational function is x = -5.