Describe the vertical asymptote(s) and hole(s) for the graph of y = (x-6)/(x^2 + 5x + 6)
The equation y = (x-6)/(x^2 + 5x + 6) can be simplified to y = (x-6)/((x+3)(x+2)).
This equation has a vertical asymptote at x = -3 and x = -2 because these are the values where the denominator becomes zero. Since the numerator does not become zero at x = -3 or x = -2, the function has a vertical asymptote at these values.
To determine if there is a hole in the graph, we can simplify the equation as y = 1/(x+2) and x cannot be equal to -3. This indicates a hole at x = -3 in the graph of the rational function.