Describe the vertical asymptote(s) and hole(s) for the graph of y = x minus six over x squared plus five times x plus six.
The function is y = (x - 6)/(x^2 + 5x + 6).
To find the vertical asymptotes, we need to look for values of x that make the denominator zero. In this case, the denominator can be factored as (x + 3)(x + 2). Therefore, the vertical asymptotes occur at x = -3 and x = -2.
To find the hole in the graph, we need to factor the numerator and denominator and see if there are common factors. In this case, the numerator can be factored as (x - 6) and the denominator can be factored as (x + 3)(x + 2). Since (x + 3) is a common factor in both the numerator and the denominator, we can simplify the function to y = (x - 6)/(x + 2) with the hole occurring at x = -3.
Therefore, the graph has a vertical asymptote at x = -3 and x = -2, and a hole at x = -3.
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I apologize for the mistake in my previous response. Let's correct that:
The function is y = (x - 6)/(x^2 + 5x + 6).
To find the vertical asymptotes, we need to look for values of x that make the denominator zero. In this case, the denominator can be factored as (x + 2)(x + 3). Therefore, the vertical asymptotes occur at x = -2 and x = -3.
To find the hole in the graph, we need to see if there are common factors that can be canceled out. In this case, there are no common factors between the numerator and the denominator that can be canceled out, so there is no hole in the graph.
Therefore, the graph has vertical asymptotes at x = -2 and x = -3, but no hole. Thank you for pointing that out.