Answer: -5, 5
Explanation:
To find the vertical asymptotes of the function f(x) = (3x^2 + 3x + 6) / (x^2 - 25), we need to determine the values of x for which the function approaches infinity or negative infinity as x approaches those values.
First, let's find the values of x that make the denominator equal to zero, as these values can potentially be vertical asymptotes. In this case, the denominator is (x^2 - 25), which can be factored as (x + 5)(x - 5). Setting this equal to zero, we have:
x + 5 = 0 or x - 5 = 0
Solving these equations, we find that x = -5 and x = 5. These are potential vertical asymptotes.
To determine if these are indeed vertical asymptotes, we need to check the behavior of the function as x approaches these values from both sides (positive and negative). We can do this by taking the limit of the function as x approaches the potential asymptotes.
Taking the limit as x approaches -5, we have:
lim(x -> -5) (3x^2 + 3x + 6) / (x^2 - 25)
By substituting -5 into the function, we get:
(3(-5)^2 + 3(-5) + 6) / ((-5)^2 - 25)
= (75 - 15 + 6) / (25 - 25)
= 66 / 0
Here, the denominator becomes zero, which means that the function approaches infinity as x approaches -5 from either side. Therefore, x = -5 is a vertical asymptote.
Now, let's take the limit as x approaches 5:
lim(x -> 5) (3x^2 + 3x + 6) / (x^2 - 25)
Substituting 5 into the function, we get:
(3(5)^2 + 3(5) + 6) / ((5)^2 - 25)
= (75 + 15 + 6) / (25 - 25)
= 96 / 0
Similarly, the denominator becomes zero, indicating that the function approaches infinity as x approaches 5 from either side. Therefore, x = 5 is also a vertical asymptote.
In summary, the vertical asymptotes of the function f(x) = (3x^2 + 3x + 6) / (x^2 - 25) are x = -5 and x = 5.
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