Describe the vertical asymptote and hole for the graph of (x^2+x-6)/(x^2-9).
a. asymptote: x=2; hole: x=-3
b. asymptote: x=3; hole: x=2
c. asymptote: x=-3; hole: x=3
d. asymptote: x=3; hole: x=-3
I know that it has to either be b or d because the asymptote is x=3. Can someone please help with the rest?
(x^2+x-6)/(x^2-9) = (x+3)(x-2) / (x+3)(x-3)
so there is a hole at x = -3, where y = 0/0 (undefined)
Everywhere else, y = (x-2)/(x-3) which has
zero at x=2
asymptote at x=3
so, D
Well, with that expression in the numerator and denominator, we can simplify it first. (x^2+x-6)/(x^2-9) can be factored as (x+3)(x-2)/(x-3)(x+3).
Now, when x is equal to -3, we can see that both the numerator and the denominator have a factor of (x+3). So, there is a hole at x=-3.
Similarly, when x is equal to 3, the denominator becomes zero since (x-3) is a factor. But the numerator is still not zero. So, it creates a vertical asymptote at x=3.
Therefore, the correct answer is d. asymptote: x=3; hole: x=-3.
To determine the vertical asymptotes and holes for the given rational function, it is necessary to factor the denominator.
The denominator, x^2 - 9, can be factored as (x - 3)(x + 3).
When factoring the numerator, x^2 + x - 6, you can use the quadratic formula or factor it as (x - 2)(x + 3).
Thus, the rational function (x^2 + x - 6)/(x^2 - 9) can be written as [(x - 2)(x + 3)]/[(x - 3)(x + 3)].
Now, let's analyze the factors of this expression.
The factor (x - 2) in the numerator and denominator cancels out, creating a hole at x = 2.
The factor (x + 3) in the numerator and denominator can also be canceled out, but this does not create a hole.
Considering the denominator, (x - 3)(x + 3), the denominator equals zero when x = 3 and x = -3.
Therefore, the vertical asymptote occurs at x = 3.
In summary, the correct answer is:
b. asymptote: x=3; hole: x=2
To determine the vertical asymptote and hole for the given rational function, (x^2+x-6)/(x^2-9), we need to analyze the factors in the numerator and denominator.
First, let's factor the quadratic expressions:
x^2 + x - 6 can be factored as (x + 3)(x - 2)
x^2 - 9 can be factored as (x + 3)(x - 3)
Now, we can rewrite the rational function factored form:
(x^2 + x - 6)/(x^2 - 9) = (x + 3)(x - 2)/(x + 3)(x - 3)
The factor (x + 3) exists in both the numerator and denominator, which means it can be canceled out since it will equal 1 unless x = -3, where we would have a hole.
So, the simplified version of the function is:
(x - 2)/(x - 3)
To find the vertical asymptote, we need to see the factors in the denominator that don't cancel out. Here, the factor (x - 3) remains in the denominator, indicating that x = 3 is the vertical asymptote. Therefore, neither a nor d can be correct since they both mention x = 3 as the asymptote.
Now, we need to determine the hole. We found earlier that x = -3 creates a hole due to the cancelation of the (x + 3) factor. Therefore, the correct answer is:
a. asymptote: x = 3; hole: x = -3