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?
To determine the vertical asymptote and hole of the given function (x^2+x-6)/(x^2-9), we need to simplify the function and factor the denominator.
First, let's factor the denominator (x^2-9) by using the difference of squares identity: a^2 - b^2 = (a+b)(a-b).
The denominator (x^2-9) can be written as (x+3)(x-3).
Next, let's simplify the function by canceling out common factors between the numerator (x^2+x-6) and the denominator (x+3)(x-3).
The simplified function is (x^2+x-6)/(x+3)(x-3).
Now, let's find the values where the function is undefined. In this case, it occurs when the denominator is equal to zero.
Setting the denominator (x+3)(x-3) equal to zero, we get:
x+3=0 or x-3=0
Solving for x, we find:
x=-3 or x=3
Therefore, the function is undefined at x=-3 and x=3.
This means that the graph of the function will have a vertical asymptote at x=3 and a hole (removable discontinuity) at x=-3.
Therefore, the correct answer is c: asymptote: x=-3; hole: x=3.
To determine the vertical asymptote and hole for the given rational function, let's begin by identifying the factors of the numerator and denominator.
The numerator is x^2 + x - 6, which can be factored as (x + 3)(x - 2).
The denominator is x^2 - 9, which can be factored as (x + 3)(x - 3).
Therefore, the simplified form of the rational function is:
[(x + 3)(x - 2)] / [(x + 3)(x - 3)]
Now, we can observe that (x + 3) appears in both the numerator and the denominator. This indicates the presence of a hole at x = -3 since (x + 3)/(x + 3) equals 1.
So, we have a hole at x = -3, which eliminates option d from the possible answers.
Now, let's look at the denominator. Since x^2 - 9 = (x - 3)(x + 3), we see that the denominator becomes zero when x = 3 and x = -3.
The vertical asymptote occurs when the denominator of a rational function equals zero, which becomes x = 3 and x = -3 in this case.
Hence, the correct answer is option b: asymptote: x = 3; hole: x = -3.
Your first step should have been to simplify the expression:
(x^2+x-6)/(x^2-9)
= (x+3)(x-2)/( (x+3)(x-3) )
= (x-2)/(x-3), x ≠ 3
take over