Where are the asymptotes for the following function located?

[tex]\[ f(x)=\frac{7}{x^2-2x-24} \][/tex]

A. [tex]\(x=-4\)[/tex] and [tex]\(x=6\)[/tex]

B. [tex]\(x=-4\)[/tex] and [tex]\(x=7\)[/tex]

C. [tex]\(x=4\)[/tex] and [tex]\(x=-6\)[/tex]

D. [tex]\(x=6\)[/tex] and [tex]\(x=7\)[/tex]



Answer :

To determine the locations of the vertical asymptotes for the function
[tex]\[ f(x) = \frac{7}{x^2 - 2x - 24} \][/tex]
we need to find the values of [tex]\( x \)[/tex] where the denominator equals zero. Vertical asymptotes occur at these points because the function approaches infinity or negative infinity.

### Step-by-Step Solution:

1. Identify the denominator of the function:
[tex]\[ x^2 - 2x - 24 \][/tex]

2. Set the denominator equal to zero to find where the function is undefined:
[tex]\[ x^2 - 2x - 24 = 0 \][/tex]

3. Solve the quadratic equation by factoring. We need to find two numbers that multiply together to give [tex]\(-24\)[/tex] and add together to give [tex]\(-2\)[/tex].

After examining various combinations, we see that [tex]\((x - 6)(x + 4) = 0\)[/tex] works because:
[tex]\[ \begin{cases} -6 \cdot 4 = -24 \quad (\text{product}) \\ -6 + 4 = -2 \quad (\text{sums}) \end{cases} \][/tex]

4. Set each factor equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ \begin{aligned} x - 6 &= 0 \quad \Rightarrow \quad x = 6 \\ x + 4 &= 0 \quad \Rightarrow \quad x = -4 \end{aligned} \][/tex]

5. Conclusion: The vertical asymptotes are at [tex]\( x = 6 \)[/tex] and [tex]\( x = -4 \)[/tex].

Therefore, the locations of the vertical asymptotes for [tex]\( f(x) = \frac{7}{x^2 - 2x - 24} \)[/tex] are [tex]\( x = 6 \)[/tex] and [tex]\( x = -4 \)[/tex].

The correct answer is:
[tex]\[ x = 6 \text{ and } x = -4 \][/tex]