To find the zeros of the function [tex]\( f(x) = \frac{x^2 + x - 6}{x^2 - x - 6} \)[/tex], we look for the points where the function equals zero, i.e., the numerator equals zero while the denominator does not equal zero.
Step-by-step approach:
1. Identify the numerator and denominator:
The function is given by
[tex]\[
f(x) = \frac{x^2 + x - 6}{x^2 - x - 6}
\][/tex]
where [tex]\( x^2 + x - 6 \)[/tex] is the numerator, and [tex]\( x^2 - x - 6 \)[/tex] is the denominator.
2. Find the zeros of the numerator:
To find where [tex]\( f(x) \)[/tex] is zero, set the numerator equal to zero:
[tex]\[
x^2 + x - 6 = 0
\][/tex]
Factor the quadratic equation:
[tex]\[
x^2 + x - 6 = (x + 3)(x - 2) = 0
\][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[
x + 3 = 0 \quad \text{or} \quad x - 2 = 0
\][/tex]
Therefore, the solutions are:
[tex]\[
x = -3 \quad \text{and} \quad x = 2
\][/tex]
These are potential zeros of [tex]\( f(x) \)[/tex].
3. Check the points where the denominator is zero:
The function is undefined where the denominator equals zero:
[tex]\[
x^2 - x - 6 = 0
\][/tex]
Factor the quadratic equation:
[tex]\[
x^2 - x - 6 = (x - 3)(x + 2) = 0
\][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[
x - 3 = 0 \quad \text{or} \quad x + 2 = 0
\][/tex]
Therefore, the solutions are:
[tex]\[
x = 3 \quad \text{and} \quad x = -2
\][/tex]
These are points where the function is not defined.
4. Confirm the zeros of the function:
From the numerator, we have potential zeros [tex]\( x = -3 \)[/tex] and [tex]\( x = 2 \)[/tex].
From the denominator, we identify that [tex]\( f(x) \)[/tex] is undefined at [tex]\( x = -2 \)[/tex] and [tex]\( x = 3 \)[/tex].
5. Validate the zeros:
Since [tex]\( -3 \)[/tex] and [tex]\( 2 \)[/tex] do not make the denominator zero, they are indeed the zeros of the function [tex]\( f(x) \)[/tex].
Hence, the zeros of the function [tex]\( f(x) \)[/tex] are [tex]\(-3\)[/tex] and [tex]\(2\)[/tex].
The correct answer is:
D. [tex]\(-3, 2\)[/tex]
