Let's analyze the function [tex]\( f(x) = \frac{x^2 - 6x + 8}{x + 2} \)[/tex] step by step.
### 1. Domain of the Function
First, we consider the domain of [tex]\( f(x) \)[/tex]. The function is defined wherever the denominator is not zero. For the denominator [tex]\( x + 2 \)[/tex]:
[tex]\[ x + 2 \neq 0 \implies x \neq -2 \][/tex]
Therefore, the domain of the function is all real numbers except [tex]\( x = -2 \)[/tex].
In interval notation, this is:
[tex]\[ (-\infty, -2) \cup (-2, \infty) \][/tex]
### 2. The [tex]\( x \)[/tex]-intercept(s)
To find the [tex]\( x \)[/tex]-intercepts, we set the numerator of [tex]\( f(x) \)[/tex] to zero and solve for [tex]\( x \)[/tex]. This is because the [tex]\( x \)[/tex]-intercepts occur when the numerator is zero (the function value [tex]\( f(x) = 0 \)[/tex]).
The numerator is:
[tex]\[ x^2 - 6x + 8 \][/tex]
We solve the equation:
[tex]\[ x^2 - 6x + 8 = 0 \][/tex]
Factoring the quadratic:
[tex]\[ (x - 2)(x - 4) = 0 \][/tex]
Setting each factor to zero:
[tex]\[ x - 2 = 0 \implies x = 2 \][/tex]
[tex]\[ x - 4 = 0 \implies x = 4 \][/tex]
Thus, the [tex]\( x \)[/tex]-intercepts are:
[tex]\[ (2, 0) \quad \text{and} \quad (4, 0) \][/tex]
So, the correct choice is:
A. The [tex]\( x \)[/tex]-intercept(s) is/are [tex]\( (2, 0), (4, 0) \)[/tex].
### 3. The [tex]\( y \)[/tex]-intercept(s)
To find the [tex]\( y \)[/tex]-intercept, we set [tex]\( x = 0 \)[/tex] and solve for [tex]\( f(x) \)[/tex].
Calculate [tex]\( f(0) \)[/tex]:
[tex]\[ f(0) = \frac{0^2 - 6 \cdot 0 + 8}{0 + 2} = \frac{8}{2} = 4 \][/tex]
Thus, the [tex]\( y \)[/tex]-intercept is:
[tex]\[ (0, 4) \][/tex]
So, the correct choice is:
A. The [tex]\( y \)[/tex]-intercept(s) is/are [tex]\( (0, 4) \)[/tex].
In summary:
1. The domain of the function is: [tex]\( (-\infty, -2) \cup (-2, \infty) \)[/tex]
2. The [tex]\( x \)[/tex]-intercepts are: [tex]\( (2, 0), (4, 0) \)[/tex]
3. The [tex]\( y \)[/tex]-intercept is: [tex]\( (0, 4) \)[/tex]
