Answer :
Certainly, let's analyze the function [tex]\( f(x) = \frac{x^2 - 2x - 3}{x + 2} \)[/tex].
1. Factoring the Numerator:
The function [tex]\( f(x) = \frac{x^2 - 2x - 3}{x + 2} \)[/tex] can be simplified by factoring the quadratic expression in the numerator.
Factor the numerator [tex]\( x^2 - 2x - 3 \)[/tex]:
[tex]\[ x^2 - 2x - 3 = (x - 3)(x + 1) \][/tex]
Substituting this back into the function, we get:
[tex]\[ f(x) = \frac{(x - 3)(x + 1)}{x + 2} \][/tex]
2. Simplifying the Function:
Notice that the numerator has factors [tex]\( (x - 3)(x + 1) \)[/tex] and the denominator is [tex]\( (x + 2) \)[/tex]. This doesn't simplify directly, so let's proceed to the next step.
3. Finding the Vertical Asymptote:
The vertical asymptote of a rational function occurs where the denominator is zero (and the numerator is not zero at that point):
[tex]\[ x + 2 = 0 \Rightarrow x = -2 \][/tex]
So, there is a vertical asymptote at [tex]\( x = -2 \)[/tex].
4. Finding the Horizontal Asymptote:
To find the horizontal asymptote, consider the degrees of the numerator and the denominator:
- Degree of the numerator ([tex]\(x^2\)[/tex]) is 2.
- Degree of the denominator ([tex]\(x\)[/tex]) is 1.
Since the degree of the numerator is greater than the degree of the denominator, this implies there is no horizontal asymptote. Instead, we will have an oblique (slant) asymptote.
5. Finding x- and y-intercepts:
For [tex]\( x \)[/tex]-intercepts, set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ \frac{(x - 3)(x + 1)}{x + 2} = 0 \Rightarrow x - 3 = 0 \text{ or } x + 1 = 0 \Rightarrow x = 3 \text{ or } x = -1 \][/tex]
For the [tex]\( y \)[/tex]-intercept, set [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{0^2 - 2 \cdot 0 - 3}{0 + 2} = \frac{-3}{2} = -\frac{3}{2} \][/tex]
6. Behavior at Intercepts and Asymptotes:
The function goes through the [tex]\( x \)[/tex]-intercepts at [tex]\( x = 3 \)[/tex] and [tex]\( x = -1 \)[/tex], and the [tex]\( y \)[/tex]-intercept at [tex]\( y = -\frac{3}{2} \)[/tex].
7. Oblique Asymptote:
Since the degree of the numerator is one more than the degree of the denominator, perform polynomial long division to find the equation of the oblique asymptote:
[tex]\[ \frac{x^2 - 2x - 3}{x + 2} \][/tex]
Performing the division yields:
[tex]\[ x - 4 + \frac{5}{x+2} \][/tex]
So, the oblique asymptote is [tex]\( y = x - 4 \)[/tex].
In summary, the graph has:
- Vertical asymptote at [tex]\( x = -2 \)[/tex],
- X-intercepts at [tex]\( x = 3 \)[/tex] and [tex]\( x = -1 \)[/tex],
- Y-intercept at [tex]\( y = -\frac{3}{2} \)[/tex],
- Oblique asymptote of [tex]\( y = x - 4 \)[/tex].
With these specifics at hand, the correct graph should reflect all these key features.
1. Factoring the Numerator:
The function [tex]\( f(x) = \frac{x^2 - 2x - 3}{x + 2} \)[/tex] can be simplified by factoring the quadratic expression in the numerator.
Factor the numerator [tex]\( x^2 - 2x - 3 \)[/tex]:
[tex]\[ x^2 - 2x - 3 = (x - 3)(x + 1) \][/tex]
Substituting this back into the function, we get:
[tex]\[ f(x) = \frac{(x - 3)(x + 1)}{x + 2} \][/tex]
2. Simplifying the Function:
Notice that the numerator has factors [tex]\( (x - 3)(x + 1) \)[/tex] and the denominator is [tex]\( (x + 2) \)[/tex]. This doesn't simplify directly, so let's proceed to the next step.
3. Finding the Vertical Asymptote:
The vertical asymptote of a rational function occurs where the denominator is zero (and the numerator is not zero at that point):
[tex]\[ x + 2 = 0 \Rightarrow x = -2 \][/tex]
So, there is a vertical asymptote at [tex]\( x = -2 \)[/tex].
4. Finding the Horizontal Asymptote:
To find the horizontal asymptote, consider the degrees of the numerator and the denominator:
- Degree of the numerator ([tex]\(x^2\)[/tex]) is 2.
- Degree of the denominator ([tex]\(x\)[/tex]) is 1.
Since the degree of the numerator is greater than the degree of the denominator, this implies there is no horizontal asymptote. Instead, we will have an oblique (slant) asymptote.
5. Finding x- and y-intercepts:
For [tex]\( x \)[/tex]-intercepts, set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ \frac{(x - 3)(x + 1)}{x + 2} = 0 \Rightarrow x - 3 = 0 \text{ or } x + 1 = 0 \Rightarrow x = 3 \text{ or } x = -1 \][/tex]
For the [tex]\( y \)[/tex]-intercept, set [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{0^2 - 2 \cdot 0 - 3}{0 + 2} = \frac{-3}{2} = -\frac{3}{2} \][/tex]
6. Behavior at Intercepts and Asymptotes:
The function goes through the [tex]\( x \)[/tex]-intercepts at [tex]\( x = 3 \)[/tex] and [tex]\( x = -1 \)[/tex], and the [tex]\( y \)[/tex]-intercept at [tex]\( y = -\frac{3}{2} \)[/tex].
7. Oblique Asymptote:
Since the degree of the numerator is one more than the degree of the denominator, perform polynomial long division to find the equation of the oblique asymptote:
[tex]\[ \frac{x^2 - 2x - 3}{x + 2} \][/tex]
Performing the division yields:
[tex]\[ x - 4 + \frac{5}{x+2} \][/tex]
So, the oblique asymptote is [tex]\( y = x - 4 \)[/tex].
In summary, the graph has:
- Vertical asymptote at [tex]\( x = -2 \)[/tex],
- X-intercepts at [tex]\( x = 3 \)[/tex] and [tex]\( x = -1 \)[/tex],
- Y-intercept at [tex]\( y = -\frac{3}{2} \)[/tex],
- Oblique asymptote of [tex]\( y = x - 4 \)[/tex].
With these specifics at hand, the correct graph should reflect all these key features.
