Answer :
To graph the rational function [tex]\( f(x) = \frac{x^2 + 9}{x - 2} \)[/tex], we will identify its key features, including vertical asymptotes, horizontal and/or oblique asymptotes, x-intercepts, and y-intercepts.
### 1. Vertical Asymptotes:
Vertical asymptotes occur where the denominator is zero and the numerator is non-zero. The denominator of [tex]\( f(x) \)[/tex] becomes zero when [tex]\( x - 2 = 0 \)[/tex], which gives us:
[tex]\[ x = 2 \][/tex]
So, the vertical asymptote is at [tex]\( x = 2 \)[/tex].
### 2. Horizontal Asymptotes:
To determine the horizontal asymptote, we compare the degrees of the numerator and the denominator. Here, the degree of the numerator [tex]\( x^2 + 9 \)[/tex] is 2, and the degree of the denominator [tex]\( x - 2 \)[/tex] is 1. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.
### 3. Oblique Asymptotes:
Since the degree of the numerator (2) is exactly one more than the degree of the denominator (1), the function has an oblique asymptote. To find the oblique asymptote, we perform polynomial division on [tex]\( x^2 + 9 \)[/tex] divided by [tex]\( x - 2 \)[/tex]:
Performing the division, we get:
[tex]\[ x^2 + 9 \div (x - 2) = x + 2 \][/tex]
with a remainder of 13. Thus, the oblique asymptote is the quotient [tex]\( y = x + 2 \)[/tex].
### 4. X-Intercepts:
The x-intercepts are located where the numerator equals zero. To find them, we solve the equation [tex]\( x^2 + 9 = 0 \)[/tex]:
[tex]\[ x^2 = -9 \][/tex]
[tex]\[ x = \pm 3i \][/tex]
Since the solutions are imaginary numbers ([tex]\( \pm 3i \)[/tex]), there are no real x-intercepts.
### 5. Y-Intercepts:
The y-intercept occurs where [tex]\( x = 0 \)[/tex]. Substituting [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = \frac{0^2 + 9}{0 - 2} \][/tex]
[tex]\[ f(0) = \frac{9}{-2} \][/tex]
[tex]\[ f(0) = -\frac{9}{2} \][/tex]
So, the y-intercept is [tex]\( \left( 0, -\frac{9}{2} \right) \)[/tex].
### Summary of Key Features:
- Vertical Asymptote: [tex]\( x = 2 \)[/tex]
- Horizontal Asymptote: None
- Oblique Asymptote: [tex]\( y = x + 2 \)[/tex]
- X-Intercepts: None (complex solutions [tex]\( \pm 3i \)[/tex])
- Y-Intercept: [tex]\( \left( 0, -\frac{9}{2} \right) \)[/tex]
### Plotting:
- Draw a dashed vertical line at [tex]\( x = 2 \)[/tex] for the vertical asymptote.
- Draw a dashed line at [tex]\( y = x + 2 \)[/tex] for the oblique asymptote.
- Mark the point [tex]\( \left( 0, -\frac{9}{2} \right) \)[/tex] for the y-intercept.
- Note that since there are no real x-intercepts, the graph does not touch the x-axis.
With these features identified, plot the function, bearing in mind that the graph will approach the asymptotes but never touch them. The function will have distinct behaviors in the regions separated by the vertical asymptote.
### 1. Vertical Asymptotes:
Vertical asymptotes occur where the denominator is zero and the numerator is non-zero. The denominator of [tex]\( f(x) \)[/tex] becomes zero when [tex]\( x - 2 = 0 \)[/tex], which gives us:
[tex]\[ x = 2 \][/tex]
So, the vertical asymptote is at [tex]\( x = 2 \)[/tex].
### 2. Horizontal Asymptotes:
To determine the horizontal asymptote, we compare the degrees of the numerator and the denominator. Here, the degree of the numerator [tex]\( x^2 + 9 \)[/tex] is 2, and the degree of the denominator [tex]\( x - 2 \)[/tex] is 1. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.
### 3. Oblique Asymptotes:
Since the degree of the numerator (2) is exactly one more than the degree of the denominator (1), the function has an oblique asymptote. To find the oblique asymptote, we perform polynomial division on [tex]\( x^2 + 9 \)[/tex] divided by [tex]\( x - 2 \)[/tex]:
Performing the division, we get:
[tex]\[ x^2 + 9 \div (x - 2) = x + 2 \][/tex]
with a remainder of 13. Thus, the oblique asymptote is the quotient [tex]\( y = x + 2 \)[/tex].
### 4. X-Intercepts:
The x-intercepts are located where the numerator equals zero. To find them, we solve the equation [tex]\( x^2 + 9 = 0 \)[/tex]:
[tex]\[ x^2 = -9 \][/tex]
[tex]\[ x = \pm 3i \][/tex]
Since the solutions are imaginary numbers ([tex]\( \pm 3i \)[/tex]), there are no real x-intercepts.
### 5. Y-Intercepts:
The y-intercept occurs where [tex]\( x = 0 \)[/tex]. Substituting [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = \frac{0^2 + 9}{0 - 2} \][/tex]
[tex]\[ f(0) = \frac{9}{-2} \][/tex]
[tex]\[ f(0) = -\frac{9}{2} \][/tex]
So, the y-intercept is [tex]\( \left( 0, -\frac{9}{2} \right) \)[/tex].
### Summary of Key Features:
- Vertical Asymptote: [tex]\( x = 2 \)[/tex]
- Horizontal Asymptote: None
- Oblique Asymptote: [tex]\( y = x + 2 \)[/tex]
- X-Intercepts: None (complex solutions [tex]\( \pm 3i \)[/tex])
- Y-Intercept: [tex]\( \left( 0, -\frac{9}{2} \right) \)[/tex]
### Plotting:
- Draw a dashed vertical line at [tex]\( x = 2 \)[/tex] for the vertical asymptote.
- Draw a dashed line at [tex]\( y = x + 2 \)[/tex] for the oblique asymptote.
- Mark the point [tex]\( \left( 0, -\frac{9}{2} \right) \)[/tex] for the y-intercept.
- Note that since there are no real x-intercepts, the graph does not touch the x-axis.
With these features identified, plot the function, bearing in mind that the graph will approach the asymptotes but never touch them. The function will have distinct behaviors in the regions separated by the vertical asymptote.
