Answer :
Sure, let's carefully analyze and sketch the rational function [tex]\( f(x)=\frac{-x^2 + 3x + 3}{x - 2} \)[/tex]. We will proceed step-by-step, starting with the asymptotes and then finding points to plot the function.
### Step 1: Find the Vertical Asymptote
A vertical asymptote occurs where the denominator is zero (i.e., where the function is undefined).
Set the denominator equal to zero:
[tex]\[ x - 2 = 0 \][/tex]
[tex]\[ x = 2 \][/tex]
So, there is a vertical asymptote at [tex]\( x = 2 \)[/tex].
### Step 2: Find the Horizontal Asymptote
For the horizontal asymptote, we compare the degrees of the numerator and the denominator:
- The numerator [tex]\( -x^2 + 3x + 3 \)[/tex] is a quadratic polynomial (degree 2).
- The denominator [tex]\( x - 2 \)[/tex] is a linear polynomial (degree 1).
Since the degree of the numerator is greater than the degree of the denominator, there isn't a horizontal asymptote. Instead, there will be an oblique (slant) asymptote.
### Step 3: Find the Oblique Asymptote
To find the oblique asymptote, we perform polynomial long division of the numerator by the denominator.
Dividing [tex]\( -x^2 + 3x + 3 \)[/tex] by [tex]\( x - 2 \)[/tex]:
[tex]\[ \frac{-x^2 + 3x + 3}{x - 2} \][/tex]
1. Divide the leading term of the numerator by the leading term of the denominator: [tex]\(-x^2 / x = -x\)[/tex].
2. Multiply [tex]\(-x\)[/tex] by [tex]\( x - 2 \)[/tex], which gives [tex]\(-x^2 + 2x\)[/tex].
3. Subtract [tex]\(-x^2 + 2x\)[/tex] from [tex]\(-x^2 + 3x + 3\)[/tex]:
[tex]\[ (-x^2 + 3x + 3) - (-x^2 + 2x) = x + 3 \][/tex]
4. Divide the new leading term of the numerator by the leading term of the denominator: [tex]\( x / x = 1 \)[/tex].
5. Multiply [tex]\( 1 \)[/tex] by [tex]\( x - 2 \)[/tex], which gives [tex]\( x - 2 \)[/tex].
6. Subtract [tex]\( x - 2 \)[/tex] from [tex]\( x + 3 \)[/tex]:
[tex]\[ (x + 3) - (x - 2) = 5 \][/tex]
So, the quotient is [tex]\( -x + 1 \)[/tex] (ignoring the remainder). Therefore, the equation of the oblique asymptote is:
[tex]\[ y = -x + 1 \][/tex]
### Step 4: Find and Plot Points Around the Vertical Asymptote
To better understand the behavior of the function around the vertical asymptote [tex]\( x = 2 \)[/tex], we choose some points to evaluate [tex]\( f(x) \)[/tex].
For [tex]\( x \)[/tex]:
- [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \frac{-1^2 + 3(1) + 3}{1 - 2} = \frac{-1 + 3 + 3}{-1} = \frac{5}{-1} = -5 \][/tex]
- [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = \frac{-3^2 + 3(3) + 3}{3 - 2} = \frac{-9 + 9 + 3}{1} = \frac{3}{1} = 3 \][/tex]
- [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{-0^2 + 3(0) + 3}{0 - 2} = \frac{3}{-2} = -1.5 \][/tex]
- [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = \frac{-4^2 + 3(4) + 3}{4 - 2} = \frac{-16 + 12 + 3}{2} = \frac{-1}{2} = -0.5 \][/tex]
### Step 5: Sketch the Graph
To sketch the graph of [tex]\( f(x) \)[/tex]:
1. Draw the vertical asymptote [tex]\( x = 2 \)[/tex].
2. Draw the oblique asymptote [tex]\( y = -x + 1 \)[/tex].
3. Plot the calculated points: [tex]\((1, -5)\)[/tex], [tex]\((3, 3)\)[/tex], [tex]\((0, -1.5)\)[/tex], and [tex]\((4, -0.5)\)[/tex].
4. Sketch the curve, ensuring it approaches the asymptotes appropriately as [tex]\( x \)[/tex] gets closer to 2 and as [tex]\( x \)[/tex] approaches [tex]\(\pm \infty\)[/tex].
By following these steps, you can draw an accurate graph of the function [tex]\( f(x) \)[/tex].
### Step 1: Find the Vertical Asymptote
A vertical asymptote occurs where the denominator is zero (i.e., where the function is undefined).
Set the denominator equal to zero:
[tex]\[ x - 2 = 0 \][/tex]
[tex]\[ x = 2 \][/tex]
So, there is a vertical asymptote at [tex]\( x = 2 \)[/tex].
### Step 2: Find the Horizontal Asymptote
For the horizontal asymptote, we compare the degrees of the numerator and the denominator:
- The numerator [tex]\( -x^2 + 3x + 3 \)[/tex] is a quadratic polynomial (degree 2).
- The denominator [tex]\( x - 2 \)[/tex] is a linear polynomial (degree 1).
Since the degree of the numerator is greater than the degree of the denominator, there isn't a horizontal asymptote. Instead, there will be an oblique (slant) asymptote.
### Step 3: Find the Oblique Asymptote
To find the oblique asymptote, we perform polynomial long division of the numerator by the denominator.
Dividing [tex]\( -x^2 + 3x + 3 \)[/tex] by [tex]\( x - 2 \)[/tex]:
[tex]\[ \frac{-x^2 + 3x + 3}{x - 2} \][/tex]
1. Divide the leading term of the numerator by the leading term of the denominator: [tex]\(-x^2 / x = -x\)[/tex].
2. Multiply [tex]\(-x\)[/tex] by [tex]\( x - 2 \)[/tex], which gives [tex]\(-x^2 + 2x\)[/tex].
3. Subtract [tex]\(-x^2 + 2x\)[/tex] from [tex]\(-x^2 + 3x + 3\)[/tex]:
[tex]\[ (-x^2 + 3x + 3) - (-x^2 + 2x) = x + 3 \][/tex]
4. Divide the new leading term of the numerator by the leading term of the denominator: [tex]\( x / x = 1 \)[/tex].
5. Multiply [tex]\( 1 \)[/tex] by [tex]\( x - 2 \)[/tex], which gives [tex]\( x - 2 \)[/tex].
6. Subtract [tex]\( x - 2 \)[/tex] from [tex]\( x + 3 \)[/tex]:
[tex]\[ (x + 3) - (x - 2) = 5 \][/tex]
So, the quotient is [tex]\( -x + 1 \)[/tex] (ignoring the remainder). Therefore, the equation of the oblique asymptote is:
[tex]\[ y = -x + 1 \][/tex]
### Step 4: Find and Plot Points Around the Vertical Asymptote
To better understand the behavior of the function around the vertical asymptote [tex]\( x = 2 \)[/tex], we choose some points to evaluate [tex]\( f(x) \)[/tex].
For [tex]\( x \)[/tex]:
- [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \frac{-1^2 + 3(1) + 3}{1 - 2} = \frac{-1 + 3 + 3}{-1} = \frac{5}{-1} = -5 \][/tex]
- [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = \frac{-3^2 + 3(3) + 3}{3 - 2} = \frac{-9 + 9 + 3}{1} = \frac{3}{1} = 3 \][/tex]
- [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{-0^2 + 3(0) + 3}{0 - 2} = \frac{3}{-2} = -1.5 \][/tex]
- [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = \frac{-4^2 + 3(4) + 3}{4 - 2} = \frac{-16 + 12 + 3}{2} = \frac{-1}{2} = -0.5 \][/tex]
### Step 5: Sketch the Graph
To sketch the graph of [tex]\( f(x) \)[/tex]:
1. Draw the vertical asymptote [tex]\( x = 2 \)[/tex].
2. Draw the oblique asymptote [tex]\( y = -x + 1 \)[/tex].
3. Plot the calculated points: [tex]\((1, -5)\)[/tex], [tex]\((3, 3)\)[/tex], [tex]\((0, -1.5)\)[/tex], and [tex]\((4, -0.5)\)[/tex].
4. Sketch the curve, ensuring it approaches the asymptotes appropriately as [tex]\( x \)[/tex] gets closer to 2 and as [tex]\( x \)[/tex] approaches [tex]\(\pm \infty\)[/tex].
By following these steps, you can draw an accurate graph of the function [tex]\( f(x) \)[/tex].
