Answer :
Let's analyze and graph the function
[tex]\[ h(x) = \frac{2x - 8}{x + 3} \][/tex]
### Step 1: Identify the Asymptotes
#### Vertical Asymptote
A vertical asymptote occurs where the denominator is equal to zero, which makes the function undefined. Set the denominator equal to zero:
[tex]\[ x + 3 = 0 \][/tex]
[tex]\[ x = -3 \][/tex]
So, there is a vertical asymptote at [tex]\( x = -3 \)[/tex].
#### Horizontal Asymptote
To determine the horizontal asymptote, compare the degrees of the numerator and denominator. Both the numerator (2x - 8) and the denominator (x + 3) are linear (degree 1). When the degrees are the same, the horizontal asymptote is found by dividing the leading coefficients:
[tex]\[ y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}} = \frac{2}{1} = 2 \][/tex]
So, there is a horizontal asymptote at [tex]\( y = 2 \)[/tex].
### Step 2: Calculate Function Values at Specific Points
To understand the behavior of the function around the vertical asymptote and to plot some points on the graph, let's choose several [tex]\( x \)[/tex]-values:
- [tex]\( x = -5 \)[/tex]
- [tex]\( x = -4 \)[/tex]
- [tex]\( x = -3.5 \)[/tex] (approaching vertical asymptote from the left)
- [tex]\( x = -2 \)[/tex] (approaching vertical asymptote from the right)
- [tex]\( x = 0 \)[/tex]
- [tex]\( x = 2 \)[/tex]
- [tex]\( x = 5 \)[/tex]
Compute the function values for these [tex]\( x \)[/tex]-values:
[tex]\[ \begin{aligned} h(-5) &= \frac{2(-5) - 8}{-5 + 3} = \frac{-10 - 8}{-2} = \frac{-18}{-2} = 9.0, \\ h(-4) &= \frac{2(-4) - 8}{-4 + 3} = \frac{-8 - 8}{-1} = \frac{-16}{-1} = 16.0, \\ h(-3.5) &= \frac{2(-3.5) - 8}{-3.5 + 3} = \frac{-7 - 8}{-0.5} = \frac{-15}{-0.5} = 30.0, \\ h(-2) &= \frac{2(-2) - 8}{-2 + 3} = \frac{-4 - 8}{1} = \frac{-12}{1} = -12.0, \\ h(0) &= \frac{2(0) - 8}{0 + 3} = \frac{-8}{3} \approx -2.67, \\ h(2) &= \frac{2(2) - 8}{2 + 3} = \frac{4 - 8}{5} = \frac{-4}{5} = -0.8, \\ h(5) &= \frac{2(5) - 8}{5 + 3} = \frac{10 - 8}{8} = \frac{2}{8} = 0.25. \end{aligned} \][/tex]
### Step 3: Plot the Points and Asymptotes
1. Plot the points:
- [tex]\( (-5, 9.0) \)[/tex]
- [tex]\( (-4, 16.0) \)[/tex]
- [tex]\( (-3.5, 30.0) \)[/tex]
- [tex]\( (-2, -12.0) \)[/tex]
- [tex]\( (0, -2.67) \)[/tex]
- [tex]\( (2, -0.8) \)[/tex]
- [tex]\( (5, 0.25) \)[/tex]
2. Draw the vertical asymptote: Draw a dashed line at [tex]\( x = -3 \)[/tex].
3. Draw the horizontal asymptote: Draw a dashed line at [tex]\( y = 2 \)[/tex].
4. Sketch the graph: Based on the points calculated and asymptotes, sketch the curve, making sure to approach the vertical asymptote without crossing it and to level out towards the horizontal asymptote as [tex]\( x \)[/tex] becomes very large or very small.
### Graph Summary
- Vertical Asymptote: [tex]\( x = -3 \)[/tex]
- Horizontal Asymptote: [tex]\( y = 2 \)[/tex]
- Points: [tex]\( (-5, 9.0), (-4, 16.0), (-3.5, 30.0), (-2, -12.0), (0, -2.67), (2, -0.8), (5, 0.25) \)[/tex]
By following these calculations and plotting accordingly, you should have an accurate graph of the function [tex]\( h(x) = \frac{2x-8}{x+3} \)[/tex].
[tex]\[ h(x) = \frac{2x - 8}{x + 3} \][/tex]
### Step 1: Identify the Asymptotes
#### Vertical Asymptote
A vertical asymptote occurs where the denominator is equal to zero, which makes the function undefined. Set the denominator equal to zero:
[tex]\[ x + 3 = 0 \][/tex]
[tex]\[ x = -3 \][/tex]
So, there is a vertical asymptote at [tex]\( x = -3 \)[/tex].
#### Horizontal Asymptote
To determine the horizontal asymptote, compare the degrees of the numerator and denominator. Both the numerator (2x - 8) and the denominator (x + 3) are linear (degree 1). When the degrees are the same, the horizontal asymptote is found by dividing the leading coefficients:
[tex]\[ y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}} = \frac{2}{1} = 2 \][/tex]
So, there is a horizontal asymptote at [tex]\( y = 2 \)[/tex].
### Step 2: Calculate Function Values at Specific Points
To understand the behavior of the function around the vertical asymptote and to plot some points on the graph, let's choose several [tex]\( x \)[/tex]-values:
- [tex]\( x = -5 \)[/tex]
- [tex]\( x = -4 \)[/tex]
- [tex]\( x = -3.5 \)[/tex] (approaching vertical asymptote from the left)
- [tex]\( x = -2 \)[/tex] (approaching vertical asymptote from the right)
- [tex]\( x = 0 \)[/tex]
- [tex]\( x = 2 \)[/tex]
- [tex]\( x = 5 \)[/tex]
Compute the function values for these [tex]\( x \)[/tex]-values:
[tex]\[ \begin{aligned} h(-5) &= \frac{2(-5) - 8}{-5 + 3} = \frac{-10 - 8}{-2} = \frac{-18}{-2} = 9.0, \\ h(-4) &= \frac{2(-4) - 8}{-4 + 3} = \frac{-8 - 8}{-1} = \frac{-16}{-1} = 16.0, \\ h(-3.5) &= \frac{2(-3.5) - 8}{-3.5 + 3} = \frac{-7 - 8}{-0.5} = \frac{-15}{-0.5} = 30.0, \\ h(-2) &= \frac{2(-2) - 8}{-2 + 3} = \frac{-4 - 8}{1} = \frac{-12}{1} = -12.0, \\ h(0) &= \frac{2(0) - 8}{0 + 3} = \frac{-8}{3} \approx -2.67, \\ h(2) &= \frac{2(2) - 8}{2 + 3} = \frac{4 - 8}{5} = \frac{-4}{5} = -0.8, \\ h(5) &= \frac{2(5) - 8}{5 + 3} = \frac{10 - 8}{8} = \frac{2}{8} = 0.25. \end{aligned} \][/tex]
### Step 3: Plot the Points and Asymptotes
1. Plot the points:
- [tex]\( (-5, 9.0) \)[/tex]
- [tex]\( (-4, 16.0) \)[/tex]
- [tex]\( (-3.5, 30.0) \)[/tex]
- [tex]\( (-2, -12.0) \)[/tex]
- [tex]\( (0, -2.67) \)[/tex]
- [tex]\( (2, -0.8) \)[/tex]
- [tex]\( (5, 0.25) \)[/tex]
2. Draw the vertical asymptote: Draw a dashed line at [tex]\( x = -3 \)[/tex].
3. Draw the horizontal asymptote: Draw a dashed line at [tex]\( y = 2 \)[/tex].
4. Sketch the graph: Based on the points calculated and asymptotes, sketch the curve, making sure to approach the vertical asymptote without crossing it and to level out towards the horizontal asymptote as [tex]\( x \)[/tex] becomes very large or very small.
### Graph Summary
- Vertical Asymptote: [tex]\( x = -3 \)[/tex]
- Horizontal Asymptote: [tex]\( y = 2 \)[/tex]
- Points: [tex]\( (-5, 9.0), (-4, 16.0), (-3.5, 30.0), (-2, -12.0), (0, -2.67), (2, -0.8), (5, 0.25) \)[/tex]
By following these calculations and plotting accordingly, you should have an accurate graph of the function [tex]\( h(x) = \frac{2x-8}{x+3} \)[/tex].
