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To graph the function [tex]\( h(x) = \frac{2x - 8}{x + 3} \)[/tex], we need to follow these steps:
### 1. Identify the Vertical Asymptote
A vertical asymptote occurs where the denominator is zero but the numerator is not zero. Set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x + 3 = 0 \][/tex]
[tex]\[ x = -3 \][/tex]
Thus, the vertical asymptote is at [tex]\( x = -3 \)[/tex].
### 2. Identify the Horizontal Asymptote
To find the horizontal asymptote for the function [tex]\( h(x) = \frac{2x - 8}{x + 3} \)[/tex], we can compare the degrees of the polynomials in the numerator and the denominator. Both the numerator and denominator are linear (degree 1), so the horizontal asymptote is found by dividing the coefficients of the leading terms:
[tex]\[ \text{Horizontal Asymptote}: \frac{\text{Leading Coefficient of Numerator}}{\text{Leading Coefficient of Denominator}} = \frac{2}{1} = 2 \][/tex]
Therefore, the horizontal asymptote is at [tex]\( y = 2 \)[/tex].
### 3. Calculate Points on Each Side of the Vertical Asymptote
To better understand the behavior of the function near the vertical asymptote [tex]\( x = -3 \)[/tex], we calculate a few points on each side of this asymptote.
#### Points to the Left of [tex]\( x = -3 \)[/tex]:
- For [tex]\( x = -5 \)[/tex]:
[tex]\[ h(-5) = \frac{2(-5) - 8}{-5 + 3} = \frac{-10 - 8}{-2} = \frac{-18}{-2} = 9.0 \][/tex]
- For [tex]\( x = -4 \)[/tex]:
[tex]\[ h(-4) = \frac{2(-4) - 8}{-4 + 3} = \frac{-8 - 8}{-1} = \frac{-16}{-1} = 16.0 \][/tex]
- For [tex]\( x = -3.5 \)[/tex]:
[tex]\[ h(-3.5) = \frac{2(-3.5) - 8}{-3.5 + 3} = \frac{-7 - 8}{-0.5} = \frac{-15}{-0.5} = 30.0 \][/tex]
Thus, the points on the left of [tex]\( x = -3 \)[/tex] are:
[tex]\[ (-5, 9.0), (-4, 16.0), (-3.5, 30.0) \][/tex]
#### Points to the Right of [tex]\( x = -3 \)[/tex]:
- For [tex]\( x = -2.5 \)[/tex]:
[tex]\[ h(-2.5) = \frac{2(-2.5) - 8}{-2.5 + 3} = \frac{-5 - 8}{0.5} = \frac{-13}{0.5} = -26.0 \][/tex]
- For [tex]\( x = -2 \)[/tex]:
[tex]\[ h(-2) = \frac{2(-2) - 8}{-2 + 3} = \frac{-4 - 8}{1} = \frac{-12}{1} = -12.0 \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ h(0) = \frac{2(0) - 8}{0 + 3} = \frac{-8}{3} = -2.67 \][/tex]
Thus, the points on the right of [tex]\( x = -3 \)[/tex] are:
[tex]\[ (-2.5, -26.0), (-2, -12.0), (0, -2.67) \][/tex]
### 4. Plot the Graph
- Vertical Asymptote: Draw a dashed vertical line at [tex]\( x = -3 \)[/tex].
- Horizontal Asymptote: Draw a dashed horizontal line at [tex]\( y = 2 \)[/tex].
- Point Plotting: Plot the points calculated above on the coordinate plane.
Finally, sketch the graph of the function [tex]\( h(x) = \frac{2x - 8}{x + 3} \)[/tex], making sure it approaches the asymptotes appropriately:
1. On the left side of [tex]\( x = -3 \)[/tex], the function values rise steeply toward [tex]\( +\infty \)[/tex].
2. On the right side of [tex]\( x = -3 \)[/tex], the function values fall steeply toward [tex]\( -\infty \)[/tex].
3. As [tex]\( x \)[/tex] moves far to the left or right, the function values will approach the horizontal asymptote [tex]\( y = 2 \)[/tex].
Your graph should reflect this behavior, smoothly approaching the horizontal asymptote while steeply diverging near the vertical asymptote.
### 1. Identify the Vertical Asymptote
A vertical asymptote occurs where the denominator is zero but the numerator is not zero. Set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x + 3 = 0 \][/tex]
[tex]\[ x = -3 \][/tex]
Thus, the vertical asymptote is at [tex]\( x = -3 \)[/tex].
### 2. Identify the Horizontal Asymptote
To find the horizontal asymptote for the function [tex]\( h(x) = \frac{2x - 8}{x + 3} \)[/tex], we can compare the degrees of the polynomials in the numerator and the denominator. Both the numerator and denominator are linear (degree 1), so the horizontal asymptote is found by dividing the coefficients of the leading terms:
[tex]\[ \text{Horizontal Asymptote}: \frac{\text{Leading Coefficient of Numerator}}{\text{Leading Coefficient of Denominator}} = \frac{2}{1} = 2 \][/tex]
Therefore, the horizontal asymptote is at [tex]\( y = 2 \)[/tex].
### 3. Calculate Points on Each Side of the Vertical Asymptote
To better understand the behavior of the function near the vertical asymptote [tex]\( x = -3 \)[/tex], we calculate a few points on each side of this asymptote.
#### Points to the Left of [tex]\( x = -3 \)[/tex]:
- For [tex]\( x = -5 \)[/tex]:
[tex]\[ h(-5) = \frac{2(-5) - 8}{-5 + 3} = \frac{-10 - 8}{-2} = \frac{-18}{-2} = 9.0 \][/tex]
- For [tex]\( x = -4 \)[/tex]:
[tex]\[ h(-4) = \frac{2(-4) - 8}{-4 + 3} = \frac{-8 - 8}{-1} = \frac{-16}{-1} = 16.0 \][/tex]
- For [tex]\( x = -3.5 \)[/tex]:
[tex]\[ h(-3.5) = \frac{2(-3.5) - 8}{-3.5 + 3} = \frac{-7 - 8}{-0.5} = \frac{-15}{-0.5} = 30.0 \][/tex]
Thus, the points on the left of [tex]\( x = -3 \)[/tex] are:
[tex]\[ (-5, 9.0), (-4, 16.0), (-3.5, 30.0) \][/tex]
#### Points to the Right of [tex]\( x = -3 \)[/tex]:
- For [tex]\( x = -2.5 \)[/tex]:
[tex]\[ h(-2.5) = \frac{2(-2.5) - 8}{-2.5 + 3} = \frac{-5 - 8}{0.5} = \frac{-13}{0.5} = -26.0 \][/tex]
- For [tex]\( x = -2 \)[/tex]:
[tex]\[ h(-2) = \frac{2(-2) - 8}{-2 + 3} = \frac{-4 - 8}{1} = \frac{-12}{1} = -12.0 \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ h(0) = \frac{2(0) - 8}{0 + 3} = \frac{-8}{3} = -2.67 \][/tex]
Thus, the points on the right of [tex]\( x = -3 \)[/tex] are:
[tex]\[ (-2.5, -26.0), (-2, -12.0), (0, -2.67) \][/tex]
### 4. Plot the Graph
- Vertical Asymptote: Draw a dashed vertical line at [tex]\( x = -3 \)[/tex].
- Horizontal Asymptote: Draw a dashed horizontal line at [tex]\( y = 2 \)[/tex].
- Point Plotting: Plot the points calculated above on the coordinate plane.
Finally, sketch the graph of the function [tex]\( h(x) = \frac{2x - 8}{x + 3} \)[/tex], making sure it approaches the asymptotes appropriately:
1. On the left side of [tex]\( x = -3 \)[/tex], the function values rise steeply toward [tex]\( +\infty \)[/tex].
2. On the right side of [tex]\( x = -3 \)[/tex], the function values fall steeply toward [tex]\( -\infty \)[/tex].
3. As [tex]\( x \)[/tex] moves far to the left or right, the function values will approach the horizontal asymptote [tex]\( y = 2 \)[/tex].
Your graph should reflect this behavior, smoothly approaching the horizontal asymptote while steeply diverging near the vertical asymptote.
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