Answer :
Certainly! Let's go through the process of graphing the rational function [tex]\( f(x) = \frac{3x^2 + 15x}{x^2 + 7x + 10} \)[/tex].
### Step 1: Identify Vertical Asymptotes
To identify vertical asymptotes, we look for the values of [tex]\( x \)[/tex] that make the denominator zero since the function tends to infinity at these points.
The denominator is [tex]\( x^2 + 7x + 10 \)[/tex]. Solving [tex]\( x^2 + 7x + 10 = 0 \)[/tex]:
[tex]\[ x^2 + 7x + 10 = (x + 5)(x + 2) = 0 \][/tex]
So, the solutions are:
[tex]\[ x = -5 \quad \text{and} \quad x = -2 \][/tex]
These are the vertical asymptotes.
### Step 2: Identify Holes
To find holes in the graph, we need to check if there are any common factors in both the numerator and the denominator.
The numerator is [tex]\( 3x^2 + 15x = 3x(x + 5) \)[/tex] and the denominator is [tex]\( x^2 + 7x + 10 = (x + 5)(x + 2) \)[/tex].
We see that [tex]\( x = -5 \)[/tex] is a common factor. This indicates a hole at [tex]\( x = -5 \)[/tex] rather than a vertical asymptote.
### Step 3: Calculate the y-Coordinate of the Hole
To find the y-coordinate of the hole at [tex]\( x = -5 \)[/tex], we cancel the common factor and substitute [tex]\( x = -5 \)[/tex] into the simplified function:
[tex]\[ f(x) = \frac{3x}{x + 2} \][/tex]
Substituting [tex]\( x = -5 \)[/tex]:
[tex]\[ f(-5) = \frac{3(-5)}{-5 + 2} = \frac{-15}{-3} = 5 \][/tex]
So, there is a hole at [tex]\( (-5, 5) \)[/tex].
### Step 4: Determine Other Points and Plot
Next, we plot the function while considering the asymptotes and the hole:
1. Vertical Asymptotes: There are vertical asymptotes at [tex]\( x = -2 \)[/tex].
2. Hole: There is a hole at [tex]\( (-5, 5) \)[/tex].
We now plot two points on each segment of the graph. We avoid the asymptotes, ensuring not to plot at [tex]\( x = -2 \)[/tex] or [tex]\( x = -5 \)[/tex].
Example points:
- For [tex]\( x < -5 \)[/tex]:
- [tex]\( x = -6 \)[/tex]
[tex]\[ f(-6) = \frac{3(-6)^2 + 15(-6)}{(-6)^2 + 7(-6) + 10} = \frac{108 - 90}{36 - 42 + 10} = \frac{18}{4} = 4.5 \][/tex]
- [tex]\( x = -7 \)[/tex]
[tex]\[ f(-7) = \frac{3(-7)^2 + 15(-7)}{(-7)^2 + 7(-7) + 10} = \frac{147 - 105}{49 - 49 + 10} = 4.2 \][/tex]
- For [tex]\( -5 < x < -2 \)[/tex]:
- [tex]\( x = -4 \)[/tex]
[tex]\[ f(-4) = \frac{3(-4)^2 + 15(-4)}{(-4)^2 + 7(-4) + 10} = \frac{48 - 60}{16 - 28 + 10} = -1.2 \][/tex]
- [tex]\( x = -3 \)[/tex]
[tex]\[ f(-3) = \frac{3(-3)^2 + 15(-3)}{(-3)^2 + 7(-3) + 10} = \frac{27 - 45}{9 - 21 + 10} = -1.8 \][/tex]
- For [tex]\( x > -2 \)[/tex]:
- [tex]\( x = -1 \)[/tex]
[tex]\[ f(-1) = \frac{3(-1)^2 + 15(-1)}{(-1)^2 + 7(-1) + 10} = \frac{3 - 15}{1 - 7 + 10} = -1.2 \][/tex]
- [tex]\( x = 0 \)[/tex]
[tex]\[ f(0) = \frac{3(0)^2 + 15(0)}{(0)^2 + 7(0) + 10} = 0.0 \][/tex]
- [tex]\( x = 1 \)[/tex]
[tex]\[ f(1) = \frac{3(1)^2 + 15(1)}{(1)^2 + 7(1) + 10} = \frac{18}{18} = 1.0 \][/tex]
### Step 5: Graph the Function
1. Draw the vertical asymptote at [tex]\( x = -2 \)[/tex].
2. Plot the points found in each section and draw the curves, ensuring to place a hollow dot at the hole [tex]\( (-5, 5) \)[/tex].
### Final Graph Description
The graph has vertical asymptotes at [tex]\( x = -2 \)[/tex] and a hole at [tex]\( (-5, 5) \)[/tex]. The segments of the graph formed by the identified points will approach these vertical asymptotes without actually touching them, and the function will be undefined at the hole.
To visualize, plot:
- Asymptote line at [tex]\( x = -2 \)[/tex].
- Hollow dot at [tex]\( (-5, 5) \)[/tex].
- Segmented curve passing through the points calculated, ensuring a break in the curve at the hole.
### Step 1: Identify Vertical Asymptotes
To identify vertical asymptotes, we look for the values of [tex]\( x \)[/tex] that make the denominator zero since the function tends to infinity at these points.
The denominator is [tex]\( x^2 + 7x + 10 \)[/tex]. Solving [tex]\( x^2 + 7x + 10 = 0 \)[/tex]:
[tex]\[ x^2 + 7x + 10 = (x + 5)(x + 2) = 0 \][/tex]
So, the solutions are:
[tex]\[ x = -5 \quad \text{and} \quad x = -2 \][/tex]
These are the vertical asymptotes.
### Step 2: Identify Holes
To find holes in the graph, we need to check if there are any common factors in both the numerator and the denominator.
The numerator is [tex]\( 3x^2 + 15x = 3x(x + 5) \)[/tex] and the denominator is [tex]\( x^2 + 7x + 10 = (x + 5)(x + 2) \)[/tex].
We see that [tex]\( x = -5 \)[/tex] is a common factor. This indicates a hole at [tex]\( x = -5 \)[/tex] rather than a vertical asymptote.
### Step 3: Calculate the y-Coordinate of the Hole
To find the y-coordinate of the hole at [tex]\( x = -5 \)[/tex], we cancel the common factor and substitute [tex]\( x = -5 \)[/tex] into the simplified function:
[tex]\[ f(x) = \frac{3x}{x + 2} \][/tex]
Substituting [tex]\( x = -5 \)[/tex]:
[tex]\[ f(-5) = \frac{3(-5)}{-5 + 2} = \frac{-15}{-3} = 5 \][/tex]
So, there is a hole at [tex]\( (-5, 5) \)[/tex].
### Step 4: Determine Other Points and Plot
Next, we plot the function while considering the asymptotes and the hole:
1. Vertical Asymptotes: There are vertical asymptotes at [tex]\( x = -2 \)[/tex].
2. Hole: There is a hole at [tex]\( (-5, 5) \)[/tex].
We now plot two points on each segment of the graph. We avoid the asymptotes, ensuring not to plot at [tex]\( x = -2 \)[/tex] or [tex]\( x = -5 \)[/tex].
Example points:
- For [tex]\( x < -5 \)[/tex]:
- [tex]\( x = -6 \)[/tex]
[tex]\[ f(-6) = \frac{3(-6)^2 + 15(-6)}{(-6)^2 + 7(-6) + 10} = \frac{108 - 90}{36 - 42 + 10} = \frac{18}{4} = 4.5 \][/tex]
- [tex]\( x = -7 \)[/tex]
[tex]\[ f(-7) = \frac{3(-7)^2 + 15(-7)}{(-7)^2 + 7(-7) + 10} = \frac{147 - 105}{49 - 49 + 10} = 4.2 \][/tex]
- For [tex]\( -5 < x < -2 \)[/tex]:
- [tex]\( x = -4 \)[/tex]
[tex]\[ f(-4) = \frac{3(-4)^2 + 15(-4)}{(-4)^2 + 7(-4) + 10} = \frac{48 - 60}{16 - 28 + 10} = -1.2 \][/tex]
- [tex]\( x = -3 \)[/tex]
[tex]\[ f(-3) = \frac{3(-3)^2 + 15(-3)}{(-3)^2 + 7(-3) + 10} = \frac{27 - 45}{9 - 21 + 10} = -1.8 \][/tex]
- For [tex]\( x > -2 \)[/tex]:
- [tex]\( x = -1 \)[/tex]
[tex]\[ f(-1) = \frac{3(-1)^2 + 15(-1)}{(-1)^2 + 7(-1) + 10} = \frac{3 - 15}{1 - 7 + 10} = -1.2 \][/tex]
- [tex]\( x = 0 \)[/tex]
[tex]\[ f(0) = \frac{3(0)^2 + 15(0)}{(0)^2 + 7(0) + 10} = 0.0 \][/tex]
- [tex]\( x = 1 \)[/tex]
[tex]\[ f(1) = \frac{3(1)^2 + 15(1)}{(1)^2 + 7(1) + 10} = \frac{18}{18} = 1.0 \][/tex]
### Step 5: Graph the Function
1. Draw the vertical asymptote at [tex]\( x = -2 \)[/tex].
2. Plot the points found in each section and draw the curves, ensuring to place a hollow dot at the hole [tex]\( (-5, 5) \)[/tex].
### Final Graph Description
The graph has vertical asymptotes at [tex]\( x = -2 \)[/tex] and a hole at [tex]\( (-5, 5) \)[/tex]. The segments of the graph formed by the identified points will approach these vertical asymptotes without actually touching them, and the function will be undefined at the hole.
To visualize, plot:
- Asymptote line at [tex]\( x = -2 \)[/tex].
- Hollow dot at [tex]\( (-5, 5) \)[/tex].
- Segmented curve passing through the points calculated, ensuring a break in the curve at the hole.
