Answer :
Alright, let's determine each feature of the graph for the given function [tex]\( f(x) = \frac{5x + 5}{2x^2 + 5x + 3} \)[/tex] step-by-step.
### Horizontal Asymptote
To find the horizontal asymptote, we compare the degrees of the numerator and the denominator:
- The degree of the numerator (5x + 5) is 1.
- The degree of the denominator (2x^2 + 5x + 3) is 2.
Since the degree of the denominator is higher than the degree of the numerator, the horizontal asymptote is [tex]\( y = 0 \)[/tex].
Horizontal Asymptote: [tex]\( y = 0 \)[/tex]
### Vertical Asymptote
To find the vertical asymptotes, we find the values of [tex]\( x \)[/tex] that make the denominator zero (since the denominator should not be zero for the function to be defined).
First, we find the roots of the denominator [tex]\( 2x^2 + 5x + 3 = 0 \)[/tex]:
[tex]\[ 2x^2 + 5x + 3 = 0 \][/tex]
By solving this quadratic equation, we get two solutions:
[tex]\[ x = -\frac{3}{2} \quad \text{and} \quad x = -1 \][/tex]
So, the function has vertical asymptotes at these values.
Vertical Asymptotes: [tex]\( x = -\frac{3}{2} \)[/tex] and [tex]\( x = -1 \)[/tex]
### x-Intercept
To find the [tex]\( x \)[/tex]-intercept, we set the numerator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 5x + 5 = 0 \][/tex]
[tex]\[ x = -1 \][/tex]
Thus, the [tex]\( x \)[/tex]-intercept is at the point [tex]\((-1, 0)\)[/tex].
x-Intercept: [tex]\((-1, 0)\)[/tex]
### y-Intercept
The [tex]\( y \)[/tex]-intercept is found by evaluating the function at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{5(0) + 5}{2(0)^2 + 5(0) + 3} = \frac{5}{3} \][/tex]
Thus, the [tex]\( y \)[/tex]-intercept is at the point [tex]\((0, \frac{5}{3})\)[/tex].
y-Intercept: [tex]\((0, \frac{5}{3})\)[/tex]
### Hole
A hole in the graph occurs if there is a common factor in both the numerator and the denominator that cancels out.
Factor the numerator:
[tex]\[ 5x + 5 = 5(x + 1) \][/tex]
Factor the denominator:
[tex]\[ 2x^2 + 5x + 3 = (2x + 3)(x + 1) \][/tex]
We see that [tex]\((x + 1)\)[/tex] is a common factor in both the numerator and the denominator. Therefore, there is a hole at the value where [tex]\( x + 1 = 0 \)[/tex]:
[tex]\[ x = -1 \][/tex]
To find the coordinates of the hole, we simplify the function by canceling the common factor [tex]\((x + 1)\)[/tex] and evaluate it at [tex]\( x = -1 \)[/tex]:
[tex]\[ f(x) = \frac{5}{2x + 3} \][/tex]
[tex]\[ f(-1) = \frac{5}{2(-1) + 3} = \frac{5}{1} = 5 \][/tex]
Thus, there is a hole at [tex]\((-1, 5)\)[/tex].
Hole: [tex]\((-1, 5)\)[/tex]
### Summary
Here is the summary of all the identified features:
1. Horizontal Asymptote: [tex]\( y = 0 \)[/tex]
2. Vertical Asymptotes: [tex]\( x = -\frac{3}{2} \)[/tex] and [tex]\( x = -1 \)[/tex]
3. [tex]\( x \)[/tex]-Intercept: [tex]\((-1, 0)\)[/tex]
4. [tex]\( y \)[/tex]-Intercept: [tex]\((0, \frac{5}{3})\)[/tex]
5. Hole: [tex]\((-1, 5)\)[/tex]
### Horizontal Asymptote
To find the horizontal asymptote, we compare the degrees of the numerator and the denominator:
- The degree of the numerator (5x + 5) is 1.
- The degree of the denominator (2x^2 + 5x + 3) is 2.
Since the degree of the denominator is higher than the degree of the numerator, the horizontal asymptote is [tex]\( y = 0 \)[/tex].
Horizontal Asymptote: [tex]\( y = 0 \)[/tex]
### Vertical Asymptote
To find the vertical asymptotes, we find the values of [tex]\( x \)[/tex] that make the denominator zero (since the denominator should not be zero for the function to be defined).
First, we find the roots of the denominator [tex]\( 2x^2 + 5x + 3 = 0 \)[/tex]:
[tex]\[ 2x^2 + 5x + 3 = 0 \][/tex]
By solving this quadratic equation, we get two solutions:
[tex]\[ x = -\frac{3}{2} \quad \text{and} \quad x = -1 \][/tex]
So, the function has vertical asymptotes at these values.
Vertical Asymptotes: [tex]\( x = -\frac{3}{2} \)[/tex] and [tex]\( x = -1 \)[/tex]
### x-Intercept
To find the [tex]\( x \)[/tex]-intercept, we set the numerator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 5x + 5 = 0 \][/tex]
[tex]\[ x = -1 \][/tex]
Thus, the [tex]\( x \)[/tex]-intercept is at the point [tex]\((-1, 0)\)[/tex].
x-Intercept: [tex]\((-1, 0)\)[/tex]
### y-Intercept
The [tex]\( y \)[/tex]-intercept is found by evaluating the function at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{5(0) + 5}{2(0)^2 + 5(0) + 3} = \frac{5}{3} \][/tex]
Thus, the [tex]\( y \)[/tex]-intercept is at the point [tex]\((0, \frac{5}{3})\)[/tex].
y-Intercept: [tex]\((0, \frac{5}{3})\)[/tex]
### Hole
A hole in the graph occurs if there is a common factor in both the numerator and the denominator that cancels out.
Factor the numerator:
[tex]\[ 5x + 5 = 5(x + 1) \][/tex]
Factor the denominator:
[tex]\[ 2x^2 + 5x + 3 = (2x + 3)(x + 1) \][/tex]
We see that [tex]\((x + 1)\)[/tex] is a common factor in both the numerator and the denominator. Therefore, there is a hole at the value where [tex]\( x + 1 = 0 \)[/tex]:
[tex]\[ x = -1 \][/tex]
To find the coordinates of the hole, we simplify the function by canceling the common factor [tex]\((x + 1)\)[/tex] and evaluate it at [tex]\( x = -1 \)[/tex]:
[tex]\[ f(x) = \frac{5}{2x + 3} \][/tex]
[tex]\[ f(-1) = \frac{5}{2(-1) + 3} = \frac{5}{1} = 5 \][/tex]
Thus, there is a hole at [tex]\((-1, 5)\)[/tex].
Hole: [tex]\((-1, 5)\)[/tex]
### Summary
Here is the summary of all the identified features:
1. Horizontal Asymptote: [tex]\( y = 0 \)[/tex]
2. Vertical Asymptotes: [tex]\( x = -\frac{3}{2} \)[/tex] and [tex]\( x = -1 \)[/tex]
3. [tex]\( x \)[/tex]-Intercept: [tex]\((-1, 0)\)[/tex]
4. [tex]\( y \)[/tex]-Intercept: [tex]\((0, \frac{5}{3})\)[/tex]
5. Hole: [tex]\((-1, 5)\)[/tex]
