Sure, let's break down the solution step by step and identify all the features of the given rational function [tex]\( f(x)=\frac{4 x^2+40 x+100}{3 x+15} \)[/tex].
### 1. Vertical Asymptotes
Vertical asymptotes occur where the denominator is zero. Set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 3x + 15 = 0 \][/tex]
[tex]\[ 3x = -15 \][/tex]
[tex]\[ x = -5 \][/tex]
Therefore, there is a vertical asymptote at [tex]\( x = -5 \)[/tex].
### 2. Horizontal Asymptote
Horizontal asymptotes are determined by the degrees of the numerator and the denominator:
- If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is [tex]\( y = 0 \)[/tex].
- If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is [tex]\( y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}} \)[/tex].
- If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.
In this case, the degree of the numerator (2) is greater than the degree of the denominator (1), so there is no horizontal asymptote.
### 3. Oblique Asymptote
Since the degree of the numerator is one more than the degree of the denominator, there is an oblique asymptote. We need to perform polynomial long division to find it.
Divide [tex]\( 4x^2 + 40x + 100 \)[/tex] by [tex]\( 3x + 15 \)[/tex]:
#### Polynomial Long Division
1. Divide the first term of the numerator by the first term of the denominator:
[tex]\[ \frac{4x^2}{3x} = \frac{4}{3}x \][/tex]
2. Multiply [tex]\( \frac{4}{3}x \)[/tex] by the entire denominator:
[tex]\[ \frac{4}{3}x (3x + 15) = 4x^2 + 20x \][/tex]
3. Subtract this from the numerator:
[tex]\[ (4x^2 + 40x + 100) - (4x^2 + 20x) = 20x + 100 \][/tex]
4. Divide the first term of the new polynomial by the first term of the denominator:
[tex]\[ \frac{20x}{3x} = \frac{20}{3} \][/tex]
5. Multiply [tex]\( \frac{20}{3} \)[/tex] by the entire denominator:
[tex]\[ \frac{20}{3} (3x + 15) = 20x + 100 \][/tex]
6. Subtract again:
[tex]\[ (20x + 100) - (20x + 100) = 0\][/tex]
So, the quotient is:
[tex]\[ \frac{4}{3} x + \frac{20}{3} \][/tex]
Thus, the oblique asymptote is:
[tex]\[ y = \frac{4}{3} x + \frac{20}{3} \][/tex]
### 4. Zeros (x-intercepts)
Zeros occur where the numerator is zero. Set the numerator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 4x^2 + 40x + 100 = 0 \][/tex]
To factor this, we can:
[tex]\[ x^2 + 10x + 25 = 0 \][/tex]
[tex]\[ (x + 5)^2 = 0 \][/tex]
[tex]\[ x = -5 \][/tex]
However, since the denominator also equals zero at [tex]\( x = -5 \)[/tex], adding the numerator doesn't lead to a zero for the whole function. Therefore, the function has no x-intercepts.
### 5. Y-intercept
The y-intercept occurs when [tex]\( x = 0 \)[/tex]. Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = \frac{4(0)^2 + 40(0) + 100}{3(0) + 15} = \frac{100}{15} = \frac{20}{3} \][/tex]
So, the y-intercept is [tex]\( \left(0, \frac{20}{3} \right) \)[/tex].
### Summary
The key features are:
- Vertical Asymptote: [tex]\( x = -5 \)[/tex]
- Horizontal Asymptote: None
- Oblique Asymptote: [tex]\( y = \frac{4}{3}x + \frac{20}{3} \)[/tex]
- Zeros: None
- Y-Intercept: [tex]\( \left(0, \frac{20}{3} \right) \)[/tex]
We can now plot these features on a graph.
