question 5
let ( f(x) = 3 cdot \frac{3x^2 - 2x - 8}{4x^2 - 19x - 5} = 3 cdot \frac{(3x + 4)(x - 2)}{(4x + 1)(x - 5)} )
find:
1) the domain in interval notation
note: use -oo for ( -infty ), oo for ( infty ), u for union.
2) the y intercept at the point
3) x intercepts at the point(s)
4) vertical asymptotes at ( x = )
5) horizontal asymptote at ( y = )
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Question

Accepted Answer

### 1) Domain in Interval Notation
# Explanation:
## Step1: Find Denominator Zeros
Set denominator $(4x + 1)(x - 5)=0$. Solve $4x + 1 = 0$ gives $x=-\frac{1}{4}$; solve $x - 5 = 0$ gives $x = 5$.
## Step2: Determine Domain
Domain excludes these zeros. So interval notation: $(-\infty, -\frac{1}{4}) \cup (-\frac{1}{4}, 5) \cup (5, \infty)$
# Answer: $(-\infty, -\frac{1}{4}) \cup (-\frac{1}{4}, 5) \cup (5, \infty)$

### 2) y - Intercept
# Explanation:
## Step1: Substitute $x = 0$
$f(0)=3\cdot\frac{(3(0)+4)(0 - 2)}{(4(0)+1)(0 - 5)}=3\cdot\frac{(4)(-2)}{(1)(-5)}=3\cdot\frac{-8}{-5}=\frac{24}{5}$
## Step2: Identify Point
y - intercept is at $(0, \frac{24}{5})$
# Answer: $(0, \frac{24}{5})$

### 3) x - Intercepts
# Explanation:
## Step1: Set Numerator Zero
Set $(3x + 4)(x - 2)=0$. Solve $3x + 4 = 0$ gives $x=-\frac{4}{3}$; solve $x - 2 = 0$ gives $x = 2$.
## Step2: Check Domain
Both $x=-\frac{4}{3}$ and $x = 2$ are in domain. So points: $(-\frac{4}{3}, 0)$ and $(2, 0)$
# Answer: $(-\frac{4}{3}, 0), (2, 0)$

### 4) Vertical Asymptotes
# Explanation:
## Step1: Use Denominator Zeros
Vertical asymptotes at $x$ where denominator is zero (and numerator non - zero). From step1 of domain, $x=-\frac{1}{4}$ and $x = 5$ (numerator non - zero at these $x$ values).
# Answer: $x=-\frac{1}{4}, x = 5$

### 5) Horizontal Asymptote
# Explanation:
## Step1: Analyze Degrees
Degree of numerator (after multiplying 3: $3(3x^{2}-2x - 8)=9x^{2}-6x - 24$) is 2. Degree of denominator ($4x^{2}-19x - 5$) is 2.
## Step2: Find Ratio of Leading Coefficients
Ratio of leading coefficients: $\frac{9}{4}$ (wait, original function: $f(x)=3\cdot\frac{3x^{2}-2x - 8}{4x^{2}-19x - 5}$, leading term of numerator: $3\cdot3x^{2}=9x^{2}$, leading term of denominator: $4x^{2}$, so horizontal asymptote $y=\frac{9}{4}$? Wait, no: original numerator inside fraction: $3x^{2}-2x - 8$, denominator: $4x^{2}-19x - 5$. When we multiply by 3, the leading term is $3\cdot3x^{2}=9x^{2}$ over $4x^{2}$, so $y = \frac{9}{4}$? Wait, no, wait: $f(x)=3\times\frac{3x^{2}-2x - 8}{4x^{2}-19x - 5}=\frac{9x^{2}-6x - 24}{4x^{2}-19x - 5}$. As $x\to\pm\infty$, $f(x)\to\frac{9}{4}$.
# Answer: $y=\frac{9}{4}$

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QUESTION IMAGE

Question

Explanation:

1) Domain in Interval Notation

Step1: Find Denominator Zeros

Step2: Determine Domain

Step1: Substitute \(x = 0\)

Step2: Identify Point

Step1: Set Numerator Zero

Step2: Check Domain

Answer:

2) y - Intercept