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Find all asymptotes, \( x \)-intercepts, and \( y \)-intercepts (if they do not exist, answer "DNE") for the graph of the rational function and sketch the graph of the function. \( f(x)=\frac{4 x}{x^{2}+8 x+16} \)

Ask by Herrera Mann. in the United States
Feb 25,2025

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Answer

Vertical asymptote at x = –4, horizontal asymptote at y = 0, x-intercept and y-intercept at (0, 0).

Solution

We are given the function   f(x) = (4x)⁄(x² + 8x + 16). Step 1. Factor the Denominator Notice that   x² + 8x + 16 = (x + 4)². So the function can be rewritten as   f(x) = 4x⁄(x + 4)². The domain is all real numbers except x = –4 (because (–4 + 4)² = 0). Step 2. Find the Vertical Asymptote A vertical asymptote occurs when the denominator is zero (and the numerator is not simultaneously zero – i.e. no cancellation occurs). Since (x + 4)² = 0 when x = –4, there is a vertical asymptote at   x = –4. Step 3. Find the Horizontal Asymptote Compare the degrees of the numerator and denominator:  • Numerator: degree 1 (4x)  • Denominator: degree 2 ((x + 4)²) Since the degree in the numerator is less than that in the denominator, we have a horizontal asymptote at   y = 0. Step 4. Find the x-intercept(s) The x-intercepts occur where f(x) = 0 (and the function is defined). Set the numerator equal to zero:   4x = 0 ⟹ x = 0. Since x = 0 is in the domain, the only x-intercept is   (0, 0). Step 5. Find the y-intercept The y-intercept is found by evaluating f(0):   f(0) = 4(0)⁄(0 + 4)² = 0⁄16 = 0. So the y-intercept is also   (0, 0). Step 6. Sketching the Graph • The domain is x ∈ (–∞, –4) ∪ (–4, ∞). • There is a vertical asymptote at x = –4. • There is a horizontal asymptote at y = 0. • The graph crosses the origin (0, 0), which is both the x– and y–intercept. • Behavior near the vertical asymptote:  – For x-values approaching –4 (from either side), the denominator (x + 4)² is very small and positive, while the numerator 4x is near –16. Thus, f(x) → –∞ on both sides of x = –4. • End behavior:  – As x → +∞, f(x) ≈ 4x/(x²) = 4/x → 0 (from above, if x is positive).  – As x → –∞, f(x) ≈ 4x/(x²) = 4/x → 0 (from below, since x is negative). The graph therefore comes in from near y = 0 on both ends, drops down sharply near the vertical line x = –4, and passes through the origin. Summary of answers:  • Vertical asymptote: x = –4.  • Horizontal asymptote: y = 0.  • x-intercept: (0, 0).  • y-intercept: (0, 0). This information is all you need to sketch the graph of f(x).

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The Deep Dive

To find the asymptotes, \( x \)-intercepts, and \( y \)-intercepts for the function \( f(x) = \frac{4x}{x^2 + 8x + 16} \), let’s go through each step: ### Asymptotes 1. **Vertical Asymptotes**: These occur where the denominator is zero (and the numerator is not zero). First, factor the denominator: \[ x^2 + 8x + 16 = (x + 4)^2 \] Setting the denominator equal to zero gives: \[ (x + 4)^2 = 0 \implies x = -4 \] Therefore, there is a vertical asymptote at \( x = -4 \). 2. **Horizontal Asymptotes**: To find horizontal asymptotes, we look at the degrees of the numerator and denominator. Here, the degree of the numerator (1) is less than the degree of the denominator (2), which means there is a horizontal asymptote at: \[ y = 0 \] ### Intercepts 1. **\( x \)-intercepts**: These occur where the function is equal to zero. To find the \( x \)-intercepts, set the numerator equal to zero: \[ 4x = 0 \implies x = 0 \] So, there is an \( x \)-intercept at \( (0, 0) \). 2. **\( y \)-intercepts**: The \( y \)-intercept occurs when \( x = 0 \): \[ f(0) = \frac{4(0)}{0^2 + 8(0) + 16} = \frac{0}{16} = 0 \] Hence, the \( y \)-intercept is also at \( (0, 0) \). ### Summary of Results - **Vertical Asymptote**: \( x = -4 \) - **Horizontal Asymptote**: \( y = 0 \) - **\( x \)-intercept**: \( (0, 0) \) - **\( y \)-intercept**: \( (0, 0) \) ### Sketch of the Graph To sketch the graph, note: - The graph approaches the vertical asymptote at \( x = -4 \) and will get infinitely close to the horizontal asymptote \( y = 0 \) as \( x \) tends to \( \infty \) or \( -\infty \). - The function has an intercept at the origin, indicating that it touches the origin. Overall, the graph will increase from negative infinity as it approaches \( x = -4 \) from the left, plunge down to infinity as it approaches \( x = -4 \) from the right, and then gradually approach the horizontal asymptote \( y = 0 \) as \( x \) moves towards infinity.

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