Question

secondary math iii // module 4
rational expressions & functions - 4.6
set
topic: predicting and sketching rational functions
7
find the vertical asymptote(s), horizontal or slant asymptote, and intercepts. then sketch the
graph. (do not use technology to get the graph. the max and mins do not need to be accurate.)
$y = \frac{(x + 4)}{(-2x - 6)}$
vertical asymptote(s)
horizontal or slant asymptote
intercepts
graph:

6.
$y = \frac{3x}{(x - 3)} \cdot \frac{(x - 4)}{(x + 1)}$
vertical asymptote(s)
horizontal or slant asymptote
intercepts
graph:
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attribution cc by 4.0
mathematic
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Explanation:

Problem 7 (Function: \( y = \frac{(x + 4)}{(-2x - 6)} \))

Step 1: Simplify the Denominator

Factor the denominator: \( -2x - 6 = -2(x + 3) \). The function becomes \( y = \frac{x + 4}{-2(x + 3)} \).

Step 2: Find Vertical Asymptote(s)

Vertical asymptotes occur where the denominator is zero (and numerator is not zero). Set \( -2(x + 3) = 0 \). Solving \( x + 3 = 0 \) gives \( x = -3 \). The numerator at \( x = -3 \) is \( -3 + 4 = 1
eq 0 \), so vertical asymptote is \( x = -3 \).

Step 3: Find Horizontal Asymptote

For a rational function \( y = \frac{ax + b}{cx + d} \) (degree of numerator = degree of denominator), the horizontal asymptote is \( y = \frac{a}{c} \). Here, numerator degree = 1, denominator degree = 1. So \( y = \frac{1}{-2} = -\frac{1}{2} \).

Step 4: Find Intercepts

• x-intercept: Set \( y = 0 \). So \( \frac{x + 4}{-2(x + 3)} = 0 \). Numerator must be zero: \( x + 4 = 0 \) → \( x = -4 \).
• y-intercept: Set \( x = 0 \). \( y = \frac{0 + 4}{-2(0 + 3)} = \frac{4}{-6} = -\frac{2}{3} \). So intercepts are \( (-4, 0) \) (x-intercept) and \( (0, -\frac{2}{3}) \) (y-intercept).

Problem 6 (Function: \( y = \frac{3x}{(x - 3)} \cdot \frac{(x - 4)}{(x + 1)} \))

Step 1: Simplify the Function

Multiply the fractions: \( y = \frac{3x(x - 4)}{(x - 3)(x + 1)} \).

Step 2: Find Vertical Asymptote(s)

Vertical asymptotes where denominator is zero (numerator not zero). Set \( (x - 3)(x + 1) = 0 \). Solutions: \( x = 3 \) and \( x = -1 \). Check numerator at \( x = 3 \): \( 3(3)(3 - 4) = -9
eq 0 \). At \( x = -1 \): \( 3(-1)(-1 - 4) = 15
eq 0 \). So vertical asymptotes: \( x = 3 \), \( x = -1 \).

Step 3: Find Horizontal Asymptote

Numerator degree: \( 3x(x - 4) = 3x^2 - 12x \) (degree 2). Denominator degree: \( (x - 3)(x + 1) = x^2 - 2x - 3 \) (degree 2). So horizontal asymptote is \( y = \frac{3}{1} = 3 \) (since leading coefficients are 3 (numerator) and 1 (denominator)).

Step 4: Find Intercepts

• x-intercept: Set \( y = 0 \). Numerator \( 3x(x - 4) = 0 \). Solutions: \( x = 0 \) or \( x = 4 \). Check denominator at these x-values: at \( x = 0 \), denominator \( (0 - 3)(0 + 1) = -3

eq 0 \); at \( x = 4 \), denominator \( (4 - 3)(4 + 1) = 5
eq 0 \). So x-intercepts: \( (0, 0) \), \( (4, 0) \).

• y-intercept: Set \( x = 0 \). \( y = \frac{3(0)(0 - 4)}{(0 - 3)(0 + 1)} = 0 \). So y-intercept is \( (0, 0) \) (already included in x-intercepts).

Answer:

s:

Problem 7:

• Vertical Asymptote(s): \( x = -3 \)
• Horizontal or Slant Asymptote: \( y = -\frac{1}{2} \)
• Intercepts: \( (-4, 0) \) (x-intercept), \( (0, -\frac{2}{3}) \) (y-intercept)

Problem 6:

• Vertical Asymptote(s): \( x = 3 \), \( x = -1 \)
• Horizontal or Slant Asymptote: \( y = 3 \)
• Intercepts: \( (0, 0) \), \( (4, 0) \) (x-intercepts), \( (0, 0) \) (y-intercept)