The given function is h(x) = (6)/(x-2) - 1.

Here are the solutions for Question 6. 6.1 The given function is h(x) = (6)/(x-2) - 1. For a rational function of the form y = (a)/(x-p) + q: The vertical asymptote is x=p. The horizontal asymptote is y=q. Step 1: Identify the vertical asymptote. The denominator cannot be zero, so x-2 ≠ 0. x-2 = 0 x = 2 The vertical asymptote is x=2. Step 2: Identify the horizontal asymptote. The constant term is -1. The horizontal asymptote is y=-1. The equations of the asymptotes are: Vertical asymptote: x=2 Horizontal asymptote: y=-1 6.2 Step 1: Determine the x-intercept. Set h(x) = 0: (6)/(x-2) - 1 = 0 (6)/(x-2) = 1 6 = x-2 x = 8 The x-intercept is (8, 0). Step 2: Determine the y-intercept. Set x = 0: h(0) = (6)/(0-2) - 1 h(0) = (6)/(-2) - 1 h(0) = -3 - 1 h(0) = -4 The y-intercept is (0, -4). 6.3 To sketch the graph of h(x) = (6)/(x-2) - 1: • Draw the vertical asymptote at x=2 as a dashed line. • Draw the horizontal asymptote at y=-1 as a dashed line. • Plot the x-intercept at (8, 0). • Plot the y-intercept at (0, -4). • Since the numerator 6 is positive, the graph will have branches in the top-right and bottom-left sections relative to the asymptotes. • Draw the branch that passes through (0, -4), approaching x=2 from the left and y=-1 from below. • Draw the branch that passes through (8, 0), approaching x=2 from the right and y=-1 from above. 6.4 Determine the values of x for which h(x) < 0 for x [0; ). Step 1: Set up the inequality. (6)/(x-2) - 1 < 0 Step 2: Combine terms to a single fraction. (6 - (x-2))/(x-2) < 0 (6 - x + 2)/(x-2) < 0 (8 - x)/(x-2) < 0 Step 3: Find the critical points. The numerator is zero when 8-x=0 x=8. The denominator is zero when x-2=0 x=2. These points divide the number line into intervals: (-, 2), (2, 8), (8, ). Step 4: Test each interval. • For x < 2 (e.g., x=0): (8-0)/(0-2) = (8)/(-2) = -4. Since -4 < 0, the inequality holds for x < 2. • For 2 < x < 8 (e.g., x=3): (8-3)/(3-2) = (5)/(1) = 5. Since 5 > 0, the inequality does not hold for 2 < x < 8. • For x > 8 (e.g., x=9): (8-9)/(9-2) = (-1)/(7). Since (-1)/(7) < 0, the inequality holds for x > 8. So, h(x) < 0 when x < 2 or x > 8. Step 5: Consider the given domain x [0; ). We need to find the intersection of (-, 2) (8, ) with [0, ). • For x < 2: The part within [0, ) is [0, 2). • For x > 8: The part within [0, ) is (8, ). The values of x for which h(x) < 0 for x [0; ) are x [0, 2) (8, ). Send me the next one 📸