Here are the solutions for Question 6. 6.1 The given function is $h(x) = \frac{6}{x-2} - 1$. For a rational function of the form $y = \frac{a}{x-p} + q$: The vertical asymptote is $x=p$. The horizontal asymptote is $y=q$. Step 1: Identify the vertical asymptote. The denominator cannot be zero, so $x-2 \neq 0$. $$x-2 = 0$$ $$x = 2$$ 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: $\boxed{x=2}$ Horizontal asymptote: $\boxed{y=-1}$ 6.2 Step 1: Determine the x-intercept. Set $h(x) = 0$: $$\frac{6}{x-2} - 1 = 0$$ $$\frac{6}{x-2} = 1$$ $$6 = x-2$$ $$x = 8$$ The x-intercept is $\boxed{(8, 0)}$. Step 2: Determine the y-intercept. Set $x = 0$: $$h(0) = \frac{6}{0-2} - 1$$ $$h(0) = \frac{6}{-2} - 1$$ $$h(0) = -3 - 1$$ $$h(0) = -4$$ The y-intercept is $\boxed{(0, -4)}$. 6.3 To sketch the graph of $h(x) = \frac{6}{x-2} - 1$: • 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 $x \in [0; \infty)$. Step 1: Set up the inequality. $$\frac{6}{x-2} - 1 < 0$$ Step 2: Combine terms to a single fraction. $$\frac{6 - (x-2)}{x-2} < 0$$ $$\frac{6 - x + 2}{x-2} < 0$$ $$\frac{8 - x}{x-2} < 0$$ Step 3: Find the critical points. The numerator is zero when $8-x=0 \implies x=8$. The denominator is zero when $x-2=0 \implies x=2$. These points divide the number line into intervals: $(-\infty, 2)$, $(2, 8)$, $(8, \infty)$. Step 4: Test each interval. • For $x < 2$ (e.g., $x=0$): $\frac{8-0}{0-2} = \frac{8}{-2} = -4$. Since $-4 < 0$, the inequality holds for $x < 2$. • For $2 < x < 8$ (e.g., $x=3$): $\frac{8-3}{3-2} = \frac{5}{1} = 5$. Since $5 > 0$, the inequality does not hold for $2 < x < 8$. • For $x > 8$ (e.g., $x=9$): $\frac{8-9}{9-2} = \frac{-1}{7}$. Since $\frac{-1}{7} < 0$, the inequality holds for $x > 8$. So, $h(x) < 0$ when $x < 2$ or $x > 8$. Step 5: Consider the given domain $x \in [0; \infty)$. We need to find the intersection of $(-\infty, 2) \cup (8, \infty)$ with $[0, \infty)$. • For $x < 2$: The part within $[0, \infty)$ is $[0, 2)$. • For $x > 8$: The part within $[0, \infty)$ is $(8, \infty)$. The values of $x$ for which $h(x) < 0$ for $x \in [0; \infty)$ are $\boxed{x \in [0, 2) \cup (8, \infty)}$. Send me the next one 📸