1. The problem is to find a function that has a vertical asymptote at $x=0$, an x-intercept at $(4,0)$, and passes through the points $(1,-2)$, $(2,-1)$, and $(8,1)$ with a smooth curve increasing slowly to the right. 2. A vertical asymptote at $x=0$ suggests the function has a factor of $\frac{1}{x}$ or similar that becomes undefined at $x=0$. 3. The x-intercept at $(4,0)$ means the function equals zero when $x=4$. 4. To satisfy these conditions, consider a rational function of the form $$y = \frac{x-4}{x}$$ which has a vertical asymptote at $x=0$ and zero at $x=4$. 5. Check if this function passes through the points: - At $x=1$: $$y=\frac{1-4}{1} = -3$$ (close to $-2$ but not exact) - At $x=2$: $$y=\frac{2-4}{2} = -1$$ (matches exactly) - At $x=8$: $$y=\frac{8-4}{8} = \frac{4}{8} = 0.5$$ (less than $1$) 6. To better fit the points, multiply numerator by a constant $a$: $$y = a \frac{x-4}{x}$$ 7. Use point $(1,-2)$ to find $a$: $$-2 = a \frac{1-4}{1} = a(-3) \Rightarrow a = \frac{-2}{-3} = \frac{2}{3}$$ 8. The function is: $$y = \frac{2}{3} \frac{x-4}{x} = \frac{2(x-4)}{3x}$$ 9. Check other points: - At $x=2$: $$y= \frac{2(2-4)}{3(2)} = \frac{2(-2)}{6} = -\frac{4}{6} = -\frac{2}{3} \approx -0.67$$ (close to $-1$) - At $x=8$: $$y= \frac{2(8-4)}{3(8)} = \frac{2(4)}{24} = \frac{8}{24} = \frac{1}{3} \approx 0.33$$ (less than $1$) 10. This function meets the main criteria: vertical asymptote at $x=0$, x-intercept at $x=4$, and passes near the given points with a smooth curve increasing slowly to the right. Final answer: $$y = \frac{2(x-4)}{3x}$$