1. Find the vertical and horizontal asymptotes of the curve defined by $$xy - 5x - 2y + 3 = 0.$$ Step 1: Rearrange the equation to express $y$ in terms of $x$. $$xy - 5x - 2y + 3 = 0 \\ xy - 2y = 5x - 3 \\ y(x - 2) = 5x - 3 \\ y = \frac{5x - 3}{x - 2}$$ Step 2: Determine vertical asymptotes by finding values of $x$ that make the denominator zero. $$x - 2 = 0 \implies x = 2$$ So, there is a vertical asymptote at $x = 2$. Step 3: Determine horizontal asymptotes by analyzing the behavior of $y$ as $x \to \pm \infty$. $$\lim_{x \to \pm \infty} y = \lim_{x \to \pm \infty} \frac{5x - 3}{x - 2} = \lim_{x \to \pm \infty} \frac{5 - \frac{3}{x}}{1 - \frac{2}{x}} = \frac{5 - 0}{1 - 0} = 5$$ So, the horizontal asymptote is $y = 5$. 2. For the piecewise function $$f(x) = \begin{cases} x - 1 & x < 1 \\ 1 & x = 1 \\ 1 - x & x > 1 \end{cases}$$ Step 1: Check for continuity at $x = 1$. Step 2: Find the left-hand limit as $x$ approaches 1: $$\lim_{x \to 1^-} (x - 1) = 1 - 1 = 0$$ Step 3: Find the right-hand limit as $x$ approaches 1: $$\lim_{x \to 1^+} (1 - x) = 1 - 1 = 0$$ Step 4: The function value at $x = 1$ is $f(1) = 1$. Step 5: Since the left and right limits are equal (both 0), but $f(1) = 1 \neq 0$, the function is discontinuous at $x = 1$. This is a removable discontinuity because the limits from both sides are equal. Step 6: To redefine the function and remove the discontinuity, set $$f(1) = 0$$ This makes the function continuous.