1. **State the problem:** We are given two functions: $$f(x) = \frac{x - 3}{x - 1}$$ and $$f(x) = \frac{x - 2}{(2x - 3)^2}$$ We want to analyze these functions, focusing on their behavior, especially vertical asymptotes and points of interest. 2. **Recall important rules:** - Vertical asymptotes occur where the denominator is zero and the numerator is not zero. - Points where the function is undefined or has special behavior should be noted. 3. **Analyze the first function:** $$f(x) = \frac{x - 3}{x - 1}$$ - The denominator is zero at $x = 1$, so there is a vertical asymptote at $x = 1$. - The numerator is zero at $x = 3$, so the function crosses the x-axis at $x = 3$. 4. **Analyze the second function:** $$f(x) = \frac{x - 2}{(2x - 3)^2}$$ - The denominator is zero when $2x - 3 = 0 \Rightarrow x = \frac{3}{2} = 1.5$. - Since the denominator is squared, the function tends to $+\infty$ or $-\infty$ on both sides of $x = 1.5$, indicating a vertical asymptote there. - The numerator is zero at $x = 2$, so the function crosses the x-axis at $x = 2$. 5. **Check the marked point WP(-1/0):** - This point is at $x = -1$, $y = 0$. - For the first function, $f(-1) = \frac{-1 - 3}{-1 - 1} = \frac{-4}{-2} = 2$, so not zero. - For the second function, $f(-1) = \frac{-1 - 2}{(2(-1) - 3)^2} = \frac{-3}{(-2 - 3)^2} = \frac{-3}{(-5)^2} = \frac{-3}{25} = -0.12$, not zero. - So the point WP(-1/0) is not on either function; it might be a special point or label on the graph. 6. **Summary:** - Vertical asymptotes at $x = 1$ for the first function and $x = 1.5$ for the second. - Zero crossings at $x = 3$ for the first and $x = 2$ for the second.