To analyze the given function $y = \frac{(x-1)(x-2)}{x^2(x-4)}$, let's address each part step by step.

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(a) Determine the symmetry of the curve

1. Test for even symmetry ($f(-x) = f(x)$):
   Substitute $-x$ into the equation: = \frac{(-x-1)(-x-2)}{x^2(-x-4)} = \frac{(x+1)(x+2)}{x^2(x+4)}.$$ Since $$y(-x) \neq y(x)$$, the function is not even.
2. Test for odd symmetry ($f(-x) = -f(x)$):
   Comparing $y(-x)$ with $-y(x)$,
   $y(-x) = \frac{(x+1)(x+2)}{x^2(x+4)}, \quad -y(x) = -\frac{(x-1)(x-2)}{x^2(x-4)}.$
   These are not equal, so the function is not odd.

Conclusion: The function has no symmetry.

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(b) Give the coordinates of the $x$- and $y$-intercepts

1. $x$-intercepts:
   These occur when the numerator equals zero, i.e., $(x-1)(x-2) = 0$.
   Solving, $x = 1$ and $x = 2$.
   The $x$-intercepts are $(1, 0)$ and $(2, 0)$.

2. $y$-intercept:
   Substitute $x = 0$:
   $y(0) = \frac{(0-1)(0-2)}{0^2(0-4)}.$
   The denominator becomes $0$, so there is no $y$-intercept (undefined).

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(c) Discuss the behavior of the curve

1. Critical points (zeros and undefined points):

   • Zeros: $x = 1, x = 2$.
   • Undefined: $x = 0, x = 4$ (vertical asymptotes).

2. End behavior:
   As $x \to \pm\infty$,
   $y \approx \frac{x^2}{x^3} = \frac{1}{x},$
   so $y \to 0$. The curve approaches the horizontal asymptote $y = 0$.

3. Behavior near vertical asymptotes:

   • As $x \to 0^+$, $y \to +\infty$.
   • As $x \to 0^-$, $y \to -\infty$.
   • As $x \to 4^+$, $y \to -\infty$.
   • As $x \to 4^-$, $y \to +\infty$.

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(d) Give the equations of asymptotes

1. Vertical asymptotes: $x = 0$ and $x = 4$.
2. Horizontal asymptote: $y = 0$.

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(e) Identify regions where the curve lies

The function changes signs based on the factors in the numerator and denominator. Analyze intervals between critical points $x = 0, 1, 2, 4$:

1. For $x \in (-\infty, 0):$ $y > 0$.
2. For $x \in (0, 1):$ $y < 0$.
3. For $x \in (1, 2):$ $y > 0$.
4. For $x \in (2, 4):$ $y < 0$.
5. For $x \in (4, \infty):$ $y > 0$.

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(f) Give coordinates of additional points

Choose test points in each region to find coordinates:

1. For $x = -1$:
   $y(-1) = \frac{((-1)-1)((-1)-2)}{(-1)^2((-1)-4)} = \frac{(-2)(-3)}{1(-5)} = \frac{6}{-5} = -\frac{6}{5}.$ $(x, y) = (-1, -1.2)$.

2. For $x = 0.5$:
   $y(0.5) = \frac{(0.5-1)(0.5-2)}{(0.5)^2(0.5-4)} = \frac{(-0.5)(-1.5)}{0.25(-3.5)} = \frac{0.75}{-0.875} \approx -0.857.$ $(x, y) = (0.5, -0.86)$.

3. For $x = 3$:
   $y(3) = \frac{(3-1)(3-2)}{3^2(3-4)} = \frac{(2)(1)}{9(-1)} = -\frac{2}{9}.$ $(x, y) = (3, -0.22)$.

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Would you like a plot of this function for further clarification?

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Related Questions:

1. How does the function's derivative relate to its critical points?
2. Can you verify vertical asymptotes using limits?
3. How would the function change if the numerator was squared?
4. Is there an inflection point in the curve?
5. What does the second derivative reveal about concavity?

Tip: Always test symmetry and sign changes in intervals to understand the curve's behavior.