Math Problem Statement

solve the 𝑦 = (π‘₯βˆ’1)(π‘₯βˆ’2) / xΒ²(π‘₯βˆ’4) a. determine the symmetry of the curve b. give the coordinates of the π‘₯- and 𝑦-intercepts c. discuss the behavior of the curve d. give the equations of asymptotes (if any) e. identify regions where the curve lies f. give coordinates of additional points (if any). just like the picture

Solution

To analyze the given function y=(xβˆ’1)(xβˆ’2)x2(xβˆ’4)y = \frac{(x-1)(x-2)}{x^2(x-4)}, let's address each part step by step.


(a) Determine the symmetry of the curve

  1. Test for even symmetry (f(βˆ’x)=f(x)f(-x) = f(x)):
    Substitute βˆ’x-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(βˆ’x)=βˆ’f(x)f(-x) = -f(x)):
    Comparing y(βˆ’x)y(-x) with βˆ’y(x)-y(x),
    y(βˆ’x)=(x+1)(x+2)x2(x+4),βˆ’y(x)=βˆ’(xβˆ’1)(xβˆ’2)x2(xβˆ’4).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.


(b) Give the coordinates of the xx- and yy-intercepts

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

  2. yy-intercept:
    Substitute x=0x = 0:
    y(0)=(0βˆ’1)(0βˆ’2)02(0βˆ’4).y(0) = \frac{(0-1)(0-2)}{0^2(0-4)}.
    The denominator becomes 00, so there is no yy-intercept (undefined).


(c) Discuss the behavior of the curve

  1. Critical points (zeros and undefined points):

    • Zeros: x=1,x=2x = 1, x = 2.
    • Undefined: x=0,x=4x = 0, x = 4 (vertical asymptotes).
  2. End behavior:
    As xβ†’Β±βˆžx \to \pm\infty,
    yβ‰ˆx2x3=1x,y \approx \frac{x^2}{x^3} = \frac{1}{x},
    so y→0y \to 0. The curve approaches the horizontal asymptote y=0y = 0.

  3. Behavior near vertical asymptotes:

    • As xβ†’0+x \to 0^+, yβ†’+∞y \to +\infty.
    • As xβ†’0βˆ’x \to 0^-, yβ†’βˆ’βˆžy \to -\infty.
    • As xβ†’4+x \to 4^+, yβ†’βˆ’βˆžy \to -\infty.
    • As xβ†’4βˆ’x \to 4^-, yβ†’+∞y \to +\infty.

(d) Give the equations of asymptotes

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

(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=0,1,2,4x = 0, 1, 2, 4:

  1. For x∈(βˆ’βˆž,0):x \in (-\infty, 0): y>0y > 0.
  2. For x∈(0,1):x \in (0, 1): y<0y < 0.
  3. For x∈(1,2):x \in (1, 2): y>0y > 0.
  4. For x∈(2,4):x \in (2, 4): y<0y < 0.
  5. For x∈(4,∞):x \in (4, \infty): y>0y > 0.

(f) Give coordinates of additional points

Choose test points in each region to find coordinates:

  1. For x=βˆ’1x = -1:
    y(βˆ’1)=((βˆ’1)βˆ’1)((βˆ’1)βˆ’2)(βˆ’1)2((βˆ’1)βˆ’4)=(βˆ’2)(βˆ’3)1(βˆ’5)=6βˆ’5=βˆ’65.y(-1) = \frac{((-1)-1)((-1)-2)}{(-1)^2((-1)-4)} = \frac{(-2)(-3)}{1(-5)} = \frac{6}{-5} = -\frac{6}{5}. (x,y)=(βˆ’1,βˆ’1.2)(x, y) = (-1, -1.2).

  2. For x=0.5x = 0.5:
    y(0.5)=(0.5βˆ’1)(0.5βˆ’2)(0.5)2(0.5βˆ’4)=(βˆ’0.5)(βˆ’1.5)0.25(βˆ’3.5)=0.75βˆ’0.875β‰ˆβˆ’0.857.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,βˆ’0.86)(x, y) = (0.5, -0.86).

  3. For x=3x = 3:
    y(3)=(3βˆ’1)(3βˆ’2)32(3βˆ’4)=(2)(1)9(βˆ’1)=βˆ’29.y(3) = \frac{(3-1)(3-2)}{3^2(3-4)} = \frac{(2)(1)}{9(-1)} = -\frac{2}{9}. (x,y)=(3,βˆ’0.22)(x, y) = (3, -0.22).


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


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.

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Math Problem Analysis

Mathematical Concepts

Rational Functions
Symmetry of Graphs
Intercepts
Asymptotes
End Behavior
Critical Points
Sign Analysis

Formulas

y = (x - 1)(x - 2) / xΒ²(x - 4)
End behavior: y β‰ˆ 1/x as x β†’ ±∞

Theorems

Symmetry Test for Even and Odd Functions
Asymptote Analysis (Vertical and Horizontal)
Sign Analysis in Intervals

Suitable Grade Level

Grades 11-12