10. find the vertical asymptotes for the following function (2pts). determine if the function approaches +∞…

10. find the vertical asymptotes for the following function (2pts). determine if the function approaches +∞ or -∞ at each of the vertical asymptotes as they are approached from the right and from the left (4pts). additionally, determine what value the function approaches as x→±∞ (2pts). y(x)=\\frac{x^{2}}{x^{2}+x - 6}\n11. object a is at rest at x = 0 m at t = 0 s. object b is at x = 100 m and moving at 20 m/s in the positive x - direction at t = 0 s. if a begins accelerating in the positive x - direction at 4 m/s² at t = 0 s, when (5pts) and where (5pts) will a catch up to b? provide a plot (5pts) of the positions x_{a}(t) of a and x_{b}(t) of b as a function of time (on the same plot), including the point where they meet in position and time.

10. find the vertical asymptotes for the following function (2pts). determine if the function approaches +∞ or -∞ at each of the vertical asymptotes as they are approached from the right and from the left (4pts). additionally, determine what value the function approaches as x→±∞ (2pts). y(x)=\\frac{x^{2}}{x^{2}+x - 6}\n11. object a is at rest at x = 0 m at t = 0 s. object b is at x = 100 m and moving at 20 m/s in the positive x - direction at t = 0 s. if a begins accelerating in the positive x - direction at 4 m/s² at t = 0 s, when (5pts) and where (5pts) will a catch up to b? provide a plot (5pts) of the positions x_{a}(t) of a and x_{b}(t) of b as a function of time (on the same plot), including the point where they meet in position and time.

Answer

Explanation:

Step1: Find vertical asymptotes

Vertical asymptotes occur where the denominator is zero (and numerator is non - zero). Set (x^{2}+x - 6=0). Factor the quadratic: (x^{2}+x - 6=(x + 3)(x - 2)=0). So (x=-3) and (x = 2) are the candidates. For (x=-3), (y(-3)=\frac{(-3)^{2}}{(-3)^{2}+(-3)-6}=\frac{9}{9 - 3 - 6}=\frac{9}{0}) (undefined) and numerator (x^{2}=9\neq0). For (x = 2), (y(2)=\frac{2^{2}}{2^{2}+2 - 6}=\frac{4}{4 + 2 - 6}=\frac{4}{0}) (undefined) and numerator (x^{2}=4\neq0). So the vertical asymptotes are (x=-3) and (x = 2).

Step2: Analyze the behavior near vertical asymptotes

  • Near (x=-3):
    • From the right ((x\rightarrow-3^{+})): Let (x=-3 + h), where (h\rightarrow0^{+}). (y(x)=\frac{(-3 + h)^{2}}{(-3 + h)^{2}+(-3 + h)-6}=\frac{9-6h+h^{2}}{9-6h+h^{2}-3 + h - 6}=\frac{9-6h+h^{2}}{h^{2}-5h}) As (h\rightarrow0^{+}), (y(x)\approx\frac{9}{-5h}\rightarrow-\infty)
    • From the left ((x\rightarrow-3^{-})): Let (x=-3 - h), where (h\rightarrow0^{+}). (y(x)=\frac{(-3 - h)^{2}}{(-3 - h)^{2}+(-3 - h)-6}=\frac{9 + 6h+h^{2}}{9+6h+h^{2}-3 - h - 6}=\frac{9 + 6h+h^{2}}{h^{2}+5h}) As (h\rightarrow0^{+}), (y(x)\approx\frac{9}{5h}\rightarrow+\infty)
  • Near (x = 2):
    • From the right ((x\rightarrow2^{+})): Let (x=2 + h), where (h\rightarrow0^{+}). (y(x)=\frac{(2 + h)^{2}}{(2 + h)^{2}+(2 + h)-6}=\frac{4+4h+h^{2}}{4 + 4h+h^{2}+2 + h - 6}=\frac{4+4h+h^{2}}{h^{2}+5h}) As (h\rightarrow0^{+}), (y(x)\approx\frac{4}{5h}\rightarrow+\infty)
    • From the left ((x\rightarrow2^{-})): Let (x=2 - h), where (h\rightarrow0^{+}). (y(x)=\frac{(2 - h)^{2}}{(2 - h)^{2}+(2 - h)-6}=\frac{4-4h+h^{2}}{4-4h+h^{2}+2 - h - 6}=\frac{4-4h+h^{2}}{h^{2}-5h}) As (h\rightarrow0^{+}), (y(x)\approx\frac{4}{-5h}\rightarrow-\infty)

Step3: Analyze (x\rightarrow\pm\infty)

Divide numerator and denominator by (x^{2}): (y(x)=\frac{x^{2}}{x^{2}+x - 6}=\frac{1}{1+\frac{1}{x}-\frac{6}{x^{2}}}) As (x\rightarrow\pm\infty), (\frac{1}{x}\rightarrow0) and (\frac{6}{x^{2}}\rightarrow0) So (y(x)\rightarrow1)

Answer:

  • Vertical asymptotes: (x=-3) and (x = 2)
  • At (x=-3):
    • As (x\rightarrow-3^{+}), (y(x)\rightarrow-\infty)
    • As (x\rightarrow-3^{-}), (y(x)\rightarrow+\infty)
  • At (x = 2):
    • As (x\rightarrow2^{+}), (y(x)\rightarrow+\infty)
    • As (x\rightarrow2^{-}), (y(x)\rightarrow-\infty)
  • As (x\rightarrow\pm\infty), (y(x)\rightarrow1)