9. (5pts) an object has an initial velocity of 40 m/s upward from a height of ( x = 0 ) m. at what time will…

9. (5pts) an object has an initial velocity of 40 m/s upward from a height of ( x = 0 ) m. at what time will the object be at a height of ( x = 20 ) m on its way upward? use ( g = 10 ) m/s² for the magnitude of the acceleration due to gravity. (4pts). additionally, determine what value the function approaches as ( x \to pminfty ) (2pts). 10. find the vertical asymptotes for the following function (2pts). determine if the function approaches ( +infty ) or ( -infty ) 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 \to pminfty ) (2pts). ( y(x)=\frac{x^{2}}{x^{2}+x - 6} )

9. (5pts) an object has an initial velocity of 40 m/s upward from a height of ( x = 0 ) m. at what time will the object be at a height of ( x = 20 ) m on its way upward? use ( g = 10 ) m/s² for the magnitude of the acceleration due to gravity. (4pts). additionally, determine what value the function approaches as ( x \to pminfty ) (2pts). 10. find the vertical asymptotes for the following function (2pts). determine if the function approaches ( +infty ) or ( -infty ) 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 \to pminfty ) (2pts). ( y(x)=\frac{x^{2}}{x^{2}+x - 6} )

Answer

Explanation:

Step1: Find vertical asymptotes

Vertical asymptotes occur where the denominator is zero. Set (x^{2}+x - 6=0). Factor: ((x + 3)(x - 2)=0). So (x=-3) and (x = 2) are vertical asymptotes.

Step2: Analyze behavior near vertical asymptotes

For (x\to - 3^{-}) (left - hand limit): Let (x=-3 - h), (h\to0^{+}). (y=\frac{x^{2}}{x^{2}+x - 6}=\frac{(-3 - h)^{2}}{(-3 - h)^{2}+(-3 - h)-6}). Denominator: ((9 + 6h+h^{2})-3 - h - 6=h^{2}+5h\to0^{+}) as (h\to0^{+}). Numerator: ((-3 - h)^{2}\to9). So (y\to+\infty) as (x\to - 3^{-}). For (x\to - 3^{+}) (right - hand limit): Let (x=-3 + h), (h\to0^{+}). Denominator: ((9-6h + h^{2})-3 + h - 6=h^{2}-5h\to0^{-}) as (h\to0^{+}). Numerator: ((-3 + h)^{2}\to9). So (y\to-\infty) as (x\to - 3^{+}). For (x\to2^{-}) (left - hand limit): Let (x = 2 - h), (h\to0^{+}). Denominator: ((4-4h+h^{2})+2 - h - 6=h^{2}-5h\to0^{-}) as (h\to0^{+}). Numerator: ((2 - h)^{2}\to4). So (y\to-\infty) as (x\to2^{-}). For (x\to2^{+}) (right - hand limit): Let (x = 2+h), (h\to0^{+}). Denominator: ((4 + 4h+h^{2})+2 + h - 6=h^{2}+5h\to0^{+}) as (h\to0^{+}). Numerator: ((2 + h)^{2}\to4). So (y\to+\infty) as (x\to2^{+}).

Step3: Analyze as (x\to\pm\infty)

Divide numerator and denominator by (x^{2}): (y=\frac{x^{2}}{x^{2}+x - 6}=\frac{1}{1+\frac{1}{x}-\frac{6}{x^{2}}}). As (x\to\pm\infty), (\frac{1}{x}\to0) and (\frac{6}{x^{2}}\to0). So (y\to1) as (x\to\pm\infty).

Answer:

  • Vertical asymptotes: (x=-3) and (x = 2).
  • As (x\to - 3^{-}), (y\to+\infty); as (x\to - 3^{+}), (y\to-\infty); as (x\to2^{-}), (y\to-\infty); as (x\to2^{+}), (y\to+\infty).
  • As (x\to\pm\infty), (y\to1).