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

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^{-}) (approaching (-3) from the left): Let (x=-3 - h), (h\to0^{+}). (y=\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}\to+\infty)

For (x\to - 3^{+}) (approaching (-3) from the right): Let (x=-3 + h), (h\to0^{+}). (y=\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}\to-\infty)

For (x\to2^{-}) (approaching (2) from the left): Let (x = 2 - h), (h\to0^{+}). (y=\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}\to-\infty)

For (x\to2^{+}) (approaching (2) from the right): Let (x = 2+h), (h\to0^{+}). (y=\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}\to+\infty)

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

Divide numerator and denominator by (x^{2}): (y=\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)

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)