Solve the problem.
-A plane suddenly experiences turbulence while in flight. At this point the change in its altitude, in feet, is represented by the following function of time: $F ( x ) = \frac { 5 ( x - 10 ) ^ { 2 } - 19 } { 20 ( x - 10 ) ^ { 2 } - 10 ( x - 10 ) + 17 }$.

Assuming that the turbulence first hits at time $x = 0$, what is the maximum relative drop the plane will experience due to turbulence? (Round the result to the nearest hundredth.)
A) $28.39 \mathrm { ft }$
B) $10.21 \mathrm { ft }$
C) $0.25 \mathrm { ft }$
D) $1.19 \mathrm { ft }$

Question

Solve the problem.
-A plane suddenly experiences turbulence while in flight. At this point the change in its altitude, in feet, is represented by the following function of time: $F ( x ) = \frac { 5 ( x - 10 ) ^ { 2 } - 19 } { 20 ( x - 10 ) ^ { 2 } - 10 ( x - 10 ) + 17 }$.

Assuming that the turbulence first hits at time $x = 0$, what is the maximum relative drop the plane will experience due to turbulence? (Round the result to the nearest hundredth.)
A) $28.39 \mathrm { ft }$
B) $10.21 \mathrm { ft }$
C) $0.25 \mathrm { ft }$
D) $1.19 \mathrm { ft }$

Quizplus · Accepted Answer

To find the maximum relative drop, evaluate the function $F(x)$ at $x = 0$. Simplifying $F(0)$ gives the maximum drop as approximately $1.19$ feet.

Solve the Problem F(x)=5(x−10)2−1920(x−10)2−10(x−10)+17F ( x ) = \frac { 5 ( x - 10 ) ^ { 2 } - 19 } { 20 ( x - 10 ) ^ { 2 } - 10 ( x - 10 ) + 17 }F(x)=20(x−10)2−10(x−10)+175(x−10)2−19​

Solve the Problem $F ( x ) = \frac { 5 ( x - 10 ) ^ { 2 } - 19 } { 20 ( x - 10 ) ^ { 2 } - 10 ( x - 10 ) + 17 }$