Answer:
Step-by-step explanation:
If the engine torque y (in foot-pounds) of one model of car is given by y=−3.75x^2+23.2x+38.8
The engine speed is at maximum if dy/dx = 0
dy/dx = -2(3.75)x+23.2
dy/dx = -7.5x + 23.2
since dy/dx = 0
0 = -7.5x + 23.2
7.5x = 23.2
x = 23.2/7.5
x = 3.093
Hence the maximum torque is 3.09 rev/min
A. x intercept is -8 ; y intercept is 4
