Question

The curve y = 1/3x^3 + 1/2x^2 - 5x has a gradient of -3 at points P and Q. Find the coordinates of P and Q

Answer 1

Answer:

y = 1/3x^3 +1/2x^2-5x

differentiate wrt x

dy/dx = x^2+x-5

gradient = -3

x^2+x-5= -3

on solving we get , x= -2 and x= 1

putting values of x in y= 1/3x^3+1/2x^2-5x

we get , at x= -2 , y= 28/3

at x= 1 , y= -25/6

so coordinates of P and Q are (-2,28/3) and (1,-25/6)