Find the stationary point on the line of y = 6x - x^2 and state whether this point is a maximum or a minimum

This question requires differentiating the equation to find where its gradient is equal to zero. Differentation is done via a simple equation -->  if y = xn then dy/dx = nxn-1

Therefore if y = 6x - x2  , then dy/dx = 6(x0) - 2x1   , so   dy/dx = 6 - 2x

The gradient is 0 at the stationary point, so 6 - 2x = 0           2x = 6        so x = 3

To find y, substitute (x=3) into the original formula to find y.     y = 6(3) - 32   = 18 - 9    = 9

The stationary point is (3,9), and to find out whether this is a maximum or minimum, x=4 can be subbed in to the formula to find the next point on the line. y = 6(4) - 42   =  24 - 16     =    8     so the next point is (4,8)

This is below the stationary point, so we can see that (3,9) is a maximum.

PW
Answered by Percy W. Maths tutor

6111 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

What's the difference between the quotient rule and the product rule?


Do the circles with equations x^2 -2x + y^2 - 2y=7 and x^2 -10x + y^2 -8y=-37 touch and if so, in what way (tangent to each other? two point of intersection?)


whats the integral of x.e^x wrt x


Find the integral of ((2(7x^(2)-xe^(-2x))-5)/x) . Given that y=27 at x=1, solve the differential equation dy/dx=((2(7x^(2)-xe^(-2x))-5)/-3x).y^(2/3) in terms of y.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning