Find the coordinates of the stationary points of the curve y = 2x^3 + 3x^2 - 12x + 1

This is a typical A/S level question, which can be broken down into three key steps.1) Differentiation: Remember that at a stationary point, the gradient of the curve is zero. Therefore, we first have to find the gradient, which means differentiating. (Multiply the coefficient of x by the power, and then reduce the power by 1)y = 2x3 + 3x2 - 12x + 1dy/dx = 6x2 + 6x - 122) Solve Quadratic: To find where the gradient is zero (dy/dx = 0), we must solve the quadratic equation,dy/dx = 6x2 + 6x - 12 = 0.This quadratic can be most easily solved by factorisation (alternative methods are completing the square and using the quadratic formula). Firstly we can take 6 out as a common factor, which results in 6(x+ x - 2) = 0. We can then factorise the terms in brackets which results in 6(x+2)(x-1) = 0.This gives the two solutions x = -2, x = 1.3) Substitution: So far, we have only found the x coordinate of the stationary point, but to answer the question fully we need the y coordinate as well. We do this by substituting x = -2, and x = 1 into the original equation of the curve, y = 2x3 + 3x- 12x + 1When x = -2, we can calcuclate that y = 21. Similarly when x = 1, y = -6.Thus the coordinates of the stationary points are (-2, 21) and (1, -6).

Answered by Barnum S. Maths tutor

49432 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Simplify (3x^2-x-2)/(x^2-1)


A curve has the equation y=3x^3 - 7x^2+52. Find the area under the curve between x=2 and the y-axis.


I don't understand how functions work. How do I decide if something is a function?


We are given y=(x^2)+3x-5. Find the derivative of y in terms of x.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy