The curve with the equation: y=x^2 - 32sqrt(x) + 20 has a stationary point P. Find the coordinates of P.

A stationary point implies that the gradient at this point will be equal to 0, it is a turning point in the graph. So we need to find the value of X at which dy/dx = 0. [where dy/dx is the gradient.]So we first differentiate the equation Y with respect to X giving us: dy/dx = 2x + 16x^(-0.5) [using simple differentiation, i.e. bring the power down, and take 1 away from the power]We would now equate this to 0, as this will tell us when the gradient is equal to 0.so 2x + 16x^(-0.5) = 0Now solve for x: equation simplifies to x^(1.5) = 8, which tells us that X=4We now have X, we need to work out the Y value at this point. So we substitue X=4 into the equation of the curvey=4^2 - 32sqrt(4) + 20, which tells us that Y= -28. So point P(4,-28)

Answered by Maninder B. Maths tutor

8952 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How do I find the equation of the normal to the curve y=x^2 at the point (x1,y1)? Where x1=2 and y1=4 .


(x+2)(x-3)


Integrate x * sin(x) with respect to x by using integration by parts


The curve C has the equation y=((x^2+4)(x-3))/2*x where x is not equal to 0 . Find the tangent to the curve C at the point where x=-1 in the form y=mx+c


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 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences