Let Curve C be f(x)=(1/3)(x^2)+8 and line L be y=3x+k where k is a positive constant. Given that L is tangent to C, find the value of k. (6 marks approx)

SO when we see the word tangent we should be thinking about rate of change. Recall that the line being a tangent means they meet and have the same derivative at this point OR we find k such that f(x)-y=0 has a double root. (We can prove that this is true!)So(1/3)x^2+8-k-3x=0 so we solve for k such that the discriminant is 0. that is 9-4(1/3)(8-k)=0 This implies k=8-27/4=5/4

GJ
Answered by Gurbir J. Further Mathematics tutor

7012 Views

See similar Further Mathematics GCSE tutors

Related Further Mathematics GCSE answers

All answers ▸

Work out the gradient of the curve y=x^3(x-3) at the point (3,17)


The function f is given by f(x) = SQRT(2x − 5). Work out x when f(x) = 1.2


What is the equation of a circle with centre (3,4) and radius 4?


(x+4)((x^2) - kx - 5) is expanded and simplified. The coefficient of the x^2 term twice the coefficient of the x term. Work out the value of k.


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