Write down three linear factors of f(x) such that the curve of f(x) crosses the x axis at x=0.5,3,4. Hence find the equation of the curve in the form y = 2(x^3) + a(x^2) + bx + c.

the curve crosses the graph at the x axis at 0.5,3 and -4 so f(0.5)=0, f(3)=0,f(-4)=0. All linear factors are the form of g(x)=(x-a) so 0=g(0.5)=0.5-a. rearranging we get that a=0.5 ie g(x)=x-0.5 similarly the other linear factors are (x-3),(x+4) so f(x) is the product of three linear factors and so f(x)=(x-0.5)(x-3)(x+4) expanding f(x)=(x-0.5)[(x-3)(x+4)] =(x-0.5)[x(x+4)-3(x+4)] =(x-0.5)[x2+4x-3x-12] =(x-0.5)(x2+x-12)  =x(x2+x-12)-0.5(x2+x-12)  =x3+x2-12x-0.5x2-0.5x+6 =x3+0.5x2-12.5x+6   now we can multiply the coeffiecient by 2 to get the desired form f(x)=2x3+x2-25x+12    

KT
Answered by Kishan T. Maths tutor

5700 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Differentiate y=(x-1)^4 with respect to x.


(i) Prove sin(θ)/cos(θ) + cos(θ)/sin(θ) = 2cosec(2θ) , (ii) draw draph of y = 2cosec(2θ) for 0<θ< 360°, (iii) solve to 1 d.p. : sin(θ)/cos(θ) + cos(θ)/sin(θ) = 3.


Differentiate a^x


Find the centre coordinates, and radius of the circle with equation: x^2 + y^2 +6x -8y = 24


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