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

6260 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Integrate the function f(x) where f(x)= x^2 +sin(x) + sin^2(x)


Suppose that you go to a party where everyone knows at least one other person, you get a bit bored and wonder whether there are at least two people which know the same number of people there.


A curve is defined by parametric equations: x = t^(2) + 2, and y = t(4-t^(2)). Find dy/dx in terms of t, hence, define the gradient of the curve at the point where t = 2.


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 .


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:

© 2025 by IXL Learning