Another Polynomial Question (1 Viewer)

[YBK]

Hey, I have another question....

i) Sketch the curve y = x^3 - 4x^2 + 3x clealry showing all intercepts with the coordinate axes

Done: y = x(x-3)(x-1)
x = 0,1,3

ii) The line y = kx cuts the curve in part (i) at the origin
Show that, if k < -1, they have no other points in common



Well, for this part I'm not sure how to put k < -1

If I put say k = -2, and then solve simultaniously then it'd only touch at the origin... but how can I prove that for all k < -1 ???

iii) Discuss the case where k = -1 , giving a geometrical interpratation

Done (i think)

What I did was simultaneously solve y = -x and y = x^3 - 4x^2 + 3x
which gives: x(x-2)(x-2)

Therefore the line y = -x cuts y = x^3 - 4x^2 + 3x at the origin and is tangent at x=2


iv) Find the set of values of k for which the line y = kx cuts the curve in three distinct points.

Well, I think it's for k > -1



That's the question... any help would be appreciated

and correct me if i'm wrong.

thanks !!!

 

[mojako]

ii
x(x-3)(x-1)=kx
x could be zero, but if it isn't,
(x-3)(x-1)=k
bu the graph of parabola y=(x-3)(x-1) can't go below y=-1
so k can't be below -1

iv
(x-3)(x-1)=k has distinct roots for k>-1
they're also not zero (except when k=3)
so x(x-3)(x-1)=kx
has three different roots
when k>-1, k isn't 3

 

[YBK]

ii
x(x-3)(x-1)=kx
x could be zero, but if it isn't,
(x-3)(x-1)=k
bu the graph of parabola y=(x-3)(x-1) can't go below y=-1
so k can't be below -1

iv
(x-3)(x-1)=k has distinct roots for k>-1
they're also not zero (except when k=3)
so x(x-3)(x-1)=kx
has three different roots
when k>-1, k isn't 3

cool thanks

i don't understand the k = 3 part though...

 

[mojako]

when u said
x(x-3)(x-1)=kx
either x=0 or x=something else
(x-3)(x-1)=k if x isn't 0
when k=3,
(x-3)(x-1)=3
x=0 or x=4,
but you mentioned that x isn't zero so the solution is x=4 only

or from original eqn,
x(x-3)(x-1)=kx
x^3 - 4x^2 + 3x = kx
when k=3,
x^3 - 4x^2 + 3x = 3x
x^3 - 4x^2 = 0
x^2 (x-4) = 0


[you get the k=3 thing by substituting x=0 into y=(x-3)(x-1)]

 

[YBK]

when u said
x(x-3)(x-1)=kx
either x=0 or x=something else
(x-3)(x-1)=k if x isn't 0
when k=3,
(x-3)(x-1)=3
x=0 or x=4,
but you mentioned that x isn't zero so the solution is x=4 only

or from original eqn,
x(x-3)(x-1)=kx
x^3 - 4x^2 + 3x = kx
when k=3,
x^3 - 4x^2 + 3x = 3x
x^3 - 4x^2 = 0
x^2 (x-4) = 0


[you get the k=3 thing by substituting x=0 into y=(x-3)(x-1)]

ohhhh! great!

I kinda get it now...

so the values for which the line y = kx cuts the graph in three distinct points is:

k > -1 and k not equal to 3

thanks again