Curve Sketching Question (1 Viewer)

[Hermes1]

could someone explain to me how to do question 3 (a) (ii) of this trial paper:
http://mathexampapers.web.fc2.com/test/4U/2008CSSATrial.pdf

man i hate curve sketching

 

[xV1P3R]

f(x) = kx
x(x-3)² = kx
(x-3)² = k, x = 0 is always a solution
Therefore find the situation where (x-3)² = k has only 1 solution

Graph y = (x-3)²

That has two solutions for y > 0 and a double root for y = 0. So no values of k work.

 

[Hermes1]

i just found the answer and it say k = 0 or 9.

 

[michaeljennings]

lol i hate curve sketching too its so boring lol

 

[deterministic]

f(x) = kx
x(x-3)² = kx
(x-3)² = k, x = 0 is always a solution
Therefore find the situation where (x-3)² = k has only 1 solution

Not necessarily... the other case is that x=0 is also a solution to (x-3)² = k, which implies k=9. k=0 works as well obviously.

 

[xV1P3R]

Goodness, what was I thinking. Yep k = 9 works, but I was under the impression that k = 0 gives you 3 roots, 2 of which are equal.