T Twickel Member Joined Dec 5, 2007 Messages 390 Gender Male HSC 2009 Nov 19, 2008 #1 Hi Q6b , how do I find the intercepts? please solve and show working.
T TCCPICHAMP New Member Joined Mar 15, 2008 Messages 14 Gender Male HSC 2009 Nov 19, 2008 #2 heres the answers (b) (i) (0,0) and (4,0). (ii) (0,0) is a horizontal pt of inflexion. (3,27) is a minimum T.P. (iii) (0,0) and (2,16). from there u should be able to work it out, if u do, it will make u remember how to do it better Last edited: Nov 19, 2008
heres the answers (b) (i) (0,0) and (4,0). (ii) (0,0) is a horizontal pt of inflexion. (3,27) is a minimum T.P. (iii) (0,0) and (2,16). from there u should be able to work it out, if u do, it will make u remember how to do it better
T Twickel Member Joined Dec 5, 2007 Messages 390 Gender Male HSC 2009 Nov 19, 2008 #3 I know the rest, I just cant do (i) please do it for me.
Chinmoku03 Ippen, Shin de Miru? Joined Nov 7, 2006 Messages 393 Location Australia >.<; Gender Male HSC 2007 Nov 19, 2008 #4 Sub x = 0: f(0) = 0 So (0,0) is the point at which the curve crosses the y axis. Sub f(x) = 0: x4 - 4x3 = 0 x3(x - 4) = 0 x = 0 or x = 4 So (0,0) and (4,0) are points at which the curve crosses the x axis.
Sub x = 0: f(0) = 0 So (0,0) is the point at which the curve crosses the y axis. Sub f(x) = 0: x4 - 4x3 = 0 x3(x - 4) = 0 x = 0 or x = 4 So (0,0) and (4,0) are points at which the curve crosses the x axis.