Stationary Points, Asymptotes and Points of Inflexion. (1 Viewer)

[Michaelmoo]

Graphmatica believes one of the solutions is x=~-0.63, im unsure how do find this value though..put it this way u wouldnt get a Q like this in the HSC if ur required to factorise such a shit expression.

Thats what I needed to hear. Thanks.

 

[Timothy.Siu]

u cud get close by using newtons method
but in the exam they'd tell u to use it if they wanted u to find it

 

[Michaelmoo]

u cud get close by using newtons method
but in the exam they'd tell u to use it if they wanted u to find it

Ah ok. I see. Thanks.

 

[jchoi]

Ok Basically the question says

Find any Stationary points, Points of Inflexion or asymptotes for:

y = 2x + 3
. . .(x^2 - 4)

Ok so I find the asymptotes, X = plus or minus 2 and y = 0

When I try to find any stationary points or Points of inflexion, I get no solution. Although Clearly, there should be a horizontal point of inflexion at (0, -3/4), but I can't get this.

This means that the first and second derivative must equal zero at this point.

Could someone please run me through this question??

Thanks in advance.

asymptotes are plusminus 2 and y = 0.
as x -> infinity, y -> 0.
this should look like

a maximum at (0, -3/4), and two curves approaching 0 as x -> plus minus infinity.

This graph should have no poi.