Graphing (1 Viewer)

[kaz1]

I can do part (i) and (ii) but can someone explain how to do part (iii)?

 

[Prashant Sallan]

I can do part (i) and (ii) but can someone explain how to do part (iii)?

Firstly,
the original asymptote is y=-1, so for part (iii) the curve approaches 4^-1 = 0.25
@ y axis (original curve), x=.5, so for part (iii) the curve = 4^.5 = (0,2)
@ x axis, x=0 (original curve), so for part (iii) the curve = 4^0 = 1 @ both (0,1) and (0,-1)
then you just join the dots...
note* nothing is below the x=0.25

 

[Prashant Sallan]

The blue curves part (iii), and the black curves the original...
y=1, is just the reference line.. because the x asymptotes from the original curve lie on this reference line..
sorry about the dodgy drawing..
just noticed that the blue line closest to the x - axis is the asymptote (y=0.25) for the new curve.. (part iii)

 

[kwabon]

sub in values matey.