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Graphing (1 Viewer)
- Thread starter Lukybear
- Start date
y = (x+1)^4 is a parabola with vertex (-1,0)
y = (x^4+1) is a parabola with vertex (0,1)
divide the parabola by subbing points or how you like it
note when you divide the vertical tangent will be x = -1
horizontal tangent at x= 1 too (lim x^4/x^4)
Btw can you draw graphs here or no?
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There's a horiz. asymptote at y=1.
Vertical asymptotes are at x=1 and x=-1
To find stat points find y' and solve y'=0. (I'd do this)
I'm not sure that it'd have inflexions, but to find them you firstly find y'' and solve y''=0 and check for concacity changes. (I wouldn't do this, unless specifically asked, or if the question implied it. Eg with marks).
you'd have to differentiate + point of intersection if you want to find the turning point
on the left side of the graph (of x= -1) : as x->-oo, y -> 1
right side (of x=-1): as x-> oo, y -> 1, BUT it goes through (0,1) and rises above y=1 and has a max turning point somewhere (differentiate to obtain)
The graph looks like a bird kinda with the wings sprouting out from (-1,0)
on the left side of the graph (of x= -1) : as x->-oo, y -> 1
right side (of x=-1): as x-> oo, y -> 1, BUT it goes through (0,1) and rises above y=1 and has a max turning point somewhere (differentiate to obtain)
The graph looks like a bird kinda with the wings sprouting out from (-1,0)
oh yeah i forgot also vertical asymptote x= -1 xD
wait wth i already said that o-o
i mean to say there is no vertical asymptote at x= 1
unless somehow you got one O_O
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tacogym27101990
Member
definitely no vertical asymptotes in this graph, as x^4+1 =/=0 for real x
oh yeah you're right, it's just a turn at (-1,0) my bad