Find all existing vertical and horizontal asymptotes,maximum and minimum and point of inflection of the graph y=(x+1)/(x^2+x+1)
HA -> check the limit for x going to +/- infinity
VA -> find the poles (zeroes of the denominator, but not of nominator) and check the limit approaching that point from both sides
min/max -> solve f'(x) = 0 and use the f" to see whether it's a min or max
inflection -> solve f"(x) = 0 and see if f" changes sign in those points
we have
vertical asymptotes occur when the denominator is zero
when
the discriminant
thus there are no real roots, hence no vertical asymptotes.
so as then from above
and as then from below
considering turning points.
from the information we know there must be two turning points, a minimum less than zero, and a maximum, greater than zero. also there cannot be more than this else it would mean there could be a horizontal line intersecting the curve in more than two places, which defies the fact our function is of second degree.
if we look at the horizontal lines
we know that the curve does not exist for values of c such that
has no real solutions.
so using this we can find the range of the curve for all values of c which there are real solutions, and thus find the turning points.
call this (1)
so this has real values when
so at y=-1/3 we have a minimum. and y = 1 we have a maximum. for which we can find corresponding x values, using (1) and c = -1/3
so
so there is a minimum at
and for the maximum y=1 we have from (1)
so x = 0 giving coordinate of maximum
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