Find all existing vertical and horizontal asymptotes,maximum and minimum and point of inflection of the graph y=(x+1)/(x^2+x+1)

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- Apr 29th 2006, 02:38 AMbobby77please help -very urgent...
Find all existing vertical and horizontal asymptotes,maximum and minimum and point of inflection of the graph y=(x+1)/(x^2+x+1)

- Apr 29th 2006, 03:32 AMTD!
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 - Apr 29th 2006, 11:42 AMbobby77please check my answer
please check my answer

- Apr 29th 2006, 05:50 PMAradesh
we have

$\displaystyle y=\frac{x+1}{x^2+x+1}$

vertical asymptotes occur when the denominator is zero

when

$\displaystyle x^2+x+1 = 0$

the discriminant $\displaystyle (1)^2-4 \cdot 1 \cdot 1 < 0$

thus there are no real roots, hence no vertical asymptotes.

$\displaystyle y = \frac{x+1}{x^2+x+1}$

so as $\displaystyle x \to \infty$ then $\displaystyle y \to 0$ from above

and as $\displaystyle x \to -\infty$ then $\displaystyle y \to 0$ 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

$\displaystyle y=c$ we know that the curve does not exist for values of c such that

$\displaystyle \frac{x+1}{x^2+x+1} = c $ 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.

$\displaystyle \Leftrightarrow x+1 = c(x^2+x+1)$

$\displaystyle \Leftrightarrow cx^2+x(c-1) + c-1 = 0$ call this (1)

so this has real values when

$\displaystyle (c-1)^2-4 c (c-1) \geq 0$

$\displaystyle \Leftrightarrow -3c^2+2c+1 \geq 0 $

$\displaystyle \Leftrightarrow 3c^2-2c-1 \leq 0$

$\displaystyle \Leftrightarrow (3c+1)(c-1) \leq 0$

$\displaystyle \Leftrightarrow -\frac{1}{3} \leq c\leq 1$

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

$\displaystyle -\frac{1}{3}x^2-\frac{4}{3}x - \frac{4}{3} = 0$

$\displaystyle \Leftrightarrow x^2 +4x+ 4= 0$

$\displaystyle \Leftrightarrow (x+2)^2 = 0$

so $\displaystyle x=-2$

so there is a minimum at $\displaystyle (-2,-\frac{1}{3})$

and for the maximum y=1 we have from (1)

$\displaystyle x^2 = 0$

so x = 0 giving coordinate of maximum $\displaystyle (0,1)$

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