# Thread: Extrema, Intercepts, Symmetry, Asymptotes

1. ## Extrema, Intercepts, Symmetry, Asymptotes

Sketch the graph of the equation. Look for extrema, intercepts, symmetry, and asymptotes as necessary.

y=(2+x)/(1-x)

Please Help

2. Originally Posted by rawkstar
Sketch the graph of the equation. Look for extrema, intercepts, symmetry, and asymptotes as necessary.

y=(2+x)/(1-x)

Please Help
all of this part is precalculus stuff ...

y-intercept is (0,2)

x-intercept is (-2,0)

vertical asymptote is x = 1

horizontal asymptote is y = -2

calculus part ...

$y' = \frac{3}{(1-x)^2} > 0$ for all $x \ne 1$

what's that tell you?

$y'' = \frac{6}{(1-x)^3}$

$y''(x) < 0$ for $x > 1$

$y''(x) > 0$ for $x < 1$

what does that tell you?

3. it's a simple one-variable function to investigate...

you can open it to $y=-x^2-x+2$

and then derivate it $y=-2x-1$.
comparing the derivative to 0 will find extrema $x=-1/2$
and since the function is well defined for all x, this is the only
extrema.

intercept with x axis is in the two roots which are conviniently
shown to you $x=-2, x=1$
and with y axis is simply $y=2$ (put 0 instead of x)

it is symmetrical, and the symmetry axis is the straight line $x=-1/2$, since it is a squre function.

and there are no asymptotes here, since the function is well defined
for all x, and no asymptote as x--->infinity because it is a square function.

that's all, you need to focus your question to the part you're having
trouble with, because i'm not sure which one i needed to answer,
and if, at all, i've answered your question.

4. Originally Posted by vonflex1
it's a simple one-variable function to investigate...

you can open it to $y=-x^2-x+2$
uhh ... the function is $y = \frac{2+x}{1-x}$ , not $y = (2+x)(1-x)$

5. Originally Posted by skeeter
all of this part is precalculus stuff ...

y-intercept is (0,2)

x-intercept is (-2,0)

vertical asymptote is x = 1

horizontal asymptote is y = -2

calculus part ...

$y' = \frac{3}{(1-x)^2} > 0$ for all $x \ne 1$

what's that tell you?

$y'' = \frac{6}{(1-x)^3}$

$y''(x) < 0$ for $x > 1$

$y''(x) > 0$ for $x < 1$

what does that tell you?
I'm not sure how you found the horizontal asymptote to be -2, i got -1 by dividing everything by x

6. Originally Posted by rawkstar
I'm not sure how you found the horizontal asymptote to be -2, i got -1 by dividing everything by x
you're right ... I was staring at the 2 in the numerator.

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