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 ...
$\displaystyle y' = \frac{3}{(1-x)^2} > 0$ for all $\displaystyle x \ne 1$
what's that tell you?
$\displaystyle y'' = \frac{6}{(1-x)^3}$
$\displaystyle y''(x) < 0$ for $\displaystyle x > 1$
$\displaystyle y''(x) > 0$ for $\displaystyle x < 1$
what does that tell you?
it's a simple one-variable function to investigate...
you can open it to $\displaystyle y=-x^2-x+2$
and then derivate it $\displaystyle y=-2x-1$.
comparing the derivative to 0 will find extrema $\displaystyle x=-1/2$
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 $\displaystyle x=-2, x=1$
and with y axis is simply $\displaystyle y=2$ (put 0 instead of x)
it is symmetrical, and the symmetry axis is the straight line $\displaystyle x=-1/2$, 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.