find the critical points and classify local extrema.
f(x)= 2-3x/2+x
Thanks for clearing that up.
If you divide the function as suggested then f(x) will be.
$\displaystyle
\frac{2-3x}{2+x} = \frac{4}{2+x}-3$
so
$\displaystyle f(x) = \frac{4}{2+x}-3$
This suggests a vertical aymptote at x = -2 and a horizontal aymptote at y = -3.
To find a y-intercept make x = 0
$\displaystyle f(0) = \frac{4}{2+0}-3 = 2-3 = -1$
To find a x-intercept make y = 0
$\displaystyle 0 = \frac{4}{2+x}-3 $
quotient rule
$\displaystyle 3(2+x) = 4 $
$\displaystyle x = \frac{-2}{3} $
To find any maxima or minima you should use the
quotient rule , set your derivative equal to zero and solve for x.
quotient rule is, if
$\displaystyle y = \frac{u}{v} $
then
$\displaystyle y' = \frac{v\times u'-u\times v'}{v^2} $
I hope this helps some!