posted by Melanie .
concave up interval
concave down interval
You should be able to do some of these yourself. If not, you are behind class level and would benefit from private tutoring.
The x intercept(s) occur where f(x) = 0. That can only happen at when the numerator is zero. In other words, at x = 0.
Vertical asymptotes occur wherever the denominator of f(x) is zero. Note that the denominator of f(x) can be written (x+2)(x-1). That should tell you where the asymptotes are.
For the derivative of
f(x) = x/(x^2 +x -2), let
u(x) = x and v(x) = x^2 +x -2
and use the rule for the derivative of the fraction u(x)/v(x), which you should know.
f'(x) = (v*u' - u*v')/v^2
= [x^2 +x -2 -x*(2x +1)]/(x^2 +x -2)^2
= (-x^2 -2)/(x^2 +x -2)^2
Use the same "u/v" rule for the second derivative.
Plotting the f(x) function yourself should help for the last four questions. "Increasing up or down" is determined by the sign of f'(x).
"Concave up or down" is determined by the sign of f''(x), the second derivative.